POWER OF MONOMIAL IDEALS
SUSAN COOPER
BRIAN JOHNSON
Contents
1. Preliminaries
2. Gröbner Bases
2.1. Motivating Problems
2.2. Ordering of Monomials
2.3. The Division Algorithm in S = k[x1 , . . . , xn ]
2.4. Dickson’s Lemma
2.5. Gröbner Bases and the Hilbert Basis Theorem
2.6. Some Further Applications of Gröbner bases
3. Hilbert Functions
3.1. Macaulay’s Theorem
3.2. Hilbert Functions of Reduced Standard Graded k-algebras
3.3. Hilbert Functions of Points in Pn
3.4. Maximal Growth of Hilbert Functions and Consequences
3.5. The Eisenbud-Green-Harris Conjecture
3.6. A combinatorial approach to EGH
3.7. Some enumeration
3.8. Sum fun with points and regular sequences
4. Lex ideals and Betti numbers
4.1. Group actions on S = k[x1 , . . . , xn ]
4.2. Generic initial ideals
4.3. Some comments on Gröbner bases and modules
4.4. Comparing lex-segment ideals and Borel-fixed ideals
5. Squarefree monomial ideals
5.1. Hilbert series
5.2. Shellable simplicial complexes and H-vectors
6. A crash course on resolutions
6.1. The graded world
7. Group Presentations
7.1. Lifting monomial ideals: A. Croll, C. Gibbons, and B. Johnson
7.2. Representations of monomial orders: R. Brase, A. Denkert,
and M. Janssen
Date: May 7, 2010.
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BRIAN JOHNSON
7.3. Associated primes of monomial ideals and odd holes: D.
Boeckner and D. Stolee
7.4. Resolutions and Betti numbers: J. DeVries and X. Yu
References
Appendix A. Problem Set 1
Appendix B. Problem Set 2
Appendix C. Problem Set 3
Appendix D. Problem Set 4
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POWER OF MONOMIAL IDEALS
SUSAN COOPER
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1. Preliminaries
We let k be a field, and define S = k[x1 , . . . , xn ] = k[x].
Definition 1.1.
(1) A monomial in S is a product xa = xa11 xa22 . . . xann , where a =
(a1 , . . . , an ) is a vector of nonnegative integers. [It should be noted
that we do not allow constants other than 1 as a coefficient for our
monomials.]
(2) An ideal I ⊆ S is monomial if it is generated by monomials.
2. Gröbner Bases
References: [7, Ch. 2] and [8, Ch. 15]
The main idea here is to reduce questions in S to questions about monomials (which are often easier to solve). Gröbner bases give rise to interesting
theoretical results and computational applications. They are the heart of
many practical algorithms used in many computer algebra programs (including Macaulay2, CoCoA)
2.1. Motivating Problems. Given an ideal I ⊆ S, a Gröbner basis is a
generating set for I with a special property. These are useful in studying:
• “Ideal description problem” Does every ideal have a finite generating
set? (Hilbert Basis Thm says yes for our S).
• “Ideal Membership Problem” Given an ideal I = (f1 , . . . , ft ) ⊆ S
and f ∈ S, determine if f ∈ I. Geometrically, we are asking if
V (f1 , . . . , ft ) lies on V (f ).
• “Solving Polynomial Equations” Find all common solutions in k n of
a system f1 (x1 , . . . , xn ) = · · · = ft (x1 , . . . , xn ) = 0. I.e., find all the
points in the affine variety V (f1 , . . . , ft ).
• Given a variety V ⊆ An , find the equations for its closure in Pn .
• Computing Hilbert functions and polynomials
• Computing syzygies; i.e., find the kernel of φ : G −→ F for F, G free
S-modules.
• Computing intersections of ideals
2.2. Ordering of Monomials. Ideal membership requires a division algorithm. Recall:
Theorem 2.1 (Division Algorithm in k[x]). Let g ∈ k[x] r {0}. Every
f ∈ k[x] can be written as f = gq + r, where q, r ∈ k[x] with r = 0 or
deg r < deg g. Moreover, q and r are unique.
Example 2.2. If g = x2 + 3x − 1 and f = x4 + 2x2 − x, polynomial long
division gives q = x2 − 3x + 12 and r = −40x + 12. So we can answer one
question: If I = (q), is f ∈ I? No, as r 6= 0.
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BRIAN JOHNSON
It turns out the algorithm works because we systematically use the ordering · · · > x3 > x2 > x > 1 whenever we remove terms. However, if we have
more than one variable, we’ll need a similar notion of ordering.
Remark 2.3. We have an obvious bijection
{monomials in S} ←→
Zn≥0
a1
a
α
n
←→ α = (a1 , . . . , an )
x = x1 , . . . , x n
So xα > xβ if and only if α > β.
Definition 2.4. A monomial ordering on S is any relation > on the monomials of S (equivalently, on Zn≥0 ) satisfying
(1) > is a total ordering. That is, for every α, β, exactly one of xα > xβ ,
xα < xβ , or xα = xβ holds.
(2) α, β ∈ Zn≥0 such that α > β implies α + γ > β + γ for all γ ∈ Zn≥0 .
(3) > is a well-ordering. That is, every nonempty subset of Zn≥0 has a
least element.
Example 2.5 (Examples of Monomial Orderings).
(1) Lex(icographic) Ordering: We have xα >lex xβ if in α−β the leftmost
nonzero entry is positive. E.g., (5, 6, 2) >lex (5, 4, 3) implies that
x51 x62 x23 >lex x51 x42 x33 . Also, it is easy to see that x1 >lex x2 >lex
· · · >lex xn .
P
(2) Graded/Homogeneous Lex Order: Define |α| =
ai . We have
xα >grlex xβ if |α| > |β| or if both |α| = |β| and xα >lex xβ . As
in the previous example, it’s easy to see that x1 >grlex x2 >grlex
· · · >grlex xn . One may also verify that x31 x2 x23 x44 >grlex x31 x2 x3 x54 .
(3) Graded Reverse Lex Order: We have xα >grevlex xβ if |α| > |β| or if
both |α| = |β| and in α − β, the rightmost nonzero entry is negative.
[It is worth noting here that there is no plain revlex, as there would
be no well-ordering.] Once again, we also have x1 >grevlex x2 >grevlex
· · · >grevlex xn .
(4) Inverse Lex Order: We have xα >invlex xβ if in α − β the rightmost
nonzero entry is positive.
Proposition 2.6. The following are monomial orders on S.
(1)
(2)
(3)
(4)
Lex (>lex )
Graded Lex (>grlex )
Graded RevLex (>grevlex )
Inverse Lex (>invlex )
Proof. For lex, see [7, Ch.2]. For grlex and grevlex, we’ll use Dickson’s
Lemma later. Invlex is a good exercise.
P
Definition 2.7. Let f = α aα xα ∈ k[x] r {0}, and let > be a monomial
order.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
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(1) The multidegree of f is
multideg(f ) := max{α|aα 6= 0},
(2) the leading coefficient of f is
lc(f ) := amultideg(f ) ,
(3) the leading monomial of f is
lm(f ) := xmultideg(f ) ,
and
(4) the leading term of f is
lt(f ) := lc(f ) · lm(f ).
[Note: the final two definitions are not standard in all books. Some
include constants in the leading monomial.]
Example 2.8. If f = −2x61 x43 + 5x71 x2 x24 + 4x51 x52 ∈ k[x1 , x2 , x3 , x4 ] and
>=>grevlex , then we see multideg(f ) = (5, 5, 0, 0), lm(f ) = x51 x52 , lc(f ) = 4,
and lt(f ) = 4x51 x52 .
Lemma 2.9. Let f, g ∈ S r {0}. Then
(1) multideg(f g) = multideg(f ) + multideg(g)
(2) multideg(f + g) ≤ max{multideg(f ) + multideg(g)}, and we have
equality if multideg(f ) 6= multideg(g).
Example 2.10. If we take >=>lex , and let f = −5x31 + 7x21 x22 and g =
5x31 + 4x1 x3 , then multideg(f ) = multideg(g) = (3, 0, 0), but multideg(f +
g) = (2, 0, 0).
2.3. The Division Algorithm in S = k[x1 , . . . , xn ]. There is a question
about what it means to divide a single polynomial f ∈ S by more than one
polynomial f1 , . . . , fs ∈ S. An answer is that we’ll think of f1 , . . . , fs as an
ordered set, and we’ll find other polynomials a1 , . . . , as , r ∈ S such that
f = a1 f1 + · · · + as fs + r.
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BRIAN JOHNSON
Example 2.11.
(1) Let >=>lex , f = x2 y+xy 2 +y 2 , and take F = {f1 , f2 } = {xy−1, y 2 −
1} (as an ordered list). The basic idea of the algorithm is at each
step to first make sure the polynomial is written with the leading
term first, and then to try, in the given order of F , to divide the the
leading term of the fi into the leading term of the polynomial. Then
it proceeds more or less like usual long division. When we reach a
point where none of the leading terms of any of the fi divide the
leading term of the polynomial (as with x + y 2 + y below), we put
the leading term (in this case x) into the remainder column at the
far right.
r
a1 : x + y
a2 : 1
f1 = xy − 1
f2 = y 2 − 1
x2 y + xy 2 + y 2
−(x2 y − x)
xy 2 + x+ y 2
−(xy 2 − y)
x + y2 + y
−→ x
y2 + y
−(y 2 − 1)
y + 1 −→ y
1 −→ 1
Thus, the result of this division would be
f = (x + y) f1 + (1) f2 + x + y + 1 .
| {z }
| {z }
|{z}
a1
a2
r
(2) Now, keep > and f the same, but reverse the order of f1 and f2 , say
F = {g1 , g2 } = {y 2 − 1, xy − 1}. Then we get the following:
r
a1 : x + 1
a2 : x
g1 = y 2 − 1
g2 = xy − 1
x2 y + xy 2 + y 2
−(x2 y − x)
xy 2 + x+ y 2
−(xy 2 − x)
2x+ y 2
−→ 2x
2
y
−(y 2 − 1)
1 −→ 1
POWER OF MONOMIAL IDEALS
SUSAN COOPER
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Thus, the result of this division would be
+ 1},
f = (x + 1) g1 + (x) g2 + 2x
| {z
| {z }
|{z}
a2
a1
r
highlighting the problem that the result of this procedure is, in general, not unique.
Theorem 2.12 (Division Algorithm). Fix a monomial order > on Zn≥0 ,
and let F = {f1 , . . . , fs } be an ordered s-tuple of polynomials in S. using
the above algorithm, every f ∈ S can be written as
f = a1 f1 + · · · + as fs + r,
where ai , r ∈ S, and either r = 0 or r is a linear combination over k of
monomials, none of which is divisible by the leading terms of f1 , . . . , fs .
Moreover, if ai fi 6= 0, then
multideg(f ) ≥ multideg(ai fi ).
Proof. See [7, Ch.2]
Corollary 2.13 (Ideal Membership Problem). If r = 0 above, then f ∈ I =
(f1 , . . . , fs ).
Example 2.14. Set >=>lex , and let f = xy 2 − x, f1 = xy + 1, and f2 =
y 2 − 1 in k[x, y]. One can check that performing the division algorithm with
F = {f1 , f2 }, we get
f = y(xy + 1) + 0(y 2 − 1) + (−x − y),
but performing the division algorithm with F 0 = {f2 , f1 }, we get
f = x(y 2 − 1) + 0(xy + 1) + 0.
That is, in one case we have a nonzero remainder, and in the other, we don’t.
The previous example raises a question: Can we find a generating set
f1 , . . . , ft for an ideal I ⊆ S that makes the corollary above an if and only
if?
2.4. Dickson’s Lemma. We note some preliminary facts before proving
Dickson’s Lemma.
Lemma 2.15. Let A ⊆ Zn≥0 and I = (xα | α ∈ A) ⊆ S. A monomial xβ is
in I if and only if xβ is divisible by xα , where α ∈ A.
Proof. (⇐=): Suppose xβ is divisible by xα . Then xβ = xα xγ for some
γ ∈ Zn≥0 . So xβ ∈ I.
(=⇒): Suppose xβ ∈ I. Then
xβ =
t
X
i=1
xα(i) gi ,
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BRIAN JOHNSON
with α(i) ∈ A and gi ∈ S. Expand the gi into linear combinations of
monomials. The right-hand side must be divisible by some xα , so the same
must be true of the left-hand side.
Remark 2.16. Note that {exponents of xβ | xβ is divisible by xα } = α +
Zn≥0 = {α + γ | γ ∈ Zn≥0 }. A graphical representation works as follows.
Consider the vectors corresponding to the generators of your ideal as points
in the appropriate-dimensional space, and then fill in a sort of “convex hull”
to see which other monomials are in the space. To see what we mean, consult
the following picture, where we consider the ideal I = (x3 y, x2 y 4 , xy 5 ) ⊆
k[x, y].
6
y
We can obtain any monomial
given by a lattice point over
here.
s
(1, 5) s
(2, 4)
s
(3, 1)
x
-
Figure 1. A graphic representation of a monomial ideal.
Lemma 2.17. Let I be a monomial ideal in S and f ∈ S. The following
are equivalent:
(1) f ∈ I
(2) Every term of f is in I
(3) f is a k-linear combination of monomials in I.
Proof. Exercise.
Corollary 2.18. Let I and J be monomial ideals in S. Then I = J if and
only if I, J contain precisely the same monomials.
Lemma 2.19 (Dickson’s Lemma). Let I = (xα | α ∈ A ⊆ Zn≥0 ) be a
monomial ideal in S. then I = (xα(1) , . . . , xα(t) ), where α(i) ∈ A for all
i = 1, . . . , t. I.e., I is generated by a finite number of the original generators.
Proof. We use induction on n. If n = 1, then any monomial ideal is principal,
as it is generated by the monomial in the single variable x1 of least degree.
POWER OF MONOMIAL IDEALS
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Assume the result is true for n − 1 variables. Define
J = (xα ∈ k[x1 , . . . , xn−1 ] | xα xan ∈ I, some a ≥ 0).
The inductive hypothesis then implies that J = (xα(1) , . . . , xα(l) ) for some
i
α(1), . . . , α(l). Now, for each i = 1, . . . , l, xα(i) xm
n ∈ I for some mi ≥ 0. Let
m = max{m1 , . . . , ml }. For each k = 0, . . . , m − 1, define
Jk = (xβ ∈ k[x1 , . . . , xn ] | xβ xn xk ∈ I).
Again, by the inductive hypothesis, we can write
Jk = (xαk (1) , . . . , xαk (lk ) ).
Now, we claim that I is generated by the following list:
α(l) m α0 (1)
xn , x
, . . . , xα0 (l0 ) ,
L = xα(1) xm
n ,...,x
. . . , xαm−1 (1) xm−1
, . . . , xαm−1 (lm−1 ) xm−1
.
n
n
To see this, consider (L), the ideal generated by the elements of L. Suppose
b = xα xpn ∈ I. If p ≥ m, then b is divisible by xα(i) xm
n for some i by the
definition of J. If p ≤ m − 1, then b ∈ Jp . Clearly (L) ⊆ I, and so we have
(L) = I. To see that we can choose the generators to be from our original
list, A, take any monomial b = xβ in our list L of generators. Since b ∈ I,
we know b is divisible by xα for some α ∈ A. Proceeding in this way, we can
replace each element of L (if necessary) by an element of A, and it is easy
to see they still generate I.
Remark 2.20. The generators produced from the Jk need not be minimal.
In particular, suppose I = (x3 y, x2 y 4 , xy 5 ) ⊆ k[x, y]. Then J = (x), J0 =
(0), J1 = J2 = J3 = (x3 ), and J4 = (x2 ), so
(L) = (xy 5 , x3 y, x3 y 2 , x3 y 3 , x2 y 4 ).
| {z }
unnecessary
And so now we’ve solved the “Ideal Membership Problem” for monomial
ideals. That is, let I ⊆ S be a monomial ideal and f ∈ S. Dickson’s Lemma
implies I = (xα(1) , . . . , xα(t) ). Thus f ∈ I if and only if
X
f=
aβ xβ
such that xβ ∈ I, but this is if and only if
f=
t
X
aα(i) xα(i) ,
i=1
and this is if and only if the division algorithm gives a remainder of 0 when
f is divided by xα(1) , . . . , xα(t) .
Corollary 2.21 (to Dickson’s Lemma). Let > be a relation on Zn≥0 . If
(1) > is a total ordering and
(2) α > β implies α + γ > β + γ for all γ ∈ Zn≥0 ,
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BRIAN JOHNSON
then > is a well-ordering if and only if α ≥ 0 for all α ∈ Zn≥0 . [It’s worth
noting here that the “≥” in Zn≥0 is different from the one in the statement
“α ≥ 0.”]
Proof. (⇐=): Let A ⊆ Zn≥0 such that A 6= ∅. We need to show A has a
least element. Now, I = (xα | α ∈ A) ⊆ S is monomial. Dickson’s Lemma
then implies I has a finite generating set of elements of A, and after possible
reordering, we may write I = (xα(1) , . . . , xα(t) ) such that xα(1) < · · · < xα(t) .
We claim that xα(1) is the smallest element of A. To see this, let β ∈ A.
Then xβ ∈ I, and so xβ is divisible by xα(i) for some i. Thus β = α(i) + γ
for some γ. But by assumption and the application of (1), we have
β = α(i) + γ ≥ α(i) + 0 = α(i) ≥ α(1).
(=⇒): If > is a well-ordering, then we can find the least element of Zn≥0 ,
call it α0 . Then all we need to show is that α0 ≥ 0. Assume α0 < 0. Then
by (2), 2α0 < α0 , contradicting the minimality of α0 . Thus α0 ≥ 0.
2.5. Gröbner Bases and the Hilbert Basis Theorem.
Definition 2.22. Let I ⊆ S, I 6= (0), be an ideal. Fix a monomial order >
on Zn≥0 . The initial ideal of I is
in(I) = (lt(I)) := (cxα | there exists f ∈ I with in(f ) := lt(f ) = cxα ) .
The notation lt(I) comes from [7] and stands for the set of leading terms
of I. We’ll more often use the in(I) notation, and will use in(f ) and lt(f )
interchangeably.
Example 2.23. Note that if I = (f1 , . . . , ft ), then
(in(f1 ), . . . , in(ft )) ⊆ in(I),
but in general this containment is strict. In particular, let >=>grevlex , and
suppose I = (f1 , f2 ) = (x2 y − 2y 2 + y, y 2 + 2x) ⊆ k[x, y]. We see that
f
= yf1 − x2 f2
= −2x3 − 2y 3 + y 2
∈ I.
So in(f ) = −2x3 ∈ in(I), but −2x3 ∈
/ (in(f1 ), in(f2 )) = (x2 y, y 2 ) (since the
latter is a monomial ideal).
Observe that in(I) is a monomial deal. Then Dickson’s Lemma implies
in(I) is finitely generated. In particular, there exist g1 , . . . , gt ∈ I such that
in(I) = (in(g1 ), . . . , in(gt )).
This motivates the next definition:
Definition 2.24. Fix a monomial order >. A finite subset G = {g1 , . . . , gt }
of an ideal I ⊆ S is a Gröbner basis if
in(I) = (in(g1 ), . . . , in(gt )).
POWER OF MONOMIAL IDEALS
SUSAN COOPER
11
Example 2.25.
(1) Set >=>grevlex and I = (f1 , f2 ) = (x2 y − 2y 2 + y, y 2 + 2x) ⊆ k[x, y].
We showed above that −2x3 ∈ in(I), but −2x3 ∈
/ (in(f1 ), ∈ (f2 )).
Thus {f1 , f2 } is not a Gröbner basis for I.
(2) Set >=>lex and I = (g1 , g2 ) = (x + 2z, y + x) ⊆ C[x, y, z]. We
claim G = {g1 , g2 } is a Gröbner basis for I. Obviously (x, y) =
(in(g1 ), in(g2 )) ⊆ in(I), so we only need to show the reverse containment. Let f ∈ I r {0}, say
f = a(x + 2z) + b(y + z),
where a, b ∈ C[x, y, z]. If in(f ) ∈
/ (x, y), then in(f ) is not divisible by
x or y. So by the definition of lex ordering, f must be a polynomial
in z alone. Since f ∈ I, f vanishes on V (I) = V (x + 2z, y + z) ⊆ C3 .
That is, f vanishes on all the points of the form (−2t, −t, t), for
t ∈ C. The only such polynomial in z alone that satisfies this is
f ≡ 0, a contradiction.
Theorem 2.26 (Hilbert Basis Theorem). Every ideal I ⊆ S is finitely
generated.
Proof. Fix a monomial order >. If I = (0), we’re done, so assume I 6= (0).
Since in(I) is monomial, Dickson’s Lemma gives g1 , . . . , gt ∈ I such that
in(I) = (in(g1 ), . . . , in(gt )). We claim I = (g1 , . . . , gt ). Clearly, we only need
to show I ⊆ (g1 , . . . , gt ). Let f ∈ I. Applying the division algorithm divide
f by {g1 , . . . , gt }. Then there exist a1 , . . . , at , r ∈ S such that
X
f=
(ai gi ) + r,
where
P no term of r is divisible by any of in(g1 ), . . . , in(gt ). Now r =
f −
ai gi ∈ I. Suppose r 6= 0. Then in(r) 6= 0, and in(r) ∈ in(I) =
(in(g1 ), . . . , in(gt )). But this implies in(r) is divisible by one of the in(gi ), a
contradiction. Therefore r = 0 and f ∈ (g1 , . . . , gt ).
Corollary 2.27 (really more of a porism...). Fix an order >. Every ideal
I 6= (0) in S has a Gröbner basis. Moreover, any Gröbner basis for I is also
a generating set for I.
We’ll now look at some properties of Gröbner bases.
Proposition 2.28. Let I ⊆ S be an ideal and F = {g1 , . . . , gt } be a Gröbner
basis for I. Let f ∈ S. Then there exists a unique r ∈ S such that
(1) no term of r is divisible by any of in(g1 ), . . . , in(gt ).
(2) there exists g ∈ I such that f = g + r. In particular, r is the
remainder upon dividing f by G no matter how the elements of G
are listed when using the division algorithm.
Proof. Using the division algorithm, we obtain a1 , . . . , at , r ∈ S such that
X
f=
(ai gi ) + r,
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BRIAN JOHNSON
P
and r satisfies the first claim. Let g =
ai gi ∈ I. Then f = g + r (the
second claim). Finally, to see that r is unique, suppose all of g, gb, r, rb ∈ S
satisfy the above conclusions. Notice rb − r = g − gb ∈ I. Assume r 6= rb.
Then in(b
r − r) ∈ in(I), and in(b
r − r) is divisible by one of the in(gi ). But
this is a contradiction. Thus rb − r = 0 = g − gb.
P
Note: g = ai gi does not necessarily have unique ai :
Example 2.29. Let >=>grlex and I = (x3 − 2xy, x2 y − 2y 2 + x) ⊆ k[x, y].
Then G = {x2 − 3xy, xy, y 2 − 21 x} is a (minimal) Gröbner basis for I (see
[7]). Let f = y 3 + x2 y + y 2 . One can show
1
1
1
f = (x + )xy + 0(x2 = 3xy) + (y + 1)(y 2 − x) + x
2
2
2
|{z}
r
and
1
1
1
f = y(x2 − 3xy) + (3y + )xy + (y + 1)(y 2 − x) + x
2
2
2
|{z}
r
Corollary 2.30. Let G = {g1 , . . . , gt } be a Gröbner basis for an ideal (0) 6=
I ⊆ S. Let f ∈ S. Then f ∈ I if and only if r = 0 when using the division
algorithm to divide f by G.
All of the above is well and good, but the question still remains: how
do we find Gröbner bases? One obstruction to the collection {f1 , . . . , ft }
being a Gröbner basis is cancelations of leading terms when working with
polynomial combinations of the fi .
Definition 2.31. Let f, g ∈ S r {0}, and fix >, a monomial order.
(1) If multideg(f ) = α and multideg(g) = β, the least common multiple
of lm(f ), lm(g) is xγ , where γi = max{αi , βi }.
(2) The S-polynomial of f, g is
xγ
xγ
f−
g,
S(f, g) :=
lt(f )
lt(g)
where xγ = lcm(lt(f ), lt(g)).
Example 2.32. Let S = k[x, y, z], >=>grevlex , f = 4xy 2 z + 7x2 z 2 − 5x3 +
4z 2 , and g = 3x4 yz + y 2 z 2 . Then α = multideg(f ) = (1, 2, 1) and β =
multideg(g) = (4, 1, 1). With notation from the previous definition, we have
γ = (4, 2, 1), and lcm(lm(f ), lm(g)) = x4 y 2 z. So one can compute
x4 y 2 z
xr y 2 z
f
−
g
4xy 2 z
3x4 yz
7 5 2 5 6
1
=
x z − x + x3 z 2 − y 3 z 2 .
4
4
3
Notice lm(f ) and lm(g) are now gone (but also, the multidegree has gone
up).
S(f, g) =
POWER OF MONOMIAL IDEALS
SUSAN COOPER
13
Lemma 2.33. Fix a monomial order >. Let
g=
t
X
ci fi ∈ S,
i=1
where ci ∈ k and multideg(fi ) = δ for all i. If multideg(g) < δ, then g is a
linear combination of the S-polynomials S(fj , fk ) for 1 ≤ j, k ≤ t. Moreover,
each S(fj , fk ) has multidegree less than δ.
Proof. Let di = lc(fi ). then lc(ci fi ) = ci di . Note that
t
X
ci di = 0 (since
i=1
multideg(cj fj ) = δ for all j, yet multideg(g) < δ). Define hi = dfii ∈ S for
i = 1, . . . , t. We have lc(hi ) = 1. Note that lt(fi ) = di xδ implies
lcm(lm(fj ), lm(fk )) = xδ
for all j, k. So
S(f, fk ) =
xδ
xδ
fj −
fk
lt(fj )
lt(fk )
xδ
xδ
f
−
fk
j
dj xδ
dk xδ
= hj − hk .
=
This gives us the following telescoping sum.
g =
=
t
X
i=1
t
X
ci fi
ci di hi
i=1
= c1 d1 (h1 − h2 ) + (c1 d1 + c2 d2 )(h2 − h3 )
+ · · · + (c1 d1 + · · · + ct−1 dt−1 )(ht−1 − ht ) + (c1 d1 + · · · + ct dt ) ht
|
{z
}
=0
= c1 d1 S(f1 , f2 ) + (c1 d1 + c2 d2 )S(f2 , f3 )
+ · · · + (c1 d1 + · · · + ct−1 dt−1 )S(ft−1 , ft ),
whence g is a linear combination of S-polynomials. Now,
multideg(hj ) = δ = multideg(hk ),
and lc(hj ) = lc(hk ) = 1. Therefore, multideg(hj − hk ) < δ.
Theorem 2.34. Let I ⊆ S be an ideal with generating set G := {g1 , . . . , gt }.
The set G is a Gröbner basis for I if and only if the remainder on division
of S(gj , gk ) by G is 0 for all j, k.
14
BRIAN JOHNSON
Proof. Fix a monomial order <.
(=⇒): Since S(gj , gk ) ∈ I, the remainder when divided by G is zero from
Corollary 2.30.
(⇐=): Since we know I = (G), we need only show
in(I) = (in(g1 ), . . . , in(gt )).
And clearly, we need only show in(I) ⊆ (in(g1 ), . . . , in(gt )). Let f ∈ I. Then
there exist h1 , . . . , ht ∈ S such that
(1)
f=
t
X
hi gi .
i=1
We know
multideg(f ) ≤ max {multideg(hi gi ) multideg} .
1≤i≤t
Let m(i) = multideg(hi gi ) and δ = max{m(1), . . . , m(t)}, so that
multideg(f ) ≤ δ.
Do this for each (possibly infinitely many times) possible combination of the
form (1). This gives possibly different δ, but we can take δ to be minimal
because > is a well-ordering.
Claim: multideg(f ) = δ.
By way of contradiction, assume multideg(f ) < δ. Rewrite f :
X
X
f =
h1 gi +
hi gi
m(i)=δ
X
=
m(i)<δ
(hi − lt(hi )gi +
m(i)=δ
m(i)=δ
|
X
lt(hi )gi +
{z
}
(†)
hi gi
m(i)<δ
{z
|
X
all have multideg < δ
}
Notice that (†) above has multidegree less than δ, as multideg(f ) < δ.
Working with (†), let lt(hi ) = ci xα(i) . Then
X
X
(†) =
lt(hi )gi =
ci xα(i) gi .
m(i)=δ
m(i)=δ
multideg(xα(i) gi )
Now
= δ. So the previous lemma implies that (†) is a
linear combinatino of the S-polynomials S(xα(j) gj , xα(k) gk ). Observe that
S(xα(j) gj , xα(k) gk ) =
xδ
xδ
α(j)
x
g
−
xα(k) gk
j
xα(j) lt(gj )
xα(k) lt(gk )
= xδ−γjk S(gj , gk ),
where γjk = lcm(lm(gj ), lm(gk )). Therefore, there exist cjk ∈ k such that
X
(†) =
lt(hi )gi
m(i)=δ
=
X
cjk xδ−γjk S(gj , gk ).
POWER OF MONOMIAL IDEALS
SUSAN COOPER
15
But when we divide S(gj , gk ) by G, we have zero remainder. This means
S(gj , gk ) can be written
S(gj , gk ) =
t
X
aijk gi ,
i=1
for some aijk ∈ S, and
multideg(aijk gi ) ≤ multideg(S(gj , gk )).
(2)
So we have
xδ−γjk S(gj , gk ) =
(3)
t
X
bijk gi ,
i=1
where bijk = aijk
(4)
xδ−γjk .
Then (2) and the lemma give
multideg(bijk gi ) ≤ multideg(xδ−γjk S(gj , gk ))
< δ.
Then
(†) =
X
lt(hi )gi
m(i)=δ
=
X
cjk xδ−γjk S(gj , gk )
j,k
=
X
cjk
=
!
bijk gi
i=1
j,k
t
X
t
X
hei gi .
i=1
By (4), multideg(hei ) < δ. So in our original expression for f all terms have
multidegree less than δ. But this contradicts the minimality of δ. Therefore,
multideg(f ) = δ = multideg(hi gi ) for some i. This implies lt(f ) is divisible
by lt(gi ), which implies lt(f ) ∈ (in(g1 ), . . . , in(gt )), which implies in(I) ⊆
(in(g1 ), . . . , in(gt )).
Example 2.35. Let S = k[x, y, z], and set I = (g1 , g2 ) = (x − z 2 , y − z 3 ),
G = {g1 , g2 }.
(1) Let >=>lex . We look at the S-polynomial(s) of the generators:
xy
xy
g1 −
g2
S(g1 , g2 ) =
x
y
= xz 3 − yz 2
= z 3 (x − z 2 ) − z 2 (y − z 3 ) + 0.
Therefore G is a Gröbner basis for I.
16
BRIAN JOHNSON
(2) Now set >=>grevlex . Then
z3
z3
g
−
g2
1
−z 2
−z 3
= −xz + y
= 0g1 + 0g2 + (−xz + y).
S(g1 , g2 ) =
So we see G is not a Gröbner basis for I with this ordering! [It turns
out that one can show G0 = {−z 2 + x, −xz + y, −x2 + yz} is in fact
a Gröbner basis for I.]
In the second example above, G is not a Gröbner basis for I, so a question
one should ask is: how do we go about computing a Gröbner basis for I?
One idea is to add (nonzero) remainders upon dividing by the S-polynomials.
E.g., since the remainder above was y −xz, consider the set G0 = {g1 , g2 , g3 },
where g3 = y −xz. Then, since S(g1 , g2 ) = g3 , we get a zero remainder when
dividing S(g1 , g2 ) by G0 . But also, we find that
S(g1 , g3 ) = −x2 + zy,
and it’s straightforward to see that this is actually the remainder when
S(g1 , g3 ) is divided by G0 . So then we set
G00 = {g1 , g2 , g3 , −x2 + zy }.
| {z }
g4
One can then compute
S(g1 , g4 ) = −yzg1 + 0g2 + 0g3 + xg4
S(g2 , g3 ) = −yg1
S(g2 , g4 ) = (−z 2 y − xy)g1
S(g3 , g4 ) = −yg1 ,
and we see that all of these have zero remainder when divided by G00 . Thus
G00 is a Gröbner basis for I = (x − z 2 , y − z 3 ) with >=>grevlex .
Let T = {t1 , . . . , tl } ⊆ S. Recall that in(T ) = (in(t1 ), . . . , in(tl )) and
lt(T ) = {lt(t1 ), . . . , lt(tl )}.
Theorem 2.36 (Buchberger’s Algorithm). Let I = (f1 , . . . , ft ) 6= (0) be an
ideal in S. Then the following algorithm gives a Gröbner basis for I:
Input F = (f1 , . . . , ft ).
Set G := F
Repeat
G0 := G
For each {p, q} ⊆ G0 , p 6= q, Do
R := remainder of S(p, q) upon division by G0
If S 6= 0, then G := G ∪ {R}
until G = G0
POWER OF MONOMIAL IDEALS
SUSAN COOPER
17
Proof. We first claim that G ⊆ I at each step. To see this, note that it is
true initially. Then, since p, q ∈ I, we have S(p, q) ∈ I. Thus the equation
(from the division algorithm)
S(p, q) = (stuff in I) + R
implies that the remainder is also in I. Thus G ∪ {R} ⊆ I. We also observe
that at each step G is a generating set for I. It is also easy to see that G
is actually a Gröbner basis for I should the algorithm terminate (since it
terminates if and only if we get all 0 remainders upon division by G = G0 ).
So all that remains to be shown is that the algorithm does, in fact, terminate.
Note that at the end of each step G0 ⊆ G, so
in(G0 ) ⊆ in(G).
But if G0 6= G, then in(G0 ) 6= in(G). Indeed, using our prior notation, lt(R)
is not divisible by any leading terms of G0 . Since S is Noetherian, this
process cannot go on indefinitely, since that would give an infinite strictly
increasing chain.
Example 2.37. Let I = (f1 , f2 ) = (x2 y − 1, xy 2 − x) ⊆ k[x, y], and suppose
>=>lex . One can show
S(f1 , f2 ) = x2 − y = 0f1 + 0f2 + (x2 − y).
Setting f3 = x2 − y, we find
S(f1 , f3 ) = y 2 − 1 = 0f1 + 0f2 + 0f3 + (y 2 − 1).
Then one can show G = {f1 , f2 , f3 , f4 } is a Gröbner basis for I (check).
It’s fairly easy to see that we don’t need all of the fi above. In fact,
f2 = xf4 . But in general, how do we tell if we need all of the generators we
find using Buchberger’s Algorithm?
Lemma 2.38. Let (0) 6= I ⊆ S have a Gröbner basis G. Let p ∈ G, where
lt(p) ∈ in(G r {p}). Then G r {p} is a Gröbner basis.
Proof. By definition, in(G) = in(I), and so it’s clear that
in(G r {p}) = in(I).
And any Gröbner basis generates the ideal, so (G r {p}) = I.
Definition 2.39. A minimal Gröbner basis for (0) 6= I ⊆ S is a Gröbner
basis G for I satisfying:
(1) lc(p) = 1 for all p ∈ G, and
(2) lt(p) ∈
/ in(G r {p}) for all p ∈ G.
Example 2.40. Let I = (x2 y − 1, xy 2 − x) ⊆ S and >=>lex . We saw above
in 2.37 that {f1 , f2 , f3 , f4 } is a Gröbner basis for I. Upon further inspection,
one sees that lt(f1 ) = y lt(f3 ) and lt(f2 ) = x lt(f4 ). So another (minimal)
Gröbner basis for I is {f3 , f4 }.
18
BRIAN JOHNSON
Example 2.41. Gröbner bases are not in general unique. Suppose I =
(x3 − 2xy, x2 y − 2y 2 + x) ⊆ k[x, y] and >=>grlex . then
1
Ga = {x2 + axy, xy, y 2 − x}
2
is a minimal Gröbner basis for every a ∈ k.
b are minimal Gröbner bases for (0) 6=
Remark 2.42. Fix >. If G and G
b And, as a consequence, |G| = |G|.
b
I ⊆ S, then lt(G) = lt(G).
Definition 2.43. Let (0) 6= I ⊆ S. A reduced Gröbner basis for I is a
Gröbner basis G for I satisfying
(1) lc(p) = 1 for all p ∈ G and
(2) no monomial of p lies in in(G r {p}) for all p ∈ G.
Example 2.44. Let I = (x2 y − 1, xy 2 − x) ⊆ k[x, y] with >=>lex . Then
G = {x2 − y, y 2 − 1} is a reduced Gröbner basis for I.
Proposition 2.45. Let (0) 6= I ⊆ S. Fix a monomial ordering >. Then I
has a unique reduced Gröbner basis.
Proof. It’s not even clear such a thing exists, so we begin with that. Let
G be a minimal Gröbner basis for I. Let g ∈ G. We say g is reduced for
G if it satisfies the second condition in the previous definition. Our goal
is to modify G (without destroying its minimality or Gröbner basis-ality)
b
so that all of its elements are reduced for G. First of all, suppose that G
is another minimal Gröbner basis for I containing g ∈ G. By the previous
b so g being reduced for G implies g is reduced for
remark, lt(G) = lt(G),
b
G. Second, observe the following: given g ∈ G, let g 0 be the remainder
obtained by dividing g by G r {g}. Set G0 = (G r {g}) ∪ {g 0 }. We claim
G0 is a minimal Gröbner basis for I. Indeed, lt(g) is not divisible by any
element of lt(G − {g}). Thus
lt(g 0 ) = lt(g).
Therefore, in(G0 ) = in(G). Since G0 ⊆ I, G0 is a minimal Gröbner basis for
I. Moreover, by the division algorithm, g 0 is reduced for G0 . Iterate.
b be reduced Gröbner bases. Then both are minTo see uniqueness, let G, G
b (by the remark). Let g ∈ G. Then there exists
imal, so lt(G) = lt(G)
b
gb ∈ G such that lt(g) = lt(b
g ). Consider g − gb. Since g − gb ∈ I, dividing by
G gives a remainder of 0. But, since lt(g) = lt(b
g ), these terms cancel, and
b
since G, G are reduced, none of the remaining terms are divisible by any of
b Therefore, when we divide g − gb by G, our
the members of lt(G) = lt(G).
remainder is precisely g − gb = 0.
2.6. Some Further Applications of Gröbner bases.
I. Affine Varieties.
Recall:
POWER OF MONOMIAL IDEALS
SUSAN COOPER
19
(1) Let {f1 , . . . , ft } ⊆ S. The affine ariety defined by f = f1 , . . . , ft
is
V (f ) = {(a) ∈ k n | fi (a) = 0 for all i}.
(2) Let I ⊆ S e an ideal. We define
V (I) = {(a) ∈ k n | f (a) = 0 for all f ∈ I}.
Proposition 2.46. Let I ⊆ S be an ideal. Then V (I) is an affine
variety (in the sense of the first definition). In particular, if I = (f ),
then
V (I) = V (f ).
Proof. By the Hilbert Basis Theorem, I has a finite generating set,
say I = (f1 , . . . , ft ). Clearly V (I) ⊆ V (f ) as each fi ∈ I. On the
other hand, suppose (a) ∈ V (f ). Then for any f ∈ I, we can write
f=
t
X
hi fi ,
i=1
where hi ∈ S. But then
X
X
hi fi (a) =
hi · 0 = 0,
f (a) =
and V (f ) ⊆ V (I).
Example 2.47. Examples of finding V (I).
(1) Let I = (xz − y, xy + 2z 2 , y − z) ⊆ C[x, y, z], where >=>lex ,
and x > y > z. A computer algebra system will give a Gröbner
basis G = {g1 , g2 , g3 } for I, where
g1 = y − z
g2 = xz − y
g3 = −2z 2 − z.
As G is still a generating set for I, we know V (I) = V (g1 , g2 , g3 ).
Setting g3 = 0, we find that either z = 0 or z = − 21 . If z = 0,
it is easy to see that we must also have y = 0, and there are no
restrictions on x. On the other hand, if z = − 12 , we find that
y = − 12 and x = 1. Therefore,
1 1
V (I) = {(1, − , − ), (a, 0, 0) | a ∈ C}.
2 2
(2) Let I = (x2 + y 2 + z 2 − 1, y 2 + z 2 − 2x, 2x − 3y − z) ⊆ C[x, y, z],
where >=>lex , and x > y > z. We get a Gröbner basis from
20
BRIAN JOHNSON
somewhere: G = {g1 , g2 , g3 }, where
g1 = −2x + y 2 + 2z 2
1301
57
2239 2
g2 =
y + z3 −
z
8
4
24
335
4055
z−
+
72
18
114 4
16 3
1376 2
g3 = −
z +
z −
z
1301
3903
11709
352
904
+
z+
.
11709
11709
One could then theoretically solve for z, then y, and then x.
Remark 2.48. When one uses lex order, a Gröbner basis simplifies
equations by eliminating variables successively. In fact, the order of
elimination corresponds to the order of the variables (for exactitudes,
look up stuff on elimination theory).
II. Affine Varieties and Implicitization.
Let f = f1 , . . . , fn ∈ k[t1 , . . . , tm ] = k[t]. Suppose we have parametric equations
x1 = f1 (t), . . . , xn = fn (t),
defining a subset Ve of an algebraic variety V in k n . Can we find
polynomial equations in k[x] that define V ?
One possible method is the following: Study the affine variety in
k n+m defined by
x1 − f1 (t) = 0, . . . , xn − fn (t).
Use lex order on k[t, x] with t1 > · · · > tm > x1 > · · · > xn . Find
a Gröbner basis for (x1 − f1 , . . . , xn − fn ) to eliminate the ti . The
polynomials remaining in the xi are “likely candidates.”
Example 2.49. Let Ve be the surface in C3 defined by
x = ut
y = 1−u
z = u + t − ut,
where u and t are parameters. Work in C[t, u, x, y, z] with >=>lex
and t > u > x > y > z. Then I := (x − ut, y + u − 1, z − u − t + ut)
has a Gröbner basis G = {g1 , g2 , g3 }, where
g1 = u + y − 1
g2 = −t + x + y + z − 1
g3 = xy + y 2 + yz − 2y − z + 1.
Now g3 defines an algebraic variety V in C3 containing Ve . On the
other hand, V may indeed be “larger” than Ve .
POWER OF MONOMIAL IDEALS
SUSAN COOPER
21
III. The Ascending Chain Condition
One may prove that S has the a.c.c. using the Hilbert Basis Theorem
which we used Gröbner bases to prove. Exercise: Prove it.
3. Hilbert Functions
References for this section include [2, Ch. 4], [16, Ch. 2], [18], and Susan
Cooper’s thesis.
Definition 3.1.
L
(1) An N-graded ring T is a commutative ring of the form T = i≥0 Ti ,
where Ti is a subgroup of T under +, and Ti Tj ⊆ Ti+1 . The elements
of Ti are homogeneous of degree i.
(2) A standard graded k-algebra is an N-graded ring T such that T is
Noetherian, T0 = k (a field), and T is generated as a T0 -algebra by
elements of degree 1. I.e., T = k[T1 ].
(3) A ring T is reduced if it has no nonzero nilpotents.
Remark 3.2. If T is a standard graded k-algebra, then
T ∼
= k[x0 , . . . , xn ]/I,
for some homogeneous ideal I ⊆ k[x0 , . . . , xn ]. Moreover, if T is reduced,
then I is radical. Note: We will assume here that I 6= (x0 , . . . , xn ).
L
Definition 3.3. Let T = i≥0 Ti be a standard graded k-algebra.
(1) The Hilbert function of T is the sequence
H(T ) := {H(T, i)}i≥0 = {HT (i)}i≥0 ,
where H(T, i) = dimk Ti .
(2) The first difference Hilbert function of T is the sequence
∆H(T ) = {∆H(T, i)}i≥0 ,
where ∆H(T, 0) = 1, and ∆H(T, i) = H(T, i) − H(T, i − 1).
Remark 3.4. One observes that
l
X
∆H(T, i) = H(T, l).
i=0
Example 3.5.
L
(1) The polynomial ring R = k[x0 , . . . , xn ] =
d≥0 Rd is a standard
graded k-algebra. A k-basis for Rd is the set of all monomials of
degree d. Therefore,
n+d
dimk Rd =
d
= H(R, d).
22
BRIAN JOHNSON
Also,
n+d
n+d−1
∆H(R, d) =
−
(for d ≥ 1)
d
d−1
n+d−1
=
.
d
(2) Let R = k[x0 , x1 , x2 ] and I = (x20 , x31 , x42 ). We see (by just listing a
basis) that
dimk I0
dimk I2
dimk I3
dimk I4
=
=
=
=
dimk I1 = 0
1
4
10.
For example, a k-basis for I4 is seen to be
{x40 x30 x1 , x30 x2 , x20 x1 x2 , x20 x22 , x20 x21 , x41 , x31 x2 , x0 x31 , x42 }.
Also, since

A := R/I ∼
=

M
d≥0
Rd 


M

d≥0
Id  ∼
=
M
(Rd /Id ),
d≥0
we see that
H(A, d) = H(R, d) − H(I, d).
Then from our earlier formula for H(R), we can compute
H(A) = {1, 3, 5, 6, 5, 3, 1, 0, 0, . . . }
∆H(A) = {1, 2, 2, 1, −1, −2, −2, −1, 0, 0, . . . }.
This raises a couple of questions.
(1) Have all the possible Hilbert functions of standard graded k-algebras
been characterized?
(2) How are these useful?
The answer to the first is yes, and is answered by Macaulay’s Theorem, and
we’ll explore the answer to the second.
Definition 3.6. A nonempty set M of monomials in the indeterminates
y1 , . . . , yn is an order ideal of monomials if whenever m ∈ M and a monomial
m0 divides m, then m0 ∈ M.
Example 3.7. Let M = {y10 y20 , y1 , y2 , y12 , y22 , y13 , y1 y2 , y1 y22 } is an order ideal
of monomials. A picture is shown in Figure 2.
Remark 3.8. An order ideal of monomials is not and ideal, nor a k-basis
for one. However if M := {monomials in k[y]}, then Mc = M r M is a
generating set for the ideal generated by monomials in Mc .
POWER OF MONOMIAL IDEALS
SUSAN COOPER
23
6
y2
In some sense this is the complement of a picure like Figure 1 on page 8.
s
s
s
s
s
s
s
s
y1
-
Figure 2. A graphical representation of an order ideal.
Theorem 3.9 (Macaulay). Let T be a standard graded k-algebra.
x1 , . . . , xn be a k-basis for T1 and
Let
π : k[y] −→ T
be a k-algebra homomorphism defined by π(yi ) = xi . Then there exists an
order ideal of monomials M in k[y] such that
π(M) = {π(m) | m ∈ M}
is a k-basis for T .
Proof. Let M = {monomials in k[y]}. Define a total order on M as follows.
If u = y α and v = y β , then u < v if and only if the last nonzero entry of
(b1 − a1 , . . . , bn − an ,
n
X
i=1
bi −
n
X
bi )
i=1
is positive. Note that for all u, v, m ∈ M , u < v implies mu < mv. Now,
define a sequence of monomials u1 , u2 , . . . as follows. Let u1 = 1 and let
ui+1 be the least element (under <) such that
π(u1 ), . . . , π(ui ), π(ui+1 )
are k-linearly independent. If no such ui+1 exists, then the sequence terminates at ui . We claim that M = {u1 , u2 , . . . } is the desired order ideal.
Certainly, π(M) is a k-basis of T (by construction). Now, assume M is not
an order ideal of monomials. Then there exist u, v ∈ M such that u ∈ M
and v ∈
/ M, yet v divides u. Since v ∈
/ M, there exists a relation of the form
X
π(v) =
ci π(ui ),
24
BRIAN JOHNSON
where ci ∈ k, and ui < v. Let w ∈ M such that u = vw. Then
π(u) = π(vw)
= π(v)π(w)
X
=
ci π(ui )π(w)
X
=
ci π(ui w).
Now, for each i, ui < v, so ui w < vw = u. But this contradicts the fact that
u ∈ M. Therefore M is as claimed.
Note: With notation from above, let I = (m | m ∈ Mc ) ⊆ k[y]. Then
H(T ) = H(k[y]/I).
Corollary 3.10. This gives us the following bijection in answer to the first
question above.
(
)
(
)
Hilbert functions of rings
Hilbert functions of standard
k[x]/I, where I is mono←→
graded k-algebras
mial
Now, above, we have π : k[y] −→ T , where yi 7→ xi , and M is an order
ideal of monomials such that π(M) is a k-basis for T . Let J = ker π. Then
Mc = {lm(f ) | f ∈ J}.
So, to understand H(k[x]/I), where I is monomial, we need some preliminary combinatorial facts.
Lemma 3.11. Let d ∈ Z, d > 0. Then any a ∈ N can be written uniquely
as
md
md−1
mj
(5)
a=
+
+ ··· +
,
d
d−1
j
where md > md−1 > · · · > mj > j ≥ 1. Equation (5) is called the dth
Macaulay representation of a, or the d-binomial expansion of a.
Lemma 3.12. Let a, a0 ∈ N such that for some fixed d
md
m1
a=
+ ··· +
d
1
and
0
a =
m0d
d
0
m1
+ ··· +
.
1
Then a > a0 if and only if (md , . . . , m1 ) >lex (m0d , . . . , m01 ).
Warning: One of our general assumptions is that x1 > x2 > · · · > xn , but
[2] uses the convention that x1 < x2 < · · · < xn , and occasionally it makes
a cosmetic difference.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
25
Definition 3.13. Let h, d be positive integers where the d-binomial expansion of h is
md
md−1
mj
,
h=
+
+ ··· +
j
d
d−1
and md > · · · > mj ≥ j ≥ 1. We define
md+1
md−1 + 1
mj + 1
hdi
h :=
+
+ ··· +
.
d+1
d
j+1
We also set 0hdi := 0.
Example 3.14. One may find it useful here to have a copy of Pascal’s
Triangle in hand’s reach.
1
1
1
1
3
1
1
1
2
4
5
6
10
6
15
20
d=0
.
d=1
.
1
d=2
.
1
d=3
.
3
1
d=4
.
4
1
d=5
.
10
5
1
d=6
.
15
6
1
(1) Find the 3-binomial expansion of 16. Since d = 3, we look at that
diagonal of Pascal’s triangle. We see that the largest number
in
5
that diagonal less than or equal to 16 is 10, and 10 = 3 . Since
16 − 10 = 6, we now look in the d = 2 diagonal and find thelargest
number less than or equal to 6. In this case, we can use 6 = 42 . (The
top number is just coming from whichever row we are in, counting
the top row as row 0.) Thus,
5
4
16 =
+
.
3
2
Then
h3i
16
6
5
=
+
= 15 + 10 = 25.
4
3
One way to think of hhdi is to think of shifting all the binomial
coefficients in the expression for h down and to the right one spot in
Pascal’s triangle.
26
BRIAN JOHNSON
(2) Find the 4-binomial expansion of 23. We find
6
4
3
1
23 =
+
+
+
,
4
3
2
1
and
7
5
4
2
23h4i +
+
+
+
5
4
3
2
= 31.
(3) If h, d ∈ Z such that d ≥ h > 0, then the d-binomial expansion of h
is
d
d−h+1
+ ··· +
,
d
d−h+1
so hhdi = h.
Definition 3.15. A sequence of nonnegative integers c = {ci }i≥0 is called
an O-sequence (“oh-sequence”) if
(1) c0 = 1, and
hii
(2) ci+1 ≤ ci for all i ≥ 1.
Also, c is called a differentiable O-sequence if both c and ∆c are O-sequences.
Example 3.16.
(1) Suppose R = k[x0 , x1 , x2 ] and I = (x20 , x31 , x42 ) ⊆ R. We showed
H(R/I) = (1, 3, 5, 6, 5, 3, 1, 0, . . . )
and
∆H(R/I) = (1, 2, 2, 1, −1, −2, −2, −1, 0, . . . ).
Letting {cd } = H(R/I) one can check (and it’s not particularly
enlightening, trust me) that {cd } is an O-sequence. It is immediately
clear that ∆H(R/I) is not an O-sequence, because it has negative
entries.
(2) Let X be the variety of the collection of points
{[1 : 0 : 0], [1 : 0 : 1], [1 : 1 : 0], [1 : 1 : 1], [1 : 2 : 0], [1 : 3 : 0]}
in P2 . Then (one can show)
I := I(X)
=
⊆
(x0 x2 − x22 , x21 x2 − x1 x22 ,
11
1
x30 x1 − x20 x21 + x0 x31 − x41 )
6
6
k[x0 , x1 , x2 ] =: R.
Then
H(X) := H(R/I) = (1, 3, 5, 6, 6, . . . ),
and ∆H(X) = (1, 2, 2, 1, . . . ). It turns out both are O-sequences, so
H(X) is a differentiable O-sequence.
POWER OF MONOMIAL IDEALS
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27
3.1. Macaulay’s Theorem.
Theorem 3.17 (Macaulay’s Theorem). Let k be a field and let H : N −→ N
be a numerical Hilbert function. Say H(i) = di and write H = {di }i≥0 .
TFAE:
(1) There exists a standard graded k-algebra T such that H(T ) = H.
(2) There exists a monomial ideal I ⊆ S = k[x1 , . . . , xd ] such that
H(S/I) = H (in fact, I can be taken to be a “lexsegment ideal”).
(3) H is an O-sequence.
(4) Let Mt be the largest dt monomials in k[x1 , . . . , xd1 ] of
S degree t using
grevlex order (with x1 > · · · > xd1 ). Set M = t≥0 Mt . The
collection M is an order ideal of monomials.
Example 3.18. Consider H = (1, 3, 5, 6, 8, 0, . . . ) = {di }i≥0 . We’ve already
h3i
seen (1, 3, 5, 6) is the start of an O-sequence. But d3 = 6h3i = 7, and
h3i
d4 = 8 > d3 , so H is not an O-sequence.
Equivalently, one could compute
M0 = {1}
M1 = {x1 , x2 , x3 }
M2 = {x21 , x1 x2 , x22 , x1 x3 , x2 x3 }
M3 = {x31 , x21 x2 , x1 x22 , x32 , x21 x3 , x1 x2 x3 }
M4 = {x41 , x31 x2 , x21 x22 , x1 x32 , x42 , x31 x3 , x21 x2 x3 , x1 x22 x3 }
It’s easy to see that M (as above) is not an order ideal of monomials, because
x1 x22 x3 ∈ M, but x22 x3 ∈
/ M. Therefore, there does not exist a standard
graded k-algebra T with H(T ) = H.
Part of proof of 3.17. We’ve already shown the first two are equivalent.
For (4) =⇒ (2), assume M is an order ideal of monomials the ring S =
k[x1 , . . . , xd1 ]. Let I ⊆ S be the ideal generated by the monomials in Mc .
Then Sd /Id = Md by construction. Therefore,
H(S/I, i) = di .
That is, H(S/I) = H (and I is monomial).
For (3) =⇒ (4) of 3.17, we’ll need to take a detour and use:
Definition 3.19. Let S = k[x1 , . . . , xn ] and u ∈ Sd a monomial. Use lex
order with x1 >lex · · · >lex xn .
(1) Lu := {v ∈ Sd | v is a monomial and v <lex u}
(2) Ru := {v ∈ Sd | v is a monomial and v ≥lex u}
Example 3.20. Suppose S = k[x1 , x2 , x3 ], and >=>lex with x1 > x2 > x3 .
Let u = x22 ∈ S2 . Then
Lu = {x2 x3 , x23 },
and
Ru = {x21 , x1 x2 , x1 x3 , x22 }.
28
BRIAN JOHNSON
Definition 3.21. Let S = k[x1 , . . . , xn ], where >=>lex , and x1 > · · · > xn .
(1) A monomial set of the form Ru for some monomial u ∈ Sd is called
a lexsegment of degree d.
(2) A lexsegment ideal is an ideal i ⊆ S such that for all d, Id is generated
as a k-vector space by a lexsegment of degree d.
Example 3.22. Let S = k[x1 , x2 , x3 ] and >=>lex .
(1) If J = (x31 , x21 x2 , x21 x3 , x1 x22 , x32 ), then J3 is the k-span of the generators (since they’re all of degree 3), and we see that x1 x22 > x1 x2 x3 >
x32 , but x1 x2 x3 ∈
/ J3 , so J is not a lexsegment ideal.
(2) If I = (x31 , x21 x2 x3 ), then we see that I4 is the k-span of the set
{x41 , x31 x2 , x31 x3 , x21 x2 x3 }. But this is not a lexsegment, as x21 x22 ∈
/ I4 ,
so I is not a lexsegment ideal.
(3) One can show (with a little hand-waving for Id when d > 4) that
I = (x21 , x1 x2 , x1 x23 , x32 ) is a lexsegment ideal.
Remark 3.23. It can be shown an ideal I ⊆ S is a lexsegment ideal if and
only if Id is generated as a k-becor space by the largest dimk Id (under >lex )
elements in the degrees d for which I has minimal generators.
Proposition 3.24. Let u ∈ S be a monomial. Then Rxn Ru = Ruxn .
Proof. Observe Rxn = {x1 , . . . , xn }. Let v ∈ Ru . Then for any i ∈
{1, . . . , n}, xi v ≥ xn v ≥ xn u. That is, xi v ∈ Ruxn (and v was arbitrary,
so this gives one containment). Conversely, let v ∈ Ruxn . By definition,
v ≥ uxn . If v = wxn for some w, it’s clear we’re done, so suppose xn - v.
bn−1
Then v > uxn , so write u = xa11 · · · xann and v = xb11 · · · xn−1
, and let i be
the smallest index such that bi > ai . If there exists j > i with bj > 0, then
w = xvj > u implies w ∈ Ru . If no such j exists, then
v
≥u
xi
implies w
e ∈ Ru . In either case, v ∈ Rxn Ru .
w
e=
We now make an observation. Let u ∈ Sd be a monomial. We can
decompose Lu as follows: Let i be the smallest index such that xi | u. Set
L0u = {monomials in Sd in the variables xi+1 , . . . , xn }
and
L00u = L xu .
i
Then Lu = L0u ∪ L00u xi (it’s worth–in fact essential–noting that this is a
disjoint union). For example, if u = x1 x22 in the ring k[x1 , x2 , x3 ], then i = 1
and w = xu1 = x22 . So
Lu = {x1 x2 x3 , x1 x23 , x32 , x22 x3 , x2 x23 , x33 }
L0u = {x32 , x22 x3 , x2 x23 , x33 }
L00u = {x2 x3 , x23 } = Lw .
POWER OF MONOMIAL IDEALS
SUSAN COOPER
29
In general, we’re going to keep decomposing the L00 part.
Introduce the following notation: write
[xi , . . . , xn ]d = {monomials of degree d in k[xc , . . . , xn ]}.
Suppose u = xj1 xj2 · · · xjd , where 1 ≤ j1 ≤ · · · ≤ jd (e.g., for u = x1 x22 , we
have j1 = 1 and j2 = j3 = 2). Then
Lu = [xj1 +1 , . . . , xn ]d ∪ L xu xj1
j1
= [xj1 +1 , . . . , xn ]d ∪ [xj2 +1 , . . . , xn ]d−1 xj1 ∪ L x
u
j1 xj2
xj1 xj2
..
.
So we get (as a disjoint union)
Lu =
d
[
[xji +1 , . . . , xn ]d−(i−1) xj1 · · · xji−1 ,
i=1
where if i = 1, we set xj1 xj0 = 1. We call this the canonical decomposition.
Note:
d
X
|Lu | =
|[xi +1 , . . . , xn ]d−(i−1) |
=
i=1
d X
i=1
n − ji + d − i
.
d−i+1
Define mi = n − jd−i+1 + (i − 1). Then
n − j1 + d − 1
n − jd
|Lu | =
+ ··· +
d
1
md
m1
=
+ ··· +
.
d
1
Since 1 ≤ j1 ≤ j2 ≤ · · · ≤ jd , we have md > md−1 > · · · ≥ 0. That is, we’ve
found the d-binomial expansion for |Lu |.
Example 3.25. Continuing from above, we see we can write
Lu = {x1 x2 x3 , x1 x23 , x32 , x22 x3 , x2 x23 , x33 }
= {x32 , x22 x3 , x2 x23 , x33 } ∪ {x2 x3 , x23 } · x1
= [x2 , x3 ]3 ∪ x1 Lx22
= [x2 , x3 ]3 ∪ x1 {x23 } ∪ x1 x2 {x3 }
= [x2 , x3 ]3 ∪ x1 [x3 ]2 ∪ x1 x2 [x3 ]1 .
Now, let w = ux3 = x1 x22 x3 . What’s the canonical decomposition of Lw ?
One can show (and compare with the result above for Lu ) that
Lw = [x2 , x3 ]4 ∪ x1 [x3 ]3 ∪ x1 x2 [x3 ]2 .
This idea will be useful in the proof of the next proposition.
30
BRIAN JOHNSON
Proposition 3.26. Let S = k[x1 , . . . , xn ], and let u ∈ Sd be a monomial.
Then
|Luxn | = |Lu |hdi .
Proof. Let
Lu =
d
[
[xji +1 , . . . , xn ]d−(i−1) xj1 · · · xji−1
i=1
be the canonical decomposition. We claim the canonical decomposition of
Luxn is
Lu =
d
[
[xji +1 , . . . , xn ]d−(i−2) xj1 · · · xji−1 .
i=1
We’ll handwave a short proof. Note that the decomposition of Lu depended
only on the ji . So if we set
(
ji 1 ≤ i ≤ d
`i =
n i = d + 1,
then the decomposition of Lw depends only on the `i (which are mostly the
ji ).
Now, we’ve already shown
d X
n − ji + d − i
|Lu | =
,
d−i+1
i=1
so it follows
|Luxn | =
d X
n − ji + d − i + 1
i=1
d−i+2
= |Lu |hdi .
Corollary 3.27. Let I ⊆ S be a lexsegment ideal. Set A = S/I. Then
H(A, d + 1) ≤ H(A, d)hdi for all d ≥ 0, with equality if and only if Id+1 =
(x1 , . . . , xn )Id .
Proof. Because I is lexsegment, Id is the k-span of Ru for some u ∈ Sd , a
monomial. Thus
H(A, d) = dimk Sd − dimk Id
= dimk Sd − |Ru |
= |Lu |.
Now Id+1 ⊇ Rxn Ru = Ruxn , so
dimk Id+1 ≥ |Ruxn |.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
31
But this implies
H(A, d + 1) = dimk Sd+1 − dimk Id+1
≤ dimk Sd+1 − |Ruxn |
= |Luxn |
= |Lu |hdi .
Further, we have Id+1 = Rxn Ru if and only if Id+1 = (x1 , . . . , xn )Id .
Proof of 3.17 continued. (3) =⇒ (4): Let k be a field and H : N −→ N such
hdi
that H(i) = di a function. By the definition of an O-sequence, dt+1 ≤ dt
for all t ≥ 1. Then
h1i d1 + 1
d1
h1i
=
.
d2 ≤ d1 =
2
1
Similarly
d1 + 2
d3 ≤
3
..
. d1 + t − 1
dt ≤
.
t
Consider two cases. 1 +t
Case 1: dt+1 = dt+1
. Then we have Mt =
. This is only if dt = d1 +t−1
t
[x1 , . . . , xd1 ]t and Mt+1 = [x1 , . . . , xd1 ]t+1 . Clearly, if u ∈ Mt+1 and xi | u,
then xui ∈ Mt (and then one can iterate for all k < t + 1).
1 +t
Case 2: dt+1 < dt+1
. Let S 0 = k[y1 , . . . , yd1 ]. Define π : S −→ S 0 by
π(xd1 − i) = yi+1 for 0 ≤ i ≤ d1 − 1. Note that if w, w
e ∈ S are monomials
of the same degree, then
(6)
w >grevlex w
e
if and only if
π(w) <lex π(w),
e
where we’ve declared x1 >grevlex · · · >grevlex xd1 and y1 >lex · · · >lex yd1 .
Now if Mt = [x1 , . . . , xd1 ]t , were done as in Case 1. If not, then by (6),
there exist monomials ut , ut+1 ∈ S such that
π
Mt+1 ∼
= Lπ(ut+1 )
π
and
Mt ∼
= Lπ(ut ) .
Now,
hti
dt+1 = |Mt+1 | + |Lπ(ut+1 ) | ≤ dt = |Lπ(ut )yn |.
Thus,
Suppose w ∈ Mt+1
Rπ(ut )yn = Ryn Rπ(ut ) ⊆ Rπ(ut+1 ) .
and xi | w. Then π(xi ) | π(w) ∈ Lπ(ut+1 ) . Set w
e=
w
xi .
π(w)
Then π(w)
e = π(x
(we want to show w
e ∈ Mt ). If π(w)
e ∈
/ Lut , then
i)
π(w)
e ∈ Ru1 . This implies
π(xi )π(w)
e ∈ Ryn Rπ(ut ) ⊆ Rπ(ut+1 ) ,
32
BRIAN JOHNSON
/ Mt+1 , a contradiction. Therewhich then implies π(w) ∈
/ Lut+1 , or that w ∈
e ∈ Mt , finishing Case 2. So now, if a ∈ M and b | a,
fore, π(w)
e ∈ Lu1 and w
then systematically dividing single powers via the previous cases, we have
b ∈ M.
We now work on some preliminaries for (1) =⇒ (3).
Definition 3.28. Let a, d be positive integers, where the d-binomial expansion of a is
md
mj
,
a=
+ ··· +
j
d
with md > md−1 > · · · > mj ≥ j ≥ 1. Then we define (Warning: not
standard,[2] only)
md − 1
md−1 − 1
mj − 1
ahdi :=
+
+ ··· +
.
d
d−1
j
Remark 3.29 (from [2]). We note some useful facts. Let a, a0 , d be positive
integers. Suppose the d-binomial expansion of a is
md
mj
m1
a=
+ ··· +
+ ··· +
,
d
j
1
where j = min{i | mi ≥ i}. Then
(1) If a ≤ a0 , then ahdi ≤ (a0 )hdi .
(2) If a ≤ a0 , then ahdi ≤ (a0 )hdi .
(3) If mj 6= j, then (a − 1)hdi < ahdi .
(
ahdi + m1 + 1 if j = 1
hdi
(4) (a + 1) =
ahdi + 1
if j > 1.
We’re now ready to begin our final push toward completing the proof of
Macaulay’s Theorem.
Remark 3.30.
(1) For any commutative ring A, we have the Zariski topology. The
closed sets are of the form
V (I) = {p ∈ Spec A | p ⊇ I},
where I ⊆ A is an ideal.
(2) Let T be a standard graded k-algebra, where k is an infinite field.
Then the affine space T1 is irreducible. So for any U ⊆ T1 , where
open
U 6= ∅, U is dense.
Definition 3.31. Let T be a standard graded k-algebra, with k an infinite
field. We say that a property P holds for a general linear form of T1 if there
exists ∅ =
6 U ⊆ T1 such that P holds for all h ∈ U .
open
POWER OF MONOMIAL IDEALS
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33
Theorem 3.32 (Green’s Theorem). Let T be a standard graded k-algebra,
where k is an infinite field. Let d ≥ 1 be an integer. Then
H(T /gT, d) ≤ H(T, d)hdi
for a general linear form.
Proof. Let r = sup{dimk (gTd−1 ) | g ∈ T1 } (this is well-defined because
T is a standard graded k-algebra...basically we have all the nice finiteness
properties we’d want). Define
U = {g ∈ T1 | dimk (gTd−1 ) = r}.
Clearly U 6= ∅. To see that U is open (and hence dense), we do some
handwaving. Choose a basis x1 , . . . , xn for T1 , as well as bases for Td−1 and
P
·h
Td . Consider the maps αk : Td−1 −→ Td , where h =
ci xi , ci ∈ k. Each
map αh can be described by a matrix with entries that are linear forms in
c1 , . . . , cn . Replace c1 , . . . , cn with indeterminates y1 , . . . , yn to get a matrix
Mh of linear polynomials in k[y1 , . . . , yn ]. Let
Ir (M ) = ideal generated by the r × r minors of M .
Define
V (Ir (M )) = {(a1 , . . . , an ) ∈ k n | f (a) = 0 for all f ∈ Ir (M )}.
Now,
g=
n
X
ai xi ∈ U
⇐⇒ rank(Mg ) = r
i=1
⇐⇒
Mg has r linearly independent
columns
there exists an r × r subma⇐⇒ trix of Mg with nonzero determinant
⇐⇒
there exists f ∈ Ir (Mg ) such
that f (a1 , . . . , an ) 6= 0
That is, U corresponds the set
X
ai xi and there exists f ∈ Ir (Mg ) with f (a) 6= 0}.
{(a) ∈ k n | g =
And we can realize this set as (V (Ir (Mg ))c (this is the handwaving part).
Thus, dimk (gTd−1 ) = r for a general linear form.
Let g ∈ U and set A = T /gT . We’ll show
H(A, d) ≤ H(T, d)hdi
by induction on min{d, dimk T1 }. If d = 1, then
H(A, 1) = dimk ((T /gT )1 )
≤ dimk T1 − 1
34
BRIAN JOHNSON
and
dimk T1
H(T, 1) =
1
implies
H(T, 1)h1i =
dimk T1 − 1
= dimk T1 − 1.
1
If dimk T1 = 1, then
H(T, 1)h1i
1
=
= 0,
1 h1i
and certainly
H(T /gT, 1) = dimk ((T /gT )1 ) = 0
(there’s actually a little more work to show for d ≥ 1, but it’s easy). Now,
assume min{d, dimk T1 } > 1, and let
V = {f ∈ A1 | dimk (f Ad−1 ) is maximal}.
Then V 6= ∅. Let φ : T −→ T /gT = A be the canonical surjection, and let
W be the open subset
W = (U r kg) ∩ φ−1 (V ).
Note that W 6= ∅: T1 is irreducible and φ−1 (V ) 6= ∅. Assume U r kg = ∅.
Then U ⊆ kg. Since U is dense and kg is closed, kg = T1 . But dimk T1 > 1.
Choose ge ∈ W . Then
(7)
H(A, d) = dimk (Ad /e
g Ad−1 ) + dimk (e
g Ad−1 )
≤ H(A, d)hdi + dimk (e
g Ad−1 )
geTd−1
≤ H(A, d)hdi + dimk
g(e
g Td−2 )
gTd−1
,
= H(A, d)hdi + dimk
ge(gTd−2 )
To see that (7) is true, note that
gTd−1
dimk
= dimk (gTd−1 ) − dimk (e
g (gTd−2 )),
ge(gTd−2 )
and
dimk
geTd−1
g(e
g Td−2 )
= dimk (e
g Td−1 ) − dimk (g(e
g Td−2 ))
= dimk (e
g Td−1 ) − dimk (e
g (gTd−2 ))
Because g, ge ∈ U , we see that
dimk (gTd−1 ) = dimk (e
g Td−1 ),
and our claim now follows. Let
f = {f ∈ T1 | dimk (f Td−2 ) is maximal}.
W
POWER OF MONOMIAL IDEALS
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35
f is nonempty and open. Further, W ∩ W
f 6= ∅ (irreducibility),
As before, W
f
so choose ge ∈ W ∩ W . Let P = T / Ann(g). Then
Pd−1 ∼
= gTd−1 .
Therefore, by the inductive hypothesis
gTd−1
≤ H(P, d − 1)hd−1i
dimk
ge(gTd−2 )
= (dimk (gTd−1 ))hd−1i
= (dimk Td − dimk ((T /gT )d ))hd−1i
= (H(T, d) − H(A, d))hd−1i .
So, what we’ve shown is this:
H(A, d) ≤ H(A, d)hdi + (H(T, d) − H(A, d))hd−1i .
And if we let b = H(A, d) and a = H(T, d), we are done by the following
proposition.
Proposition 3.33. Given integers 0 < b < a such that
b ≤ bhdi + (a − b)hd−1i ,
one has b ≤ ahdi .
Proof. Suppose not. That is, suppose b > ahdi . Write
md
md−1
mj
b=
+
+ ··· +
.
d
d−1
j
If b > ahdi , then
a<
md + 1
mj + 1
+ ··· +
.
d
j
Then
md + 1
mj + 1
md
md−1
mj
a−b <
+···+
−
+
+ ··· +
.
d
j
d
d−1
j
x
Recalling the combinatorial identity xx + y−1
= x+1
and rearranging
y
terms, we get
md
mj
a−b<
+ ··· +
.
d−1
j−1
Then
md − 1
mj − 1
(a − b)hd−1i <
+ ··· +
d−1
j−1
and
md − 1
mj − 1
bhdi =
+ ··· +
.
d
j
36
BRIAN JOHNSON
Thus, b ≤ bhdi + (a − b)hd−1i , and this implies
md − 1
mj − 1
md − 1
mj − 1
+ ··· +
+
+ ··· +
d
j
d−1
j−1
md
mj
=
+ ··· +
j
d
= b,
a contradiction (one may have to worry about the special case j = 1, but
as long as you’re careful and define your binomial coefficients correctly, it
shouldn’t be a problem).
Finally finishing 3.17. For (1) =⇒ (3), let T be a standard graded k-algebra
with H(T ) = {di }i≥0 . We want to show H(T ) is an O-sequence. We may
assume k is infinite: if not, replace T by ` ⊗k T , where ` is an infinite
(in cardinality) field extension of k. Let g ∈ T1 be a linear form and set
A = T /gT . The exact sequence
0 −→ gTd −→ Td+1 −→ Ad+1 −→ 0
gives
H(t, d + 1) ≤ H(T, d) +H(A, d + 1).
| {z } | {z }
b
a
We want to show b ≤ ahdi . Green’s Theorem (3.32) implies
b ≤ a + bhd+1i .
(8)
Suppose the (d + 1)-binomial expansion of b is
md+1
m1
b=
+ ··· +
.
d+1
1
then
bhd+1i
md+1 − 1
m1 − 1
=
+ ··· +
.
d+1
1
But (8) implies
md+1 − 1
m1 − 1
a≥
+ ··· +
.
d
0
Let j = min{i | mi ≥ i}.
Case 1: j > 1. Then m1 = 0 implies m11 = m10−1 = 0. Thus
md+1
m2
ahdi ≥
+ ··· +
d+1
2
= b.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
37
m2 − 1
m1 − 1
m2
, we have
+
with
1
1
0
hdi
md+1 − 1
m3 − 1
m2
≥
+ ··· +
+
1
d
2
m2 + 1
md+1
m3
+
≥
+ ··· +
3
2
d+1
m2
md+1
m3
+ m2
+
=
+ ··· +
2
3
d+1
> b (because m2 > m1 )
Case 2: j = 1. Then, replacing
ahdi
Lemma 3.34. Let T be a standard graded k-algebra, where k is a field.
Then
H(T, d + 1) = H(T, d)hdi
for d 0.
Proof. We know T ∼
= S/I, where S = k[x1 , . . . , xn ] and I is a homogeneous ideal. By Macaulay’s Theorem (3.17), there exists an order ideal of
monomials
[
M=
Md
d≥0
such that H(T, d) = |Md | = H(S/J, d), where J = (Mc ). By construction
(
all monomials in Sd
if Id = 0
Md =
Lud for some monomial ud if Id 6= 0.
So there is a correspondence
{monomials in Jd } ↔ Rud .
Since J is finitely generated, there exists p ∈ N such that for all d ≥ p,
{monomials in Jd+1 } ↔ Rud+1 = Rxn Rud = Rxn ud .
Thus, for d ≥ p,
|Md+1 | = |Lud xn | = |Lud |hdi = |Md |hdi .
3.2. Hilbert Functions of Reduced Standard Graded k-algebras. A
good reference for most of this material is [11].
Lemma 3.35. Let T be a reduced standard graded k-algebra. Then H(T )
is a differentiable O-sequence.
38
BRIAN JOHNSON
Proof. Macaulay’s Theorem (3.17) implies H(T ) is an O-sequence. Now let
g ∈ T1 be a nonzerodivisor (the existence of which we’ll not prove, but still
use). Define the map αg : T −→ gT via a 7→ ag. Note that T /gT is a
standard graded k-algebra, and Macaulay’s Theorem implies H(T /gT ) is
an O-sequence. Moreover,
dimk (gT )d = dimk Td−1 .
Therefore,
H(T /gT, d) = dimk Td − dimk (gT )d
= dimk Td − dimk Td−1
= ∆H(T, d).
Remark 3.36. The converse of the lemma is also true. It can be obtained
through “lifting” of monomial ideals.
Definition 3.37. Let R = k[x0 , . . . , xn ] and S = k[x1 , . . . , xn ], where k is an
algebraically closed field of characteristic 0. Let K ⊆ S be a homogeneous
ideal. We say that K lifts to an ideal I ⊆ R if
√
(1) I = I,
(2) x0 + I is a nonzerodivisor on R/I, and
(3) under the canonical isomorphism R/(x0 ) −→ S, we have (I, x0 )/(x0 )
mapped isomorphically to K.
Lemma 3.38. Let K ⊆ S be a homogeneous ideal. Suppose K lifts to
I ⊆ R. Then
∆H(R/I, t) = H(S/K, t)
for all t ≥ 0.
Proof. We have the isomorphisms
S/K ∼
= R/(I, x0 ) ∼
= (R/I)/(x0 + I),
so H(S/K, t) = H((R/I)/(x0 + I), t) for all t ≥ 0. but x0 + I is not a
zerodivisor on R/I, so using the same argument as in the proof of Lemma
3.35,
H((R/I)/(x0 + I)) = ∆H(R/I).
Theorem 3.39 (Hartshorne, 1966). Any monomial ideal in k[x1 , . . . , xn ]
can be lifted to an ideal in k[x0 , . . . , xn ], where k is of sufficient transcendence
degree over its prime field.
We won’t use the original proof, but we need to record some preliminary
notation and facts for the alternate proof.
• k is a field, k = k, char k = 0 (in particular k is infinite)
• N = {0, 1, 2, . . . } (we include 0)
POWER OF MONOMIAL IDEALS
SUSAN COOPER
39
• M is the collection of all monomials in S = k[x1 , . . . , xn ] (including
1)
• R = k[x0 , . . . , xn ] = S[x0 ]
• M ↔ Nn via xα ↔ α
• We have a partial ordering on Nn , where (a1 , . . . , an ) ≤ (b1 , . . . , bn ) if
and only if ai ≤ bi for all i. This corresponds to the former monomial
dividing the latter.
P
• If α = (a1 , . . . , an ) ∈ Nn , then deg α = deg(xα ) = ai .
Associations: For each j ∈ {1, . . . , n}, choose an infinite set of distinct
elements tj,i ∈ K, i ≥ 0. Here, it is possible that if ` 6= m, we may have
t`,a = tm,b for some a, b. To each α = (a1 , . . . , an ) ∈ Nn , associate the point
α = [1 : t1,a1 : t2,a2 : · · · : tn,an ] ∈ Pnk .
Define
To each monomial f =
polynomial
xα
f=
deg α = deg α.
= xa11 · · · xann ∈ S, associate the homogeneous
n
Y

Y

j=1

aj −1
(xj − tj,i x0 ) ∈ R.
i=0
Thus, f is the product of distinct linear forms, and
deg f = deg α = deg α.
Example 3.40. Set n = 2, k = C, and let tj,i = i for j = 1, 2, i ≥ 0. If
α = (1, 4), then
α = [1 : t1,1 : t2,4 ] = [1, 1, 4],
4
and f = x1 x2 , so
"a −1
# "a −1
#
1
2
Y
Y
f =
(x1 − t1,i x0 )
(x2 − t2,i x0 )
i=0
i=0
= x1 x2 (x2 − x0 )(x2 − 2x0 )(x2 − 3x0 ),
and
deg α = 5 = deg α = deg f .
Lemma 3.41 (2.1 in [11]). Let f = xα ∈ S. Then
(1) f (β) = 0 if and only if α 6≤ β.
(2) f (γ) = 0 for all γ such that deg γ ≤ deg α (except for α itself ).
Lemma 3.42 (2.3 in [11]). Let f ∈ Rd . If f (α) = 0 for all α such that
deg α ≤ d, then f = 0.
Proof. Set
Pd = {α | deg α ≤ d}
= {[1 : t1,a1 : · · · : tn,an ] |
X
ai ≤ d}.
40
BRIAN JOHNSON
Now,
|Pd | =
d
X
dimk Si
i=0
d X
n−1+i
=
i
i=0
n+d
=
.
d
We’ll use induction on n and d. If n = 1, then Pd has d + 1 distinct points.
Since f ∈ Rd and f vanishes at d + 1 distinct points, f = 0. If d = 0, then
Pd has only 1 point, and certainly f = 0. Suppose d > 0 and n > 1. If f
vanishes on Pd ⊆ Pn , then f vanishes on the d+n−1
points of Pd lying on
n−1
the hyperplane
x1 − t1,0 x0 = 0
(i.e., a1 = 0).
Write
f = (x1 − t1,0 x0 )g + r,
where g ∈ Rd−1 and r is homogeneous of degree d and is in only
the variables
points as well.
x0 , x2 , . . . , xn . But then r must also vanish on the d+n−1
n−1
By induction on n, r = 0. Hence g must vanish on the points of Pd not lying
on the hyperplane x1 − t1,0 x0 (where a1 6= 0). After a change of variables,
these points are one of the Pd−1 (in the appropriate ring, etc.). By induction
on d, g = 0, and hence f = 0.
Theorem 3.43 (2.2 of [11]). Let J ⊆ S be a monomial ideal with minimal
generating set f1 = xα1 , . . . , f` = xα` . Then J lifts to I = (f1 , . . . , f` ) ∈ R.
Proof. Let N = {monomial in J} ⊆ M, and let M = M r N . Then the
elements of M are representatives of a k-basis for S/J. Let
M = {α ∈ Pn | xα ∈ M }.
Let L be the ideal generated by the homogeneous elements f ∈ R such that
f (α) = 0 for all α ∈ M . We claim I = L. By Lemma 3.41(1), I ⊆ L. So
let f ∈ Rd such that f (α) = 0 for all α ∈ M . Let {β 1 , β 2 , . . . } (a finite set)
be the points of Pd r M , where the βi are arranged in increasing order of
degree (when equal, arrange however you’d like).
Then h1 := xβ1 ∈ J, implying h1 = xγ fi for some fi , 1 ≤ i ≤ `. Thus h1
is a multiple of fi . Using Lemma 3.41(1) again, we get
h1 (β1 ) 6= 0,
which implies there exists λ1 ∈ k such that f − λ1 h1 xt01 vanishes at β1
(t1 = d − deg h1 ). Similarly, h2 := xβ2 ∈ J implies h2 is a multiple of some
fj . Also, h2 (β2 ) 6= 0. So there exists λ2 ∈ k such that
(f − λ1 h1 xt01 ) − λ2 h2 xt02
POWER OF MONOMIAL IDEALS
SUSAN COOPER
41
vanishes at β2 . But Lemma 3.41(2) implies h2 (β1 ) = 0. Rinsing and repeating, we obtain f − G for some G ∈ I. By construction f − G vanishes
on Pd r M . By assumption, f vanishes on M . Further, I ⊆ L implies G
vanishes on M . Therefore f − G vanishes on Pd . Then Lemma 3.42 implies
f = G, so f ∈ I. We conclude that I is the ideal of points in M . Thus
(1) I is a radical ideal
(2) x0 doesn’t vanish at any point of M , which implies x0 is a nonzerodivisor on R/I.
(3) (I, x0 )/(x0 ) ∼
= J by the definition of the fi .
Example 3.44 (2.4 in [11]). Let k = Q, n = 2, and
J = (x1 x42 , x21 x22 , x51 ) ⊆ S = k[x1 , x2 ].
| {z } | {z } |{z}
f1
f2
f3
If we set tj,i = i for j = 1, 2 and i ≥ 0, then
M = {α = [1 : a : b] | xa1 xb2 ∈ M }.
The affine picture of M is the same as the picture of M above (by choice of
tj,i ). Also,
f1 = x1 x42 = x1 x2 (x2 − x0 )(x2 − 2x0 )(x2 − 3x0 )
f2 = x21 x22 = x1 (x1 − x0 )x2 (x2 − x0 )
f3 = x51 = x1 (x1 − x0 )(x1 − 2x0 )(x1 − 3x0 )(x1 − 4x0 )
6
x2
s
The open circles denote generators of the
ideal, and the closed ones denote monomials of M (as defined above in the proof of
Theorem 3.43).
s
s
s
d
s
s
s
s
d
s
s
s
s
s
s
s
s
s
s
d
x1
-
Figure 3. A graphical representation of monomials in and
not in J.
42
BRIAN JOHNSON
Then J lifts to I, where the key step is that
f is homogeneous and f vanishes on x1 = 0
I = f ∈ k[x0 , x1 , x2 ]
.
and the 10 points in Figure 3 not on x1 = 0
Note: f1 vanishes at all α of degree less than or equal to 5, except for
[1 : 1 : 4]. Similar statements hold for f2 and [1 : 2 : 2] and f3 and [1 : 5 : 0].
Remark 3.45. In [11] the authors prove any monomial ideal in k[x1 , . . . , xn ]
minimally generated by f1 , . . . , f` lifts to I = (f1 , . . . , f` ) ⊆ R[x1 , . . . , xn ]
when:
(1) k is infinite
(2) |k| ≥ e, where
e := max{aji | 1 ≤ j ≤ n, 1 ≤ i ≤ `},
with fi = xαi and αi = (a1i , . . . , ani ).
They do not assume char k = 0.
Corollary 3.46 (2.5 in [11]). Let {bi }i≥0 be a differentiable O-sequence,
and let k be a field with |k| ≥ e (as above). Then there is a reduced standard
graded k-algebra T such that
H(T, i) = bi .
Proof. Let H = {bi }i≥0 . By assumption ∆H = {ci }i≥0 is an O-sequence.
So, by Macaulay’s Theorem (3.17), there exists a standard graded k-algebra
A∼
= k[x1 , . . . , xn ]/J, where J is a monomial ideal and
H(A) = {ci }i≥0 = ∆H.
Let e be as above from J, and let J be minimally generated by f1 , . . . , f` .
Then J lifts to I = (f1 , . . . , f` ) ⊆ R[x1 , . . . , xn ]. Then T = R/I is a reduced
standard graded k-algebra. Since J lifts to I, H(S/J, t) = ∆H(R/I, t), and
H(S/J, t) = ct = bt − bt−1 , we have
H(R/I) = H(T ) = H.
To summarize, we have a correspondence between differentiable O-sequences
and not just standard graded k-algebras, but reduced standard graded kalgebras.
3.3. Hilbert Functions of Points in Pn . Suppose k is a field of characteristic 0, k = k, and R = k[x0 , . . . , xn ]. Let X = {P1 , . . . , Ps } be a set
of distinct points in Pnk (note that |X| < ∞, unlike in the proof of Theorrem 3.43). We know that R/I(X) is a redued standard graded k-algebra, so
H(X) := H(R/I(X)) is a differentiable O-sequence. But can we say more
about H(X)? The answer is yes, but we
S need some notation first.
There exists a linear form F not in si=1 I(Pi ) (prime avoidance). Therefore, F is not a zerodivisor in A = R/I(X). Make a linear change of variables
so that F = x0 . Let B = A/(x0 ). Observe that
POWER OF MONOMIAL IDEALS
SUSAN COOPER
43
(1) We have
B∼
= R/(I, x0 ) ∼
= S/J = k[x1 , . . . , xn ]/J,
where J is the homogeneous ideal generated by setting x0 = 0 in the
generators of I.
(2) B is a standard graded k-algebra.
(3) √
ht(I(X)) = n implies that dim B = 0 (Krull dimension).
(4) J = (x1 , . . . , xn ).
Lemma 3.47. Bt = 0 for all t 0.
√
Proof. First note that J = (x1 , . . . , xn ) implies that (x1 , . . . , xn )t ⊆ J.
But
M
Sd = (x1 , . . . , xn )t
d≥t
for each t. So for each j ≥ t
(S/(x1 , . . . , xn )t )j = 0
implies (S/J)j = 0.
Lemma 3.48. Bt = 0 implies Bt+1 = 0.
Proof. If Bt = 0, then (S/J)t = 0. But this implies J ⊇ (x1 , . . . , xn )t (and
that J ⊇ (x1 , . . . , xn )s for all s ≥ t). Thus Bt+1 = 0.
Recall that if x0 is not a zerodivisor on A = R/I(X), then
H(A/(x0 )) = H(B) = ∆H(A) = ∆H(R/I(X)).
This gives:
Corollary 3.49. There exists d ∈ Z such that
(1) H(X, t) = d for all t 0,
(2) H(X, t) = d implies H(X, t + 1) = d, and
(3) if H(X, t) = c = H(X, t + 1), then c = d.
Lemma 3.50. Let X = {P1 , . . . , Ps } ⊆ Pn be a set of distinct points.
(1) H(X, d) ≤ s for all d ≥ 0,
(2) H(X, s − 1) = s, and
(3) H(X, d) = s for all d ≥ s.
Proof. (1): Let F ∈ (I(X))d . Then F (Pi ) = 0 for 1 ≤ i ≤ s imposes a
linear condition on the coefficients of F. This gives a homogeneous system
of coefficients with s equations in n+d
unknowns. The number of linearly
d
independent solutions is
n+d
− rank(coeff.
matrix}).
|
{z
d
size x×(n+d
d )
44
BRIAN JOHNSON
So dimk (I(X)d ) ≥
n+d
d
− min{s,
n+d
d
}. Therefore,
H(X, d) = dimk Rd − dimk (I(X)d )
n+d
n+d
n+d
≤
−
− min s,
d
d
d
n+d
= min s,
d
≤ s.
(2): Let Hi define a hyperplane in Pn such that Hi (Pi ) = 0 and Hi (Pi ) 6= 0
for all j 6= i (|k| = ∞ ensures Hi exists). Let
Fi =
s
Y
Hj ,
j=1
j6=i
for 1 ≤ i ≤ s. Then F1 , . . . , Fs ∈ Rs−1 and Fi ∈
/ I(X). We claim that
F1 , . . . , Fs ∈ As−1 (where A = R/I(X)) are linearly independent.
Indeed,
P
suppose
there
exist
scalars
c
,
.
.
.
,
c
∈
k
such
that
c
F
=
0.
Then
1
s
i i
P
ci Fi ∈ I(X), which gives
X
ci Fi (P1 ) = 0.
Without loss of generality, suppose c1 = 0. By construction, we know
Fi (P1 ) = 0 for i 6= 1 and F1 (P1 ) 6= 0, so we have a contradiction. Then
(1) implies H(X, s − 1) = s.
(3): Let F ∈ R1 such that F is not a zerodivisor on A = R/I(X). Then
·F
Ad −→ Ad+1 is one-to-one. Therefore dimk Ad+1 ≥ dimk Ad . Then (1) and
(2) finish the proof.
Theorem 3.51 (Geramita-Maroscia-Roberts). Let H = {ht }t≥0 be a sequence of nonnegative integers. The following are equivalent:
(1) H = H(X) for some finite set of s points in Pn .
(2) H is a differentiable O-sequence such that
h1 ≤ dimk (k[x0 , . . . , xn ]1 ) = n + 1,
and ht = s for t ≥ s − 1.
Example 3.52. It can be shown H = (1, 3, 4, 5, 6, 6, . . . ) is a differentiable
O-sequence. We see that
∆H = (1, 2, 1, 1, 1, 0, . . . ).
Now, h1 ≤ 2 + 1 so we see we’re looking for s points in P2 . But also, ht = 6
for t ≥ 6 − 1 = 5, so s = 6. But by considering ∆H, we see we need
1 monomial (point) of degree 0, 2 monomials (points) of degree 1, and 1
monomial in each of the degrees 2, 3, and 4. Figure 4 illustrates this and
the generators for the ideal J, which we’re going to lift (cf. Figure 3 on
page 41). Lifting J (as in 3.44) gives the 6 points
POWER OF MONOMIAL IDEALS
SUSAN COOPER
45
6
x2
Points on the diagonal x2 =
−x1 + d are monomials in degree d.
d
s
d
s
s
s
s
s
d
x1
-
Figure 4. Finding the generators for J ⊆ k[x1 , x2 ].
X = {[1 : 0 : 0], [1 : 1 : 0], [1 : 2 : 0], [1 : 3 : 0], [1 : 4 : 0], [1 : 0 : 1]},
and H(X) = H.
3.4. Maximal Growth of Hilbert Functions and Consequences. The
main reference for this material is [1], and many results will be given without
proof.
Definition 3.53 (also a theorem). Let A = k[x0 , . . . , xn ]/I, where I is
homogeneous. Then there exists a unique PA (x) ∈ Q[x] such that PA (n) =
H(A, n) for all n 0. This is called the Hilbert polynomial.
Suppose R = k[x0 , . . . , xn ] and W ⊆ Rd is a subspace. Let J = (W ),
B = R/J, and H(B, d) = bd . Macaulay’s Theorem implies
hdi
bd+1 = H(B, d + 1) = dimk Rd+1 − dimk Jd+1 ≤ bd .
But this then gives
dimk Jd+1
n+d+1
hdi
= dimk (R1 Jd ) ≥
− bd .
n
hdi
That is, when bd+1 = bd , W “grows least.”
Theorem 3.54 (Gotzmann’s Persistence Theorem). With R, W , J, and bd
as above, write
md
mj
bd =
+ ··· +
.
d
j
hdi
If bd+1 = bd , then for any ` ≥ 1,
md + `
mj + `
bd =
+ ··· +
.
d+`
j+`
46
BRIAN JOHNSON
6
x2
s
s
s
s
s
s
s
s
x1
-
Figure 5. Computing H(X) and ∆H(X).
That is,
PB (x) =
md + x − d
mj + x − d
+ ··· +
.
x
x−d+j
Definition 3.55. Let R = k[x0 , . . . , xn ], I ⊆ R be a homogeneous ideal,
A = R/I, and J ⊆ A a homogeneous ideal.
(1) H(R/I) = {bt }t≥0 has maximal growth in degree d if
hdi
bd+1 = bd .
(2) The saturation of J is
`
J sat := {f ∈ A | f · (x0 , . . . , xn ) ⊆ J for some `}.
Example 3.56. Let I ⊆ R be a radical ideal, I 6= (x0 , . . . , xn ). Then in
A = R/(0), J sat = J. I.e., J is saturated. So if X is a finite set of points in
Pn , we know I(X) is saturated.
Lemma 3.57 (1.1 in [1]). Let I ⊆ R = k[x0 , . . . , xn ] be saturated. Let
L ∈ R1 be a nonzerodivisor on A = R/I. Let
J = (I, L)/(L) ⊆ S = R/(L).
Set B = S/J. If H(A) has maximal growth in degree d, then H(B) has
maximal growth in degree d.
Example 3.58. The converse of the previous lemma is not true (Example
1.3 in [1]). Let A = R/I(X) where
[1 : 0 : 0],[1 : 1 : 0],[1 : 0 : 1],[1 : 2 : 0],
X=
.
[1 : 3 : 0],[1 : 4 : 0],[1 : 1 : 1],[1 : 0 : 2]
One can then compute
POWER OF MONOMIAL IDEALS
SUSAN COOPER
47
H(X) = (1, 3, 6, 7, 8 , 8, . . . )
|{z}
∆H(X) = (1, 2, 3, 1, 1 , 0, . . . ),
|{z}
but we see that ∆H(X) exhibits maximal growth in degree 3, but H(X)
does not, because 7h3i = 9 6= 8. [Here we are computing these as in 3.44,
and the associated picture is Figure 5.]
Definition 3.59. Fix r, k ≥ 1. Define fr,k : N −→ N by
x+r
x−k+r
fr,k (x) =
−
,
r
r
for x ≥ k. Also define fr,0 (x) = 0 for all r, x.
Remark 3.60 (2.1, 2.2 in [1]).
(1) For x ≥ k, fr,k is the Hilbert polynomial for a degree k hypersurface
in Pr .
(2) If k1 < k2 ≤ x, then fr,k1 (x) < fr,k2 (x).
Definition 3.61. Let I ⊆ R = k[x0 , . . . , xn ] be homogeneous such that
Id 6= 0. The potential GCD (or PGCD) of Id is
max{k | fn,k (d) ≤ H(R/I, d)}.
Remark 3.62. If there exists F ∈ Rt such that F | G for all G ∈ Id , then
Id ⊆ (F )d . Thus,
fn,t (d) ≤ H(R/I, d).
That is, P GCD(Id ) is the largest degree possible for a common divisor of
Id . But P GCD(Id ) > 0 does not imply that Id has a GCD of degree equal
to P GCD(Id ).
Proposition 3.63 (2.7 in [1]). Let I ⊆ R = k[x0 , . . . , xn ] be homogeneous
and suppose Id 6= 0. Assume P GCD(Id ) = c > 0 and H(R/I) has maximal
growth in degree d. Then both Id and Id+1 have a GCD, F , of degree c.
Corollary 3.64 (2.8 in [1]). Let I ⊆ R be homogeneous. Let H(R/I, d) =
fn,k (d) and H(R/I, d + 1) = fn,k (d + 1), where d ≥ k. Then Id = (F )d and
Id+1 = (F )d+1 for some F ∈ Rk .
What are the geometric consequences of all of the previous results?
Theorem 3.65 (3.4 in [1]).
Let X ⊆ Pn be a finiteset of distinct points
d+m−1
with ∆H(X, d) =
and ∆H(X, d + 1) = d+m
d
d+1 . Then X contains a
subset X1 lying on a subspace Pm and
m+1
d+m−1
d+m
∆H(X1 ) = (1, m,
,...,
,
, ∆H(X, d+2), . . . ).
2
d
d+1
Theorem 3.66 (3.6 in [1]). Let Y ⊆ Pn be a finite set of distinct points.
Assume that ∆H(Y, d) = ∆H(y, d + 1) = s for d ≥ s. Then (I(Y )≤d ) is the
48
BRIAN JOHNSON
saturated ideal of a curve V of degree s. Further, there exists Y1 ⊆ Y such
that
(
s,
s≤t≤d+1
∆H(Y1 , t) =
.
∆H(Y, t), t ≥ d
Corollary 3.67 (3.10 in [1]). Let X ⊆ Pn be a finite set of distinct points
with ∆H(X, d) = fm−1,k (d) and ∆H(X, d+1) = fm−1,k (d+1), where m ≥ 3,
d ≥ k. Then there exists X1 ⊆ X lying on a (reduced) hypersurface Y of
degree k in Pm . Further,
(
∆H(Y, t), 0 ≤ t ≤ d + 1
∆H(X1 , t) =
.
∆H(X, t), t ≥ d
Example 3.68 (0.7–0.9 in [1]).
(1) If X ⊆ P5 , |X| = 30, and H(X) = (1, 6, 21, 25, 30, 30, . . . ), then
{bt } := ∆H(X) = (1, 5, 15, 4, 5, 0, 0, . . . ).
Note b3 = 4 and 4h3i = 5, so ∆H(X) has maximal growth in degree
3. But
3 + (2 − 1)
b3 = 4 =
3
and
3+2
b4 = 5 =
,
3+1
(i.e., d = 3, m = 2) so Theorem 3.65 implies there exists X1 ⊆ X
such that
∆H(X1 ) = (1, 2, 3, 4, 5, 0, 0, . . . )
=⇒ H(X1 ) = (1, 3, 6, 10, 15, 15, . . . ).
That is, X1 consists of 15 points on a P2 ⊆ P5 .
(2) If X ⊆ P2 , |X| = 21, and H(X) = (1, 3, 6, 10, 14, 16, 18, 20, 21, 21, . . . ),
then
{bt } := ∆H(X) = (1, 2, 3, 4, 4, 2, 2, 2, 1, 0, 0, . . . ).
So b5 = 2 = b6 , and Theorem 3.66 implies (note the 4s do not allow
us to apply 3.66 because d 6≥ s) there exists X1 ⊆ X such that
(
2,
2≤t≤6
∆H(X1 , t) =
.
∆H(X, t), t ≥ 5
Therefore, ∆(1, a, 2, 2, 2, 2, 2, 2, 1, 0, 0, . . . ), and a = 2 because it has
to be an O-sequence. Then
H(X1 ) = (1, 3, 5, 7, 9, 11, 13, 15, 16, 16, . . . ),
so X1 has 16 points lying on a conic in P2 .
POWER OF MONOMIAL IDEALS
SUSAN COOPER
49
(3) If X ⊆ P3 , |X| = 18, and H(X) = (1, 4, 10, 15, 16, 17, 18, 18, . . . ),
then
∆H(X) = (1, 3, 6, 5, 1, 1, 1, 0, 0, . . . ),
and the sequence of 1s in ∆H(X) automatically implies maximal
growth. Then either 3.65 or 3.66 implies there exists X1 ⊆ X with
H(X1 ) = (1, 2, 3, 4, 5, 6, 7, 7, . . . )
with 7 points of the 18 lying on a P1 ⊆ P3 .
3.5. The Eisenbud-Green-Harris Conjecture. A reference for this material is [10]. We begin with some preliminary notation and definitions.
Definition 3.69. Let S = k[x1 , . . . , xn ], where k is a field.
(1) f1 , . . . , fr is a regular sequence in S if (f1 , . . . , fr ) 6= S and fi is a
nonzerodivisor on S/(f1 , . . . , fi−1 ) for all i.
(2) Assume f1 , . . . , fn ∈ S is a regular sequence where deg fi = ai . We
say that f1 , . . . , fn is an (a1 , . . . , an )-sequence. If A = (a1 , . . . , an ),
then f1 , . . . , fn is called an A-sequence.
(3) An ideal I ⊆ S is an (a1 , . . . , an )-sequence if I contains a subideal
(f1 , . . . , fn ) where f1 , . . . , fn is an (a1 , . . . , an )-sequence.
Example 3.70.
(1) xa11 , . . . , xann (where ai > 0) is an (a1 , . . . , an )-sequence.
(2) x1 , x2 (1 − x1 ), x3 (1 − x1 ) is a regular sequence in k[x1 , x2 , x3 ], but
x2 (1 − x1 ), x3 (1 − x1 ), x1 is not a regular sequence.
(3) Suppose F, G ∈ k[x1 , . . . , xn ], both homogeneous. Then F, G is a
regular sequence if and only if F, G have no common divisors.
(4) I = (x31 , x32 , x33 , x21 x2 , x1 x22 , x2 x23 ) ⊆ k[x1 , x2 , x3 ]. Then I is a (3, 3, 3)ideal and also a (3, 3, 4)-ideal.
A question one might ask is the following: Fix positive integers a1 ≤
· · · ≤ an . What are all the possible Hilbert functions of quotients of the
form k[x1 , . . . , xn ]/I, where I is a homogeneous (a1 , . . . , an )-ideal? Could
we use an analogue (and what would it be) to lex ideals?
Definition 3.71. Let A = (a1 , . . . , an ) where 1 ≤ a1 ≤ · · · ≤ an , ai ∈ Z.
An ideal l ⊆ S = k[x1 , . . . , xn ] is an A-lex-plus-powers ideal (or an A-LPP
ideal ) if
(1) L is minimally generated by xa11 , . . . , xann and monomials m1 , . . . , m` ,
and
(2) if r is a monomial in S such that deg r = deg mi and r >lex mi , then
r ∈ L (using x1 >lex · · · ).
Example 3.72. Let S = k[x1 , x2 , x3 ].
(1) We claim L = (x21 , x32 , x33 , x1 x22 , x1 x2 x3 ) ⊆ S is a (2, 3, 3)-LPP ideal.
Now, x21 , x32 , x33 are minimal generators of L. Also, the degree 3
50
BRIAN JOHNSON
monomials of S are (big–small under lex):
x31 , x21 x2 , x21 x3 , x1 x22 , x1 x2 x3 , x1 x23 , x32 , x22 x3 , x2 x23 , x33 .
|
{z
}
m1 and m2
Since all (three) of the monomials bigger than m1 and m2 are in L,
this is LPP.
(2) Let J = (x21 , x32 , x33 , x1 x22 , x1 x2 x3 , x22 x3 ). This is not a (2, 3, 3)-LPP
ideal as x1 x23 >lex x22 x3 , but x1 x23 ∈
/ J.
Proposition 3.73 (Francisco-Richert). Let L be an (a1 , . . . , an )-LPP ideal
in S = k[x1 , . . . , xn ]. Then the smallest H(S/L, d) monomials in Sd (under
lex with x1 >lex · · · ) not divisible by any xai i form a basis for (S/L)d .
Proof. Suppose xe = xe11 · · · xenn ∈ L is not divisible by any xai i . Let xd =
xd11 · · · xdnn be a monomial not divisible by any xai i such that deg(xe ) =
deg(xd ) and xd >lex xe . Our goal is to show xd ∈ L (then we will have
shown the proposition).
If xe is a minimal generator of L, then xd ∈ L by definition of LPP ideals.
Otherwise, there exists a minimal generator f of L such that
f = xe11 −b1 · · · xenn −bn
for bi ∈ N, where bj > 0 for some j. Let
d
i−1 ri
g = xd11 · · · xi−1
xi ,
where deg g = deg f and i is such that d1 + · · · + di−1 < deg f ≤ d1 + · · · + di .
We now show that g ≥lex f . If d1 = e1 − b1 , . . . , di−1 = ei−1 − bi−1 ,
then since deg g = deg f , we must have ri ≥ ei − bi . Therefore g ≥lex f
(by the definition of lex). On the other hand, if there exists j < i such
that dj 6= ej − bj , we claim we must have dj > ej − bj . If not, then we
have xd <lex xe , a contradiction. Thus dj > ej − bj , and we have g ≥lex f .
So, since deg g = deg f , f is a minimal generator of L, and g ≥lex f , the
definition of LPP ideals says g ∈ L. But, by construction, g | xd , which
implies xd ∈ L.
Corollary 3.74. Suppose L1 , L2 ⊆ S are two (a1 , . . . , an )-LPP ideals such
that H(S/L1 , d) = H(S/L2 , d). Further, suppose that all minimal generators
of L1 , L2 that are not pure powers have degree less than or equal to d. Then
(L1 )d+1 = (L2 )d+1 .
Proof. We have (L1 )d = (L2 )d . The only possible new generators in degree
d + 1 are of the form xd+1
.
i
Theorem 3.75 (Richert). Let a1 ≤ · · · ≤ an , b1 ≤ · · · ≤ bn , c1 ≤ · · · ≤
cn such that ai , bi , ci ∈ Z, with ai ≤ bi ≤ ci for all i. If there exists an
(a1 , . . . , an )-LPP ideal I and a (c1 , . . . , xn )-LPP ideal J in S such that
H(S/I) = H(S/J) = H,
then there exists a (b1 , . . . , bn )-LPP ideal L such that H(S/L) = H.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
51
Example 3.76 (Francisco-Richert). Let S = k[x1 , x2 , x3 ] and suppose
H = (1, 3, 3, 1, 0, 0, . . . ).
Suppose A = (2, 2, 2) and C = (2, 3, 4). Then I = (x21 , x22 , x23 ) is an A-LPP
ideal and
J = (x21 , x32 , x43 , x1 x2 , x1 x3 , x22 x3 , x2 x23 )
is a C-LPP ideal (also a lex ideal). One can show that H(S/I) = H(S/J) =
H. So there exist (2, 2, 4)-, (2, 2, 3)-, and (2, 3, 3)-LPP ideals with Hilbert
function H.
Conjecture 3.77 (Eisenbud-Green-Harris Conjecture). Let L ⊆ S be an
(a1 , . . . , an )-LPP ideal. Define Lhdi to be the ideal generated by L≤d and
(xa11 , . . . , xann ). Fix positive integers 1 ≤ a1 ≤ · · · ≤ an , and set A =
{a1 , . . . , an }. Let I ⊆ S be a homogeneous ideal. Suppose there exists an
A-LPP ideal L ⊆ S such that
H(S/I, d) = H(S/L, d).
Then
H(S/I, d + 1) ≤ H(S/Lhdi , d + 1).
Remark 3.78. With the above notation, such an A-LPP ideal may not
exist. For example, let A = (3, 3), S = k[x1 , x2 ], and I = (x21 , x22 ). Then I
is a (3, 3)-ideal since x31 , x32 ∈ I. So
H(S/I) = (1, 2, 1, 0, . . . ),
yet there does not exist a (3, 3)-LPP ideal L with H(S/L, 2) = 1. Indeed,
if there were, H(S/L, 2) = 1, and the smallest (under lex) monomial in S2
not divisible by x31 , x32 forms a basis for (S/L)2 . This is x22 . That is, x21 , x1 x2
are in L, but x31 is part of a minimal generating set for L, a contradiction.
Conjecture 3.79 (Equivalent form of EGH Conjecture). Suppose I ⊆
S = k[x1 , . . . , xn ] is an (a1 , . . . , an )-ideal which is homogeneous, and let
H(S/I) = H. Assume I is not a (b1 , . . . , bn )-ideal, where bi ≤ ai and bj < aj
for some j ∈ {1, . . . , n}. Then there exists an (a1 , . . . , an )-LPP ideal L such
that H(S/L) = H.
3.6. A combinatorial approach to EGH. In what follows, we outline a
combinatorial approach to the EGH Conjecture (3.77). References include
[4], [12], and [5].
3.6.1. Results of Clements-Lindström. Let k be a field such that k = k and
char k = 0. Fix integers 1 ≤ a1 ≤ · · · ≤ an . Set
V = {(b1 , . . . , bn ) ∈ Nn | 0 ≤ bi ≤ ai }.
Then V is in one-to-one correspondence with the monomials of the ring
k[x1 , . . . , xn ]
.
a1 +1
(x1 , . . . , xann +1 )
52
BRIAN JOHNSON
6
x2
s
s
s
s
s
s
s
s
h
s Q
s
s
s
x1
-
Figure 6.
Let W ⊆ V , and define ` = a1 + · · · + an . Set
o
n
X
bj = j
Vj = (b1 , . . . , bn ) ∈ V and Wj = W ∩ Vj .
Definition 3.80. With above notation (lex with x1 >lex x2 >lex · · · ):
(1) The set of the |Wj | smallest elements of Vj is the compression of Wj ,
denoted CWj .
S
(2) Define CW = `j=0 CWj . If CW = W , we say W is compressed.
(3) The set of the |Wj | largest elements of Vj is the last compression of
Wj , denoted LWj .
(4) Define Γ : V −→ {subsets of V } by
(b1 , . . . , bn ) 7→ {(b1 − 1, b2 , . . . , bn ), . . . , (b1 , . . . , bn−1 , bn − 1)} ∩ V.
S
Set Γ(W ) = w∈W Γ(w).
(5) If Γ(W ) ⊆ W , then W is closed.
(6) Define P : V −→ {subsets of V } by
(b1 , . . . , bn ) 7→ {(b1 + 1, b2 , . . . , bn ), . . . , (b1 , . . . , bn−1 , bn + 1)} ∩ V.
S
Set P (W ) = w∈W P (w).
Example 3.81. The set V = {(b1 , b2 ) ∈ N2 | 0 ≤ b1 ≤ 2, 0 ≤ b2 ≤ 3} is
in one-to-one correspondence with the collection of monomials of the ring
B := k[x1 , x2 ]/(x31 , x42 ). Let Q = (2, 1) ∈ V . Then we have a picture like
Figure 6. It’s straightforward to show that Γ(Q) = {(1, 1), (2, 0)}, which is
the set of degree two monomials that divide x21 x2 . Also,
P (Q) = {(3, 1), (2, 2)} ∩ V = {(2, 2)},
POWER OF MONOMIAL IDEALS
SUSAN COOPER
53
which is the set of degree four monomials of B divisible by x21 x2 . Let
W = {(0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), (2, 1), (1, 2)}.
Then Γ(W ) ⊆ W , so W is closed. Observe that
CW3 = {(0, 3), (1, 2)} * W3 .
(That is, the compressions need not stay in the original Wj .)
Theorem 3.82 (from [4]). Γ(CWj ) ⊆ C(Γ(Wj )).
Corollary 3.83 (from [4]). P (LWj ) ⊆ L(P (Wj )).
Corollary 3.84 (from [4]). If W is closed, CW is closed.
3.6.2. Order Ideals. Let 1 ≤ a1 ≤ · · · ≤ an , V , and W be as above. Fix
z1 = an , z2 = an−1 , . . . , zn = a1 . Let z = {z1 , . . . , zn }, and
Sz =
k[x1 , . . . , xn ]
.
z1 +1
(x1 , . . . , xznn +1 )
Let Mz be the collection of monomials in Sz . Reverse the order of coordinates. Then V is in one-to-one correspondence with Mz . Now, if W ⊆ V ,
then
CWj ↔ {largest |Wj | monomials of Sz of degree j using revlex}.
Definition 3.85. An order ideal of monomials of Sz is a nonempty set
M ⊆ Mz such that if m ∈ M , n ∈ Mz , and m | n, then n ∈ M .
Remark 3.86. Let M ⊆ Mz .
(1) If M is an order ideal of monomials, then Mz r M is the k-basis for
some monomial ideal in Sz .
(2) There is a correspondence M ↔ W ⊆ V , and M is an order ideal of
monomials if and only if W is closed (in the Γ(W ) ⊆ W sense).
Theorem 3.87 (from [5]). Let M be an order ideal of monomials in Sz .
f be the set consisting of the largest |Mj | elements in (Sz )j for each
Let M
f is an order ideal of monomials.
j ≥ 0 (using revlex). Then M
Proof. M ↔ W . M is an order ideal if and only if W is closed. By Corolf. Therefore, M
f is an order ideal of
lary 3.84, CW is closed. Also, CW ↔ M
monomials.
Theorem 3.88. Let I ⊆ Sz be a homogeneous ideal. Then there exists
M ⊆ Mz , an order ideal of monomials, whose canonical image in Sz /I
forms a k-basis for Sz /I.
Proof. With modifications, the proof of Theorem 3.9 should work.
Theorem 3.89 (from [5]). Let H = {ci }i≥0 . TFAE:
(1) There exists I ⊆ Sz homogeneous such that H(Sz /I) = H.
54
BRIAN JOHNSON
S
(2) let M = j≥0 Mj ⊆ Mz , where Mj consists of the cj largest monomials of Mz of degree j (using revlex). Then M is an order ideal of
monomials in Sz .
Proof. Apply the previous two theorems.
Remark 3.90. We really need z1 ≥ · · · ≥ zn (to have the theorem formulated in precisely this way). For example, suppose z1 = 1 ≤ z2 = 3. That
is,
k[x1 , x2 ]
Sz =
.
(x21 , x42 )
Then H(Sz ) = (1, 2, 2, 2, 1, 0, 0, . . . ). If I = (x1 x2 ) ⊆ Sz (I think this is a
lifting), then
H(Sz /I) = (1, 2, 1, 1, 0, 0, . . . ).
And M from the previous theorem would be:
{1, x1 , x2 , x1 x2 , x1 x22 },
which is not an order ideal of monomials.
3.7. Some enumeration.
Theorem 3.91 (from [5]). Let I ⊆ Sz be a homogeneous ideal such that
H(Sz /I) = H. Fix H(d) = c. Let J be the c largest monomials in (Mz )d
using (g)revlex. Let U be the set of all degree d + 1 monomials of M, all of
whose degree d factors are in J. Then H(d + 1) ≤ |U |.
Proof. Straightforward application of Corollary 3.83.
In order to answer the question, “what is |U |?” we introduce notation
j
from the theory of “multi-sets.” Let e1 , e2 , . . . ∈ N. We define e1 ,...,e
to
d
d
be the coefficient of x in the expansion of
j
Y
(1 + x + x2 + · · · + xei ),
i=1
which is the number of monomials of degree d in x1 , . . . , xj such that the
exponent of xi does not exceed ei .
Example 3.92. We see that
(1 + x + · · · + x4 )(1 + x + x2 ) = x6 + 2x5 + 3x4 + 3x3 + 3x2 + 2x + 1,
5 3
so 4,2
3 = 3, which is also the number of monomials in k[x1 , x2 ]/(x1 , x2 ).
4,2
Also, we define 4,2
= 0, −1
= 0, and the like, but there is one un8
fortunate ambiguity in this notation. In some cases we may have 42 = 1
(because the coefficient of x2 in 1 + x + x2 + x3 + x4 is 1). A key observation
we now make is that
! k[x1 , . . . , xn ]
e1 , . . . , e n
H
=
.
i
(xe11 +1 , . . . , xnen +1 )
i≥0
POWER OF MONOMIAL IDEALS
SUSAN COOPER
55
Definition 3.93. Let xα ∈ Sz , where deg(xα ) = d. We define
L(xα ) = {monomials of degree d in Sz which are ≥revlex xα }.
Lemma 3.94 (from [5]). Let xα ∈ (Mz )d such that α = (c1 , . . . , cs , 0, . . . , 0),
where cs > 0 for some s ∈ {1, . . . , n}. Then
(1) If xα is the smallest monomial of degree d in Mz 0 , where z 0 =
{z1 , . . . , zs }, then
z1 , . . . , z s
α
|L(x )| =
.
d
Otherwise
(2)
α
|L(x )| =
cX
s −1 i=0
z1 , . . . , z s
d−i
+ P,
where the first term comes from the number of monomials in L(xα )
with x0s , x1s , x2s , . . . , xcss −1 , and P is counting the number of monomials in L(xα ) with factor xcss .
We omit the (apparently unenlightening) proof of the previous lemma,
and instead consider the following example.
Example 3.95. Let z = {4, 3, 3, 2}, so that
Sz =
k[x1 , x2 , x3 , x4 ]
.
(x51 , x42 , x43 , x34 )
Consider L(x1 x24 ). We have
|L(x1 x24 )|
4, 3, 3
4, 3, 3
4
=
+
+
3
2
1
= 10 + 6 + 1.
So searching through lemma, we see there are supposed to be, respectively,
10 monomials involving only x1 , x2 , and x3 (corresponding to x04 ), 6 monomials of the form x`11 x`22 x`33 x4 , and 1 monomial of the form x`11 x`22 x`33 x24 .
Theorem 3.96 (from [5]). Let J, U be as in Theorem 3.91. Then |J| has a
unique decomposition
z1 , . . . , z b d
z1 , . . . , zbd−1
z 1 , . . . , z b`
|J| =
+
+ ··· +
,
d
d−1
`
and
|U | =
z1 , . . . , zbd
z 1 , . . . , z b`
+ ··· +
.
d+1
`+1
56
BRIAN JOHNSON
Remark 3.97. Binomial expansions and Pascal’s Table: We have an Osequence-like characterization of H(Sz /I) by replacing the diagonals in Pascal’s Triangle with rows: the ith row is
! z1 , . . . , z i
k[x1 , . . . , xi ]
=
.
H
d
(xz11 +1 , . . . , xzi i +1 )
d≥0
Example 3.98. Much like in Example 3.14, it is useful to have a copy
of Pascal’s Table handy in order to compute the unique decompositions
1 ,x2 ,x3 ]
described above. Let z1 = 6, z2 = z3 = 3. Then Sz = k[x
.
(x7 ,x4 ,x4 )
1
degree
1]
H k[x
7
(x1 ) 1 ,x2 ]
H k[x
7 4
(x1 ,x2 ) 1 ,x2 ,x3 ]
H k[x
(x7 ,x4 ,x4 )
1
2
2
3
0 1 2
3
4
5
6
7
8
9
10 11 12 13
1 1 1
1
1
1
1
0
0
0
0
0
0
0
1 2 3
4
4
4
4
3
2
1
0
0
0
0
1 3 6 10 13 15 16 15 13 10
6
3
1
0
3
Now, suppose I ⊆ Sz such that H(Sz /I, 8) = 7. To decompose 7, we
first look in the column for degree 8 (since that’s the degree of the Hilbert
function we’re looking for). We see 2 is the largest number smaller than 8.
Compute 7 − 2 = 5, back up to degree 7, and we see that 3 is the largest
number smaller than 5. Continuing this process, we find
6, 3
6, 3
6
6
7 =
+
+
+
8
7
6
5
= 2 + 3 + 1 + 1.
Note that the row of the table in which we find our numbers dictates which
numbers appear in the top of our “multinomial” coefficients. Then
6, 3
6, 3
6
6
(8)
7
:=
+
+
+
9
8
7
6
= 1+2+0+1
= 4
So Theorem 3.91 says H(Sz /I, 9) ≤ 7(8) = 4.
Conjecture 3.99 (combinatorial form of EGH). Fix integers 1 ≤ a1 ≤
· · · ≤ an . Let H = {di }i≥0 be such that
k[x1 , . . . , xn ]
di ≤ H
,i
(xa1n , . . . , xan1 )
for i ≥ 0. For each i, decompose di as in the previous example. Then there
exists a homogeneous (a1 , . . . , an )-ideal I with H(k[x1 , . . . , xn ]/I) = H if
(i)
and only if di+1 ≤ di for all i.
Remark 3.100. Some known cases of the EGH conjecture:
(1) When the LPP-ideal L is an almost complete intersection (Francisco).
POWER OF MONOMIAL IDEALS
SUSAN COOPER
57
(2) When I is a monomial ideal containing xa11 , . . . , xann (Clements-Lindström).
(3) When I is any homogeneous ideal containing xa11 , . . . , xann (CooperRoberts).
(4) When n = 2.
j−1
X
(5) When aj >
(ai − 1) for all j > 1 (Caviglia-Maclagan).
i=1
(6) When I = I(X)/(x0 ), for X a finite set of distinct points in Pn if
(a) n = 2 completely
(b) n = 3 with (a1 ≤ a2 ≤ a3 , a1 = 2, a1 = 3) or (a1 ≥ 4 and
a3 ≥ a1 + a2 − 1).
3.8. Sum fun with points and regular sequences.
Definition 3.101. Suppose Y = {P1 , . . . , P` } ⊆ Pn is a set of distinct
points. We say Y is complete intersection (or just c.i.) of type {d1 , . . . , dn } if
I(Y ) = (F1 , . . . , Fn ) and F1 , . . . , Fn is a regular sequence in R = k[x0 , . . . , xn ]
of homogeneous polynomials with deg(Fi ) = di (and 1 ≤ d1 ≤ · · · ≤ dn ).
Now observe that if F is homogeneous in R and deg F = `, we have
dimk (F R)t = dimk Rt−`
(this is because multiplication by F is one-to-one). So
H(R/F R, t) = dimk Rt − dimk (F R)t
= dimk Rt − dimk Rt−`
= H(R, t) − H(R, t − `).
Example 3.102. The process above can be iterated. Let
R = k[x0 , x1 , x2 , x3 , x4 ]
and Y ∈ C.I.(2, 2, 3, 5) ⊆ P4 (i.e., Y is c.i. of type {2, 2, 3, 5}). Then
I(Y ) = (F1 , F2 , F3 , F4 ).
And so knowing H(R) (through multinomial coefficients), we shift and subtract repeatedly to get H(R/(F )) := H(R/(F1 , F2 , F3 , F4 )):
H(R, t):
−H(R, t − 2):
H(R/(F1 ), t):
−H(R/(F1 ), t − 2):
H(R/(F1 , F2 ), t):
..
.
1
0
1
0
1
..
.
5 15 35 70 126 210 330
0 1 5 15 35 70 126
5 14 30 55 91 140 204
0 1 5 14 30 55 91
5 13 25 41 61 85 113
H(R/(F ), t):
1 5 13 24 36 47 55
This process is contained in the following theorem.
59
495
210
285
140
145
715
330
385
204
181
60
60
58
BRIAN JOHNSON
Theorem 3.103. If Y ∈ C.I.(d1 , . . . , dn ) ⊆ Pn , then
H(Y, t) = H(R, t) −
n
X
i=1
n
H(R, t − di ) +
X
H(R, t − di − dj )
1≤di <dj ≤n
− · · · + (−1) H(R, t − d1 − d2 − · · · − dn ).
[It should be noted it is not a coincidence that this has a Koszul flavor to it.]
We close this section with some final facts about complete intersections.
Let Y ∈ C.I.(d1 , . . . , dn ) ⊆ Pn .
(1) If n = 2, then

t+1
if 0 ≤ t ≤ d1 − 1



d
if d1 ≤ t ≤ d2 − 1
1
∆H(Y, t) =

d1 + d2 − 1 − t if d2 ≤ t ≤ d1 + d2 − 1



0
if t ≥ d1 + d2 − 1.
(2) ∆H(Y ) = H(k[x1 , . . . , xn ]/(xd11 , . . . , xdnn )).
(3) ∆H(Y, t) = ∆H(Y, d1 + · · · + dn − n − t); i.e., ∆H(Y ) is symmetric.
(4) If X ⊆ Pn and H(X) = H(Y ), it is not true in general that X ∈
C.I.(d1 , . . . , dn ). For example, if Y ∈ C.I.(2, 3) ⊆ P2 , and we set
[1 : 0 : 0],[1 : 1 : 0],[1 : 0 : 1],[1 : 1 : 1],
X=
.
[1 : 2 : 0],[1 : 3 : 0]
Using our results on lifting, we find that
∆H(X) = (1, 2, 2, 1, 0, . . . )
= H(k[x1 , x2 ]/(x21 , x32 )).
So H(X) = H(Y ) = (1, 3, 5, 6, 6, . . . ) (notice H(X, 1) = 3 implies
there’s no linear form in I(X)). Then Bézout’s Theorem implies
there exists a unique degree 2 form in I(X), namely f = x2 (x2 − x0 )
(again from lifting). Suppose there exists a homogeneous G ∈ I(X)
of degree 3. Bézout’s Theorem implies x2 | G, which implies f and
G share a common factor. Therefore, X ∈
/ C.I.(2, 3).
(5) |Y | = d1 · · · dn .
Theorem 3.104 (Cayley-Bacharach Theorem). If Y ∈ C.I.(d1 , . . . , dn ) and
X ⊆ Y , then
∆H(Y, t) = ∆H(X, y) + ∆H(Y r X, d1 + · · · + dn − n − t).
4. Lex ideals and Betti numbers
A major reference for this material is [16]. Also, the reader who isn’t
comfortable with free resolutions and syzygies may want to refer to Section 6
when appropriate. Assume k is a field with char k = 0. All ideals are
assumed to be homogeneous.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
59
Theorem 4.1 (Bigatti-Hulett, K. Pardue). For every i ∈ {0, 1, . . . , n} and
d ≥ 0, the lex-segment ideal L ⊆ k[x1 , . . . , xn ] has the most degree d ith
syzygies among all (monomial) ideals I ⊆ k[x1 , . . . , xn ] with the same fixed
Hilbert function H.
To prove this theorem, we’ll need to develop the notion of “Borel-fixed”
ideals.
4.1. Group actions on S = k[x1 , . . . , xn ]. Recall that GLn (k) is the collection of n × n invertible matrices with entries from k. Let G ∈ GLn (k),
say G = (gij ). We can define an action of GLn (k) on S by setting
G · xi =
n
X
gij xj .
j=1
For F = F (x1 , . . . , xn ) ∈ S, we define
G · F = F (G · x1 , . . . , G · xn ).
For I ⊆ S, we set
G · I = {G · F | F ∈ I}
(cf. [16, p. 22]). One might want to convince oneself that G · I is an ideal.
Example 4.2. It should be noted that G · I can be much more complicated
than I itself. For example, let


1 0 1
G =  2 −1 3  ,
2 0 1
and set I = (x1 x2 − x23 ). Note then that
G · x1 = x1 + x3
G · x2 = 2x1 − x2 + 3x3
G · x3 = 2x1 + x3 .
This gives
G · (x1 x2 − x23 ) = (−2x21 − x1 x2 + 3x1 x3 − x2 x3 − 4x23 )
(after multiplying and simplifying, of course).
A natural question is to ask which ideals I ⊆ S are fixed under the action
of GLn (k). We’ll use monomial ideals and look at subgroups of GLn (k) to
obtain an answer.
Definition 4.3. We define the Borel group, Bn (k), to be the collection of
upper triangular invertible n × n matrices with entries in k. The algebraic
torus group, Tn (k), is the collection of diagonal invertible n × n matrices
with entries in k. It’s clear that
Tn (k) ⊆ Bn (k) ⊆ GLn (k).
60
BRIAN JOHNSON
Proposition 4.4 (2.1 in [16]). A nonzero ideal I ⊆ S is fixed under Tn (k)
if and only if I is a monomial ideal.
Proof. (⇐=): If I is a monomial ideal, then Tn (k) maps each xi to a constant
multiple of xi . The same is then true of any monomial, so I is fixed under
the action of Tn (k).
P
(⇐=): Suppose I is fixed. Let f =
cα xα ∈ I. Then for any diagonal
matrix T = diag(t1 , . . . , tn ) (whose diagonal entries are t1 , . . . , tn ), we have
T · f ∈ I. Define P = {T (1) , . . . , T (`) } ⊆ Tn (k) to be a “generic” set of
(j)
(j)
diagonal matrices, where T (j) = diag(t1 , . . . , tn ) and ` is the number of
monomials in f with cα 6= 0. Now for each xα appearing in f and each T (j)
there is a corresponding monomial obtained by
(j)
αn
=: [T (j) ]α .
xα1 1 · · · xαnn 7→ (t1 )α1 · · · (t(j)
n )
Construct the `×` matrix M columns are indexed by the monomials appearing in f and whose rows are indexed by the [T (j) ]. Then, Mj,i = [T (j) ]α ,
where xα is the ith monomial appearing in f . Observe that det M is a
(j)
(j)
polynomial in {t1 , . . . , tn | j = 1, . . . , `}. Because the T (j) are “generic”
det M 6= 0. Consider the vector
 (1)

T ·f


..
v=
.
.
T (`) · f
Then one can probably convince oneself that M −1 v is a vector whose entries
are precisely cα xα for the monomials in f . I.e., each term in f is a linear
combination of polynomials of the form T ·f ∈ I. Therefore, I is a monomial
ideal.
Corollary 4.5 (2.2 in [16]). A nonzero ideal I ⊆ S is fixed under GLn (k)
if and only if I = md for some d > 0, where m = (x1 , . . . , xn ).
Proof. ⇐=: Let V be the vector space Sd . Clearly V is fixed under GLn (k).
Thus, (V ), the ideal generated by md is fixed under GLn (k).
=⇒: Let I be fixed by GLn (k). Let 0 6= f ∈ I be of minimal degree,
say d. Let G ∈ GLn (k) be a “general” matrix. Then G · f ∈ I involves
all monomials in Sd . Since Tn (k) ⊆ GLn (k), I must be a monomial by
the previous proposition. Thus, every monomial in Sd is in I. Since d was
chosen minimally, there does not exist h ∈ I with deg(h) < d. Therefore,
I = md .
Now the natural question, since we’ve determined which ideals are fixed
by the general linear group and the torus group, is what Borel-fixed ideals
look like. We already know they are monomial, but can we determine more
about them?
Proposition 4.6. Let I ⊆ S be a monomial ideal. TFAE:
(1) I is Borel-fixed.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
61
(2) Let m be a monomial divisible by xj . Then m ∈ I implies m xxji ∈ I
for any i < j.
Proof. Suppose I is Borel-fixed. Let m ∈ I be any monomial such that
xj | m. Fix i < j. let M ∈ Bn (k) be the elementary matrix such that
M xj →
7
xj + xi
M x` →
7
x`
for ` 6= j.
Then the polynomial M · m is in M · I ⊆ I. Furthermore, m xxji appears as
a term of M · m ∈ I, so m xxji ∈ I.
Conversely, suppose (2) holds, and let m ∈ I be a monomial. Let G ∈
Bn (k). Suppose xα is a term in G · m. Then we can perform a sequence
of transformations starting with xα and replacing some xj with xi where
i < j to end up with m. Repeatedly applying the contrapositive of (2) gives
xα ∈ I. Thus G · m ∈ I.
The next example illustrates the idea of the second part of the proof.
Example 4.7. Let I = (x31 , x21 x2 , x1 x2 x3 , x32 , x1 x22 , x22 x3 ) ⊆ k[x1 , x2 , x3 ].
One can see from the previous proposition that (the generators of) I is (are)
Borel-fixed. Let


1 2 3
G =  0 5 1  ∈ B3 (k)
0 0 2
and suppose m = x1 x2 x3 . Then
G · m = (x1 + 2x2 + 3x3 )(5x2 + x3 )(2x3 ).
We see that
have
6x33
is a term of G · m, and iterating the idea of the proof, we
6x33 −→ 6x2 x23 −→ 6x1 x2 x3 = 6m.
4.2. Generic initial ideals. Fix a monomial order > and suppose x1 >
x2 > · · · > xn . We begin with some preliminaries:
(1) Suppose S = k[x1 , . . . , xn ], and let y1 , . . . , y` be variables different
from the xi . Let φ : k[y1 , . . . , y` ] −→ k be a k-algebra homomorphism. This gives rise to another homomorphism:
φS : k[x1 , . . . , xn , y1 , . . . , y` ] −→ k[x1 , . . . , xn ]
which maps xi 7→ xi and yi 7→ φ(yi ). Let J ⊆ k[x1 , . . . , xn , y1 , . . . , y` ]
be a (homogeneous) ideal. Then
(a) φS (J) is an ideal of S.
(b) Given a monomial order < on S, there exists a finite set C of
polynomials in J such that φS (C) is a Gröbnerbasis for φS (J)
for all φ as above (this is quite a nontrivial fact). This C is called
a comprehensive Gröbner basis for J, due to Weispfenning.
(2) Let g11 , g12 , . . . , gnn be indeterminates and G = (gij )n×n ∈ GLn (k).
62
BRIAN JOHNSON
Definition 4.8. A Zariski closed set in GLn (k) (or k n ) is the zero
set of some ideal in k[g11 , . . . , gnn ] (or S). A Zariski open set is the
complement of a closed set.
(3) Fix a monomial order < on S.
Definition 4.9. Let I ⊆ S be an ideal. Let A, B ∈ GLn (k). We say
A, B are equivalent if in(A · I) = in(B · I).
(4) Motivating lemma:
Lemma 4.10 (2.6 in [16]). Fix an ideal I ⊆ S and an order <.
Then the number of equivalence classes in GLn (k) is finite. One of
these classes is a nonempty Zariski open subset U inside of GLn (k)
(U is unique).
Proof. Fix variables x = {x1 , . . . , xn } and g = {g11 , . . . , gnn }. Let
G = (gij ) and insist that det(G) 6= 0. Denote the generators of I by
f1 (x), . . . , f` (x) ∈ S.
Let J = (G · f1 , . . . , G · f` ) ⊆ k[x, g]. Let C be a comprehensive
Gröbner basis for J. We can think of the coefficients of the polynomials in C as polynomials in k[g]. Consider all the possible equations
(and inequations) from setting these “polynomial coefficients” equal
to 0 or not equal to 0. This gives rise to all possible in(G · I). Now,
|C| < ∞, so the number of distinct in(G · I) when you consider
all G ∈ GLn (k) is finite. To obtain U : impose the not equal to 0
condition.
(5) A definition and an example:
Definition 4.11. Fix a monomial order on S. Let I ⊆ S be an
ideal. The equivalence class of in(G · I) that, as a function of G, is
constant on a Zariski open subset U ⊆ GLn (k) is called the generic
initial ideal of I with respect to <, denoted gin< (I).
Example 4.12. Let I = (x21 , x2 x2 ) ⊆ S = k[x1 , x2 ]. Consider the
matrix
g11 g12
G=
,
g21 g22
and suppose <=<lex . Then
J
= (G · x21 , G · x1 x2 )
2 2
2 2
= (g11
x1 + 2g11 g12 x1 x2 + g12
x2 ,
g11 g21 x21 + (g11 g22 + g12 g21 )x1 x2 + g12 g22 x22 )
⊆ k[x, g].
Also, J has a comprehensive Gröbner basis:
C = {G · x21 , G · x1 x2 , f },
POWER OF MONOMIAL IDEALS
SUSAN COOPER
63
where
2
f = g11 (g11 g22 − g12 g21 )x1 x2 + (g11 g12 g22 − g12
g21 )x22 .
The “polynomial coefficients” described above are:
2
g11
2g11 g12
2
g12
g11 g21
g11 g22 + g12 g21
g12 g22
2 g −g g g
g11
22
11 12 21
2 g .
g11 g12 g22 − g12
21
It turns out that GLn (k) decomposes into two equivalence classes. If
g11 = 0, we have in< (G·I) = (x22 , x1 x2 ). If g11 6= 0, then in< (G·I) =
(x21 , x1 x2 ). Thus
gin< (I) = (x21 , x1 x2 ).
Remark 4.13. One should note that gin< (I) depends on <. For example,
if f, g ∈ k[x1 , x2 , x3 , x4 ] are generic cubics and I = (f, g), then (it turns
out) ginlex (I) has 26 generators, and gingrevlex (I) has four. Also, gingrevlex is
usually “nicer” than ginlex , because there will be fewer generators of lower
degree.
Theorem 4.14. gin< (I) is Borel-fixed.
Proof. See [8, Thm 15.20].
4.3. Some comments on Gröbner bases and modules. Let M = S β
be a free S-module with basis e1 , . . . , eβ . A monomial order on M totally
orders {mei | m ∈ S is a monomial}. Order e1 > · · · > eβ .
Definition 4.15. Fix a monomial order < on S, and let m, m0 ∈ S be
monomials.
(1) The position-over-term (POT) order on M is given by
mei > m0 ej
(if i < j) or (if i = j and m > m0 ).
(2) The term-over-position (TOP) order on M is given by
mei > m0 ej
(if m > m0 ) or (if m = m0 and i < j).
4.4. Comparing lex-segment ideals and Borel-fixed ideals. Let m ∈
S = [x1 , . . . , xn ] be a monomial. Define
max(m) = max{j | xj divides m}
min(m) = min{j | xj divides m}.
Lemma 4.16 (2.11 in [16]). Let I be a Borel-fixed ideal of S with minimal
generating set {m1 , . . . , mr }. Then each monomial m ∈ I can be written
uniquely as m = mi m0 with max(mi ) ≤ min(m0 ).
64
BRIAN JOHNSON
Proof. For i = 1, . . . , r, let ai = max(mi ). To show existence, suppose
we can write m = mj m0 for some j, but aj > a := min(m0 ). Since I is
Borel-fixed,
xa
mj
∈ I.
xaj
So there exists mi which divides mj xxaa , say mi m
b = mj xxaa . Then
j
j
xa
b j m0 .
m = mi m
xa
| {z }
call this m
e
By construction, ai ≤ aj , min(m)
e ≥ a. If ai = aj , then we must have that
the degree of xai in mi is strictly less than the degree of xai in mj . This
means that we can only replace the mj with the mi as above only finitely
many times. Performing such replacements repeatedly gives existence.
For uniqueness, suppose m = mi m0i = mj m0j , where ai ≤ min(m0i ) and
aj ≤ min(m0j ). Without loss of generality, assume ai ≤ aj . Then mi and mj
agree in all variables with index less than ai .
Case 1: xai | m0j . Then ai ≤ aj ≤ min(m0j ) ≤ ai . This implies that
ai = aj , and so mi | mj or mj | mi . Therefore, i = j.
Case 2: xai - m0j . Then the degree of xai in mi is at most the degree of
xai in mj . This latter integer equals the degree of xai in m. Thus mi | mj ,
which implies mi = mj .
We introduce some new notation: Suppose >=>lex with x1 > · · · > xn ,
S is as usual, W is a finite set of monomials in S, and Wd = W ∩ Sd . For
i ∈ {1, . . . , n},
µi (W ) = |{m ∈ W | max(m) = i}|
µ≤i (W ) = |{m ∈ W | max(m) ≤ i}|.
We say W is a Borel set if m ∈ W and xj | m implies that m xxji ∈ W for all
i < j. Finally, we say W is a lex-segment if m ∈ W and m0 >lex m implies
m0 ∈ W .
Remark 4.17. Let W be a Borel set.
(1) Every monomial m in {x1 , . . . , xn } · W factors uniquely as m = xi m
e
for some m
e ∈ W such that max(m)
e ≤ i.
(2) The previous fact implies that
µi ({x1 , . . . , xn } · W ) = µ≤i (W ).
Lemma 4.18 (2.25 in [16]). Let L be a lex-segment in Sd and B a Borel
set in Sd . If |L| ≤ |B|, then
µ≤i (L) ≤ µ≤i (B)
for all i.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
65
Proof. Use induction on n (the number of variables). The n = 1 case is
trivial. Suppose n > 1 and fix i. Note that L is a Borel set.
Case 1: i = n. Then clearly
µ≤n (L) = |L| ≤ |B| = µ≤n (B).
Case 2: Suppose i = n − 1. Partition B as:
B = B[0] ∪ (xn B[1]) ∪ · · · ∪ (xdn B[d]),
where there are no powers of xn in any of the B[`]. Note that each B[`]
is a Borel set in k[x1 , . . . , xn−1 ]d−` . Partition L similarly. Then each L[`]
is a lex-segment in k[x1 , . . . , xn−1 ]d−` . Define C[`] to be a lex-segment in
k[x1 , . . . , xn−1 ]d−` such that
|C[`]| = |B[`]|.
Note that C[`] is a Borel set. Define
C = C[0] ∪ (xn C[1]) ∪ · · · ∪ (xdn C[d]).
By induction,
µ≤j (C[`]) ≤ µ≤j (B[`])
for all `, j.
We claim C is a Borel set. Indeed, B is a Borel set, so
(9)
{x1 , . . . , xn } · B[`] ⊆ B[` − 1].
So it’s enough to show (9) for C. Thus,
|{x1 , . . . , xn−1 } · C[`]| =
n−1
X
µj ({x1 , . . . , xn−1 } · C[`])
j=1
(Remark 4.17.2) =
n−1
X
µ≤j (C[`])
j=1
≤
n−1
X
µ≤j (B[`])
j=1
(Remark 4.17.2) =
n−1
X
µj ({x1 , . . . , xn−1 } · B[`])
j=1
= |{x1 , . . . , xn−1 } · B[`]|
(Equation (9)) ≤ |B[` − 1]|
= |C[` − 1]|.
Since C[` − 1] and {x1 , . . . , xn−1 } · C[`] are lex-segments in the same degree,
we have
{x1 , . . . , xn−1 } · C[`] ⊆ C[` − 1].
Therefore C is Borel. Now, L is a lex-segment and
|L| ≤ |B| = |C|.
66
BRIAN JOHNSON
So
min(C) ≤lex min(L),
lex
lex
(where by minlex , we mean the smallest monomial in the set under <lex ).
Since L and C are Borel, we thus must have
min(C[0]) ≤lex min(L[0]).
lex
lex
This implies L[0] ⊆ C[0]. So
µ≤n−1 (L) =
≤
=
=
|L[0]|
|C[0]|
|B[0]|
µ≤n−1 (B).
Case 3: i ≤ n − 2. Partition L, B and define C as in Case 2. We have
|L[0]| ≤ |B[0]|.
Apply the claim inductively on B[0] and L[0] to obtain
µ≤i (L) = µ≤i (L[0]) ≤ µ≤i (B[0]) = µ≤i (B),
for 1 ≤ i ≤ n − 2 (where we’ve used induction at the middle inequality). Let W be any finite set of monomials in S. Define
X max(m) − 1
βi (W ) =
.
i
m∈W
A fact we will use is the following, from section 2.3 of [16]. Let I ⊆ S
be a Borel-fixed ideal. If W is a minimal generating set for I, then the
“Eliahou-Kervaire Formula for Betti Numbers” gives:
X max(m) − 1
βi (W ) =
i
m∈W
X
=
(βi (Id ) − βi ({x1 , . . . , xn } · Id−1 ))
d>0
= # of minimal ith syzygies of I.
Lemma 4.19 (2.26 in [16]). Let B be a Borel set in Sd . Then
n−1
X
n−1
j−1
βi (B) =
· |B| −
µ≤j (B)
.
i
i−1
j=1
POWER OF MONOMIAL IDEALS
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67
Proof. By definition and the previous fact, we have a long string of equalities
(some of which the reader may want to verify):
X max(m) − 1
βi (B) =
i
m∈B
n
X
j−1
=
µj (B)
i
j=1
n
X
j−1
=
(µ≤j (B) − µ≤j−1 (B))
i
j=1
X
X
n
n
n−1
j−1
j−1
= µ≤n (B)
+
µ≤j (B)
−
µ≤j−1 (B)
i
i
i
j=1
j=2
n−1
X
n−1
j−1
j
= |B|
+
µ≤j (B)
−
i
i
i
j=1
n−1
X
n−1
j−1
= |B|
−
µ≤j (B)
.
i
i−1
j=1
Lemma 4.20 (2.27 in [16]). Let L be a lex-segment in Sd and B a Borel
set in Sd with |L| = |B|. Then
(1) βi (L) ≥ βi (B)
(2) βi ({x1 , . . . , xn } · L) ≤ βi ({x1 , . . . , xn } · B)
Proof. (1): L is a Borel set. Thus, by the previous lemma
βi (L) =
n−1
X
j−1
n−1
· |L| −
µ≤j (L)
i−1
i
j=1
n−1
X
n−1
j−1
(2.25 in [16]) ≥
· |B| −
µ≤j (B)
i
i−1
j=1
= βi (B).
(2): We have the following:
X
βi ({x1 , . . . , xn } · L) =
=
m∈{x1 ,...,xn }·L
n
X
j=1
max(m) − 1
i
j−1
µj ({x1 , . . . , xn } · L)
i
68
BRIAN JOHNSON
Applying Remark 4.17.2 to the last term gives
βi ({x1 , . . . , xn } · L) =
n
X
j=1
j−1
µ≤j (L)
.
i
But then 2.25 in [16] gives
n
X
j=1
µ≤j (L)
X
n
j−1
j−1
≤
µ≤j (B)
,
i
i
j=1
and then reversing the previous equalities on B instead of L finishes the
proof.
Theorem 4.21 (2.22 in [16]). Let I ⊆ S be a homogeneous ideal. Let
L ⊆ S be the lex-segment ideal with H(S/I) = H(S/L). Then L has at least
as many degree d generators as I does for all d ≥ 0.
Proof. Let > be any monomial order on S and fix d ≥ 0. Note:
(1) For any G ∈ GLn (k),
H(S/I) = H(S/(G · I)).
(2) |{deg d min. gen. of in< (I)}| ≥ |{deg d min. gen. of I}|
(3) H(S/I) = H(S/ in< (I)) (Corollary 4.2.4 in [2])
These three facts imply that
|{deg d min. generators of B := gin< (I)}| ≥ |{deg d min. generators of I}|
and H(S/B) = H(S/I) = H(S/L). Recall that gin< (I) is Borel-fixed (for
any <). Thus, we may as well assume <=<lex . Now, we claim
(10)
|{x1 , . . . , xn } · Ld | ≤ |{x1 , . . . , xn } · Bd | ≤ |Bd+1 | = |Ld+1 |.
Indeed, we have |Ld | = |Bd | and Ld is a Borel set. Thus
|{x1 , . . . , xn } · Ld | =
n
X
µj ({x1 , . . . , xn } · Ld )
j=1
=
≤
n
X
j=1
n
X
µ≤j (Ld )
µ≤j (Bd )
j=1
= |{x1 , . . . , xn } · Bd |
≤ |Bd+1 |
= |Ld+1 |.
The number of minimal generators of L in degree d + 1 is
|Ld+1 | − |{x1 , . . . , xn } · Ld |
POWER OF MONOMIAL IDEALS
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69
and for B is
|Bd+1 | − |{x1 , . . . , xn } · Bd |.
By (10),
|Bd+1 | − |{x1 , . . . , xn } · Bd | ≤ |Ld+1 | − |{x1 , . . . , xn } · Ld |.
Therefore
|{deg d min. generators of I}| ≥ |{deg d min. generators of B}|
≥ |{deg d min. generators of L}|.
Theorem 4.22 (Bigatti-Hulett-Pardue; 2.24 in [16]). Let I ⊆ S be a homogeneous ideal. Let L ⊆ S be the lex-segment ideal with H(S/I) = H(S/L).
For every i ∈ {0, 1, . . . , n} and d ≥ 0, L has more degree d minimal ith
syzygies.
Proof. Let >=>lex , and let B = gin< (I), a Borel-fixed ideal. As above,
H(S/I) = H(S/L) = H(S/B).
From Theorem 8.29 in [16]
βi (I) ≤ βi (in< (I))
for any <. Thus, it suffices to prove that
βi (L) ≥ βi (B).
Write gens(J) for a minimal set of generators of an ideal J. The EliahouKervaire formula gives
# of minimal ith = βi (gens(B))
syzygies of gens(B)
X
=
[βi (Bd ) − β( {x1 , . . . , xn } · Bd−1 )].
d>0
We get a similar formula for the number of minimal ith syzygies of gens(L).
By Lemma 4.20, since |Ld | = |Bd | for all d ≥ 0,
βi (gens(B)) ≤ βi (gens(L))
Example 4.23. Suppose S = k[x1 , x2 , x3 ]. Let
I = (x31 , x32 , x33 , x1 x3 , x2 x3 )
L = (x21 , x1 x2 , x1 x23 , x22 x3 , x2 x33 , x53 )
J
= (x51 , x22 , x23 , x21 x2 , x21 x3 ).
Notice that L is a lex-segment. Also, it turns out
H(S/L) = H(S/I) = H(S/J) = (1, 3, 4, 2, 1, 0, 0, . . . ).
70
BRIAN JOHNSON
On the other hand,
follows:
S/I
total 1 5 6
0
1 - 1
- 2 1
2
- 3 4
3
- - 4
- - 1
the Betti diagrams (see Section 6) for these rings are as
2
1
1
S/L
1 7 10 4
1 - - - 2 1 - 3 5 2
- 1 2 1
- 1 2 1
total
0
1
2
3
4
total
0
1
2
3
4
S/J
1 5
1 - 2
- 2
- - 1
6
4
2
2
- .
1
1
As this example shows, you can have many different sets of graded Betti
numbers for a fixed Hilbert function.
Conjecture 4.24 (Lex-plus-powers Conjecture; Charalambous-Evans). Let
I ⊆ S be a homogeneous ideal containing a homogeneous regular sequence
in degrees a1 , . . . , an . Let L ⊆ S be an (a1 , . . . , an )-LPP ideal such that
H(S/I) = H(S/L). Then
βi,j (S/L) ≥ βi,j (S/I).
Example 4.25. Suppose S = k[x1 , x2 , x3 ], with L = (x21 , x32 , x23 , x1 x22 x3 )
and I = (x21 , x32 , x33 , x22 x23 ) (in fact one can take any I that is generated by
generic homogeneous polynomials in degrees 2, 3, 3, and 4). Then we have
the following Betti diagrams:
total
0
1
2
3
4
S/I
1 4
1 - 1
- 2
- 1
- -
5
4
1
2
2
total
0
1
2
3
4
S/L
1 4
1 - 1
- 2
- 1
- -
6
4
2
3
- .
1
2
Remark 4.26.
(1) The LPP Conjecture implies the EGH Conjecture. [Similar to how
Bigatti-Hulett-Pardue (4.22) implies “Macaulay’s Theorem.”]
(2) The LPP Conjecture is known to be true in very few cases:
(a) If I contains x21 , . . . , x2n (Mermin-Peeva-Stillman).
(b) If I has a regular sequence given by monomials in the degrees
(a1 , . . . , an ) (Mermin-Murai).
(c) In S = k[x1 , x2 ].
(d) If I is a monomial ideal in k[x1 , x2 , x3 ].
(e) If L = (xa11 , . . . , xann , m), where m ∈
/ (xa11 , . . . , xann ) (L is “almost
c.i.”; Francisco).
(3) To prove the LPP Conjecture, it is natural to compare L and gin(I)
(rather than L and I). Passing to gin(I) does not affect the Hilbert
functions, but the degrees of the regular sequences in I and gin(I)
may differ.
POWER OF MONOMIAL IDEALS
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71
(4) A reference for this material is [10].
5. Squarefree monomial ideals
References for this brief foray into squarefree monomial ideals are [16, Ch.
1] and [2, Ch. 5].
Definition 5.1.
(1) A simplicial complex, or s.c., ∆ on the vertex set {1, . . . , n} is a
collection of subsets called faces (or simplices) such that if σ ∈ ∆
and τ ⊆ σ, then τ ∈ ∆.
(2) An i-face of a s.c. ∆ is a face σ ∈ ∆ such that |σ| = i + 1. We say
σ has dimension i.
(3) The dimension of a s.c. ∆ is
dim(∆) := max{dim σ | σ ∈ ∆}.
We define dim({}) = −∞. Note that {} is the “void complex,”
which has no faces. This is not to be confused with the “irrelevant
complex” {∅}.
(4) A facet of a s.c. ∆ is a maximal (proper?) face with respect to
inclusion.
Example 5.2. Let ∆ be a s.c., ∆ 6= {}.
(1) ∅ is a face of ∆ and
dim(∅) = −1.
(2) If ∆ = {∅}, then its only face is ∅, so
dim(∆) = −1.
(3) Consider the vertex set {1, 2, 3, 4, 5, 6}. Let ∆ be the s.c. defined by
all subsets of the sets
{1, 2}, {1, 4}, {6}, {3, 4, 5}.
Note that it’s sufficient to define a s.c. by its facets. One sees that
dim(∆) = 2.
Introduce the following notation: for a subset σ ⊆ {1, . . . , n}, we can
associate a monomial
Y
xi ∈ S.
σ 7→ xσ =
i∈σ
For example, if σ = {1, 3, 5} ⊆ {1, 2, 3, 4, 5, 6}, then xσ = x1 x3 x5 . These are
what we call squarefree monomials (all powers are only 1 or 0).
Definition 5.3. Let ∆ be a s.c. on {1, . . . , n}. The Stanley-Reisner ideal
of ∆ is
I∆ = (xγ | γ ∈
/ ∆).
The Stanley-Reisner ring of ∆ is k[∆] := S/I∆ . Note that (by definition)
I∆ is a squarefree monomial ideal.
72
BRIAN JOHNSON
Some more notation we will use is the following: if σ ⊆ {1, . . . , n} is a
subset, define
mσ = (xi | i ∈ σ).
This is a prime (monomial) ideal in S.
Theorem 5.4 (1.7 in [16], 5.1.4 in [2]). The correspondence ∆ ↔ I∆ is a
bijection from a simplicial complexes on {1, . . . , n} to squarefree monomial
ideals in S. Furthermore,
\
I∆ =
mσ ,
σ∈∆
where σ = {1, . . . , n}rσ [one can actually just intersect over facets]. Finally,
dim(k[∆]) = dim(∆) + 1
(the left-hand dimension is Krull dimension).
Example 5.5. With the same s.c. as in 5.2.3, we have
I∆ = (x1 , x2 , x6 ) ∩ (x2 , x3 , x5 , x6 ) ∩ (x3 , x4 , x5 , x6 ) ∩ (x1 , x2 , x3 , x4 , x5 ).
Our goal now is to find the Hilbert series (not function) of k[∆]. To do
this, we define a related item called the f -vector of a s.c. ∆.
Definition 5.6. Let ∆ be a s.c. with vertex set V = {1, . . . , n} such that
dim(∆) = d − 1 ≥ 0.
Let fi be the number of i-dimensional faces of ∆ for i ≥ −1. The f -vector
of ∆ is
f (∆) = (f−1 , f0 , . . . , fd−1 ) = (1, n, . . . , fd−1 ).
Example 5.7.
(1) We use the same s.c. as in 5.2.3 again. Then one can verify that
f (∆) = (1, 6, 5, 1).
(2) Consider the vertex set {1, 2, 3, 4, 5}, and suppose ∆ is all subsets of
{1, 2, 3, 5} and {1, 2, 4, 5}. One can check that f (∆) = (1, 5, 9, 7, 2).
The following notation comes from [2, p.209]. Suppose the d-binomial
expansion for a is:
md
mj
a=
+ ··· +
.
d
j
We define
md
mj
a[d] =
+ ··· +
.
d+1
j+1
Theorem 5.8 (Kruskal, Katona). Let f = (f−1 , f0 , . . . , fd−1 ) ∈ Zd+1 . Then
f is the f -vector of some (d − 1)-dimensional s.c. if and only if
[i+1]
0 < fi+1 ≤ fi
for 0 ≤ i ≤ d − 2.
POWER OF MONOMIAL IDEALS
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73
5.1. Hilbert series.
Definition 5.9. Let S = k[x1 , . . . , xn ].
L
(1) An S-module M is Nn -graded if M = α∈Nn Mα and xβ Mα ⊆ Mα+β
for all α, β ∈ Nn .
(2) Let M be an Nn -graded S-module. If dimk (Mα ) is finite for all
α ∈ Nn , then the formal power series
X
H(M ; x1 , . . . , xn ) =
dimk (Mα )xα
α∈Nn
Nn -graded
is called the
or finely graded Hilbert series of M . Setting
x1 = · · · = xn = t gives the Z-graded or coarse Hilbert series,
H(M ; t, . . . , t).
Example 5.10. The ring S = k[x1 , . . . , xn ] has a natural Nn -grading. Then
the Nn graded ideals are precisely the monomial ideals. If I ⊆ S is a
monomial ideal, then S/I is also Nn -graded. Now,
X
H(S; x1 , . . . , xn ) =
xα
=
=
α∈Nn
∞
X
xk11 · · ·
k1 =0
n
Y
i=1
∞
X
xknn
kn =0
1
,
1 − xi
which is really just an expression for the sum of all monomials in S. So it’s
easy to see then that H(S/I; x1 , . . . , xn ) is just the sum of all monomials in
S r I.
Definition 5.11. If the Hilbert series of an Nn -graded S-module M can
expressed as a rational function
K(M ; x1 , . . . , xn )
H(M ; x1 , . . . , xn ) =
,
(1 − x1 )(1 − x2 ) · · · (1 − xn )
then the numerator K is called the K-polynomial of M .
It is a fact that if I ⊆ S is monomial, then M = S/I has a K-polynomial,
but an arbitrary module need not have a K-polynomial.
Theorem 5.12 (1.13 in [16]). The Stanley-Reisner ring S/I∆ has the following K-polynomial:


X Y Y
 xi
K(S/I∆ ; x1 , . . . , xn ) =
(1 − xj ) .
σ∈∆
i∈σ
j ∈σ
/
Proof. Suppose ∆ has vertices {1, . . . , n}. Let α ∈ Nn . We define the support
of α to be
supp(α) = {i ∈ {1, . . . , n} | αi 6= 0}.
74
BRIAN JOHNSON
Now, I∆ is generated by squarefree monomials. So xα ∈
/ I∆ precisely when
xsupp(α) ∈
/ I∆ , where
Y
xi .
xsupp(α) =
i∈supp(α)
Therefore, since I∆ is defined by the non-faces of ∆,
X
H(S/I∆ ; x1 , . . . , xn ) =
{xα | α ∈ Nn , supp(α) ∈ ∆}
XX
=
{xα | α ∈ Nn , supp(α) = σ}
σ∈∆
XY
=
σ∈∆ i∈σ
xi
.
1 − xi
Now, bring the terms over a common denominator of (1 − x1 ) · · · (1 − xn )
Q
1−xj
by multiplying the summand for each face by j ∈σ
/ 1−xj .
Example 5.13. For the vertex set {1, 2, 3, 4, 5, 6}, use the example from
5.2.3. We’ve already seen the expression for I∆ in Example 5.5. One can
then see that
x2
x6
x1
+
+ ··· +
H(k[∆]; x1 , . . . , x6 ) = 1 +
1 − x1 1 − x2
1 − x6
x1 x2
x1 x4
+
+
(1 − x1 )(1 − x2 ) (1 − x1 )(1 − x4 )
x3 x5
x3 x4
+
+
(1 − x3 )(1 − x4 ) (1 − x3 )(1 − x5 )
x4 x5
x3 x4 x5
+
+
.
(1 − x4 )(1 − x5 ) (1 − x3 )(1 − x4 )(1 − x5 )
Corollary 5.14 (1.15 in [16]). Let ∆ be a s.c. with f -vector f (∆) =
(f−1 , f0 , . . . , fd−1 ). Then k[∆] has the coarse Hilbert series
H(S/I∆ ; t, . . . , t) =
d
X
1
fi−1 ti (1 − t)n−i
(1 − t)n
i=0
(here, d = dim(∆) + 1, n = number of vertices = f0 ).
Definition 5.15. Let ∆ be a s.c. with f -vector f (∆) = (f−1 , f0 , . . . , fd−1 ).
Then
H(S/I∆ ; t, . . . , t) =
d
X
1
fi−1 ti (1 − t)d−i
(1 − t)d
i=0
=
h0 + h1 + · · · + hd td
.
(1 − t)d
We call the numerator the H-polynomial of ∆; the h-vector is h(∆) =
(h0 , . . . , hd ).
POWER OF MONOMIAL IDEALS
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75
Remark 5.16. We can read off the Hilbert function of k[∆] from its Hilbert
series. Certainly k[∆] is N-graded:
M
k[∆]α .
k[∆]i =
|α|=i
Then H(k[∆]) = {ct }t≥0 , where ct = dimk (k[∆]t ). But from 5.14, setting
xi = t for all i (also see [2, p.212]), we find c0 = 1 and
d−1 X
t−1
ct =
fi
i
i=1
for t > 0.
5.2. Shellable simplicial complexes and H-vectors.
Definition 5.17. Let ∆ be a simplicial complex.
(1) ∆ is pure if all of its facets have the same dimension.
(2) ∆ is Cohen-Macaulay over k if k[∆ is CM.
Definition 5.18. A pure s.c. is shellable if one of the following equivalent
conditions is satisfied: The facets of ∆ can be ordered as F1 , . . . , Fm such
that
(1) hFi i ∩ hF1 , . . . , Fi−1 i is generated by a nonempty set of maximal
proper faces of hFi i for all i ∈ {2, . . . , m} (where h−i is the smallest
s.c. containing −).
(2) The set {F | F ∈ hF1 , . . . , Fi i, F ∈
/ hF1 , . . . , Fi−1 i} has a unique
minimal element for all i ∈ {2, . . . , m}.
(3) For all i, j 1 ≤ j < i ≤ m, there exists v ∈ Fi r Fj and ` ∈
{1, . . . , i − 1} with Fi r F` = {v}.
Such an ordering on the facets is called a shelling.
Example 5.19. Consider the following two simplicial complexes, each on
5 vertices. Let F have facets {1, 2, 3}, {2, 3, 4}, and {3, 4, 5}, and suppose
G has facets {1, 2, 3} and {3, 4, 5}. It turns out that F is shellable and G is
not.
Theorem 5.20 (5.1.13 in [2]). A shellable s.c. is CM over every field.
Corollary 5.21 (5.1.14 in [2]; McMullen-Walkup). Let ∆ be a (d − 1)dimensional shellable s.c. with shelling F1 , . . . , Fm and h-vector h(∆) =
(h0 , . . . , hd ). For j = 2, . . . , m, let rj be the number of facets of hFj i ∩
hF1 , . . . , Fj−1 i and set r1 = 0. Then
hi = |{j | rj = i}|
for 0 ≤ i ≤ d. In particular, up to the order, the numbers rj do not depend
on the shelling.
Theorem 5.22 (5.1.15 in [2]; Stanley). Let h = (h0 , . . . , hd ) be a sequence
of integers. TFAE:
76
BRIAN JOHNSON
hii
(1) h0 = 1, and 0 ≤ hi+1 ≤ hi for all i ∈ {1, . . . , d − 1}
(2) h is the h-vector of a shellable s.c.
(3) h is the h-vector of some s.c. ∆ such that k[∆] is CM for some field
k.
6. A crash course on resolutions
A reference for this material is [6]. Suppose R is a commutative Noetherian ring.
Definition 6.1. Let M be an R-module.
(1) A base for M is a family of elements {eλ }λ∈Λ ⊆ M such that
(a) {eλ }λ generates M .
P
(b) Each m ∈ M can be written uniquely as M rλ eλ , where rλ ∈
R and rλ = 0 for all but finitely many λ.
(2) M is free if it has a base.
L
Proposition 6.2. An R-module M is free if and only if M ∼
=
λ∈Λ Rλ ,
where Λ indexes a base and Rλ = R for all λ.
Definition 6.3. A sequence of R-modules and R-module homomorphism
φi+1
φi
· · · −→ Mi+1 −→ Mi −→ Mi−1 −→ · · ·
is exact if ker φi = im φi+1 .
Example 6.4. Let R = k[x1 , x2 , x3 ], I = (x1 , x2 , x3 ), and suppose M =
R/I. Define φ1 : R3 −→ R by ei 7→ xi for i = 1, 2, 3, and where ei is the
standard basis vector with a 1 in the ith spot and 0 elsewhere. Then φ1 is
represented by the matrix B1 = [x1 , x2 , x3 ]. Now,
syz1 (x1 , x2 , x3 ) =
=
=
=
“relations among x1 , x2 , x3 ”
ker φ1
ker B1 (⊆ R3 )
module generated by f1 , f2 , f3 ,
where the fi are column vectors given by






x2
x3
0
f1 =  −x1  , f2 =  0  , f3 =  x3  .
0
−x1
−x2
Define a second map φ2 : R3 −→ R3 by ei 7→ fi for i = 1, 2, 3. This map is
then given by the matrix


x2
x3
0
0
x3  .
B2 =  −x1
0
−x1 x2
POWER OF MONOMIAL IDEALS
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77
Then
syz2 (x1 , x2 , x3 ) = 2nd syzygies of I
=
=
=
=
=
syz1 (f1 , f2 , f3 )
ker φ2
ker B2
module generated by g1 := (x3 , −x2 , x1 )t
relations on f1 , f2 , f3 .
Finally, define φ3 : R −→ R3 by e1 7→ g1 . This map is represented by the
matrix B3 = [x3 , −x2 , x1 ]t . So we can take
φ3
φ2
φ1
φ0
0 −→ R −→ R3 −→ R3 −→ R −→ R/I −→ 0
to be exact.
Definition 6.5. Let M be a finitely generated R-module. A free resolution
of M is an exact sequence of the form
φ2
φ1
φ0
· · · −→ F2 −→ F1 −→ F0 −→ M −→ 0,
where Fi ∼
= Rri for each i.
Exercise 6.6. The module M = I = (x2 −x, xy, y 2 −y) has a free resolution
over R = k[x, y] given by
φ1
φ0
0 −→ R2 −→ R3 −→ I −→ 0,
where φ0 is represented by A = [x2 − x, xy, y 2 − y] and φ1 is represented
by


y
0
B =  −x + 1 y − 1  .
0
−x
I.e., im B = ker A = syz1 (x2 − x, xy, y 2 − y).
Theorem 6.7 (Hilbert Syzygy Theorem). Let R = k[x1 , . . . , xn ]. Then
every finitely generated R-module M has a free resolution of the form
0 −→ Fd −→ · · · −→ F0 −→ M −→ 0,
where d ≤ n.
L
6.1. The graded world. Let S ∼
= n∈N Si be an N-graded commutative
Noetherian ring.
Definition 6.8. A graded S-module is an S-module of the form M =
L
n∈Z Mi , where each Mj is an additive subgroup of M and Si Mj ⊆ Mi+j
for all i ∈ N, j ∈ Z.
An entire class of easy examples is given as follows. If S = k[x1 , . . . , xn ],
and I ⊆ S is a homogeneous ideal, then I and S/I are graded S-modules.
78
BRIAN JOHNSON
Definition 6.9. Let M be a graded S-module. Fix a ∈ Z. We define the
left shift of M to be the graded module
M
M (a) :=
M (a)t ,
t∈Z
where M (a)t = Ma+t .
Definition 6.10. Let M, N be finitely generated S-modules.
(1) M is a graded free S-module if M is of the form
M
M=
S(−dλ ),
λ∈Λ
where Λ is an index set and dλ ∈ Z.
(2) An S-module homomorphism φ : M −→ N is graded of degree d if
φ(Mi ) ⊆ Ni+d for all i ∈ Z.
Definition 6.11. A graded free resolution of a finitely generated graded
S-module M is a resolution of M of the form
φ2
φ1
φ0
· · · −→ F2 −→ F1 −→ F0 −→ M −→ 0,
where each Fi is a finitely generated free graded S-module, and each φi is
graded of degree 0.
Example 6.12. Suppose R = k[x, y, z, w], and let
M = I = (g1 , g2 , g3 , g4 ),
where g1 = z 3 − yw2 , g2 = zy − xw, g3 = y 3 − x2 z, and g4 = xz 2 − y 2 w.
Define
φp : R(−3) ⊕ R(−2) ⊕ R(−3) ⊕ R(−3) −→ M
by (a, b, c, d) 7→ [g1 , g2 , g3 , g4 ]. Then
syz1 (g1 , g2 , g3 , g4 ) =
=
=
=
ker φ0
ker B0 (representing matrix)
relations among the gi
matrix with columns s1 , s2 , s3 , s4 ,
where



0
−x
 xz 
 wy


s1 = 
 −w  , s2 =  0
−y
z




0
−y

 y2 
 z2
 , s3 = 



 −z  , s4 =  0
−y
w


.

POWER OF MONOMIAL IDEALS
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79
Define φ1 : R(−4)4 −→ R(−3) ⊕ R(−2) ⊕ R(−3) ⊕ R(−3) by
 
a
 b 

(a, b, c, d) 7→ [s1 , s2 , s3 , s4 ] 
 c 
d


−xb − yd
 xza + ywb + y 2 c + z 2 d 
.
= 


−wa − zc
−ya + zb − xc + wd
Starting with the map, we see we needed the shift of −4 to make the homomorphism have degree 0. Now,
syz2 (I) = ker φ1 = module generated by [z, y, −w, −x]t .
Define φ2 : R(−5) −→ R(−4)4 by e1 7→ [z, y, −w, −x]t (here we shift by
−5 because each entry has degree 1). Thus, our resolution is
φ2
φ1
φ0
0 −→ R(−5) −→ R(−4)4 −→ R(−2) ⊕ R(−3)3 −→ M −→ 0.
Theorem 6.13 (Graded Hilbert Syzygy Theorem). Let S = k[x1 , . . . , xn ].
Then every finitely generated graded S-module has a graded free resolution
0 −→ Fd −→ · · · −→ F1 −→ F0 −→ M −→ 0,
where d ≤ n.
Definition 6.14. Let R = k[x1 , . . . , xn ], and suppose M = R/I for some
homogeneous ideal I ⊆ R. We define the (i, j)-graded Betti number of R/I,
βi,j (R/I), to be the number of minimal syzygies of degree j at step i of
a minimal graded resolution of R/I. (In a minimal resolution,
P all nonzero
entries of all matrices are of degree > 0.) We define βi (R/I) = j βi,j (R/I).
Example 6.15. Suppose R = k[x, y], and let I = (x2 , y 2 ), L = (x2 , xy, y 3 ).
Then H(R/I) = H(R/L), so L is a lex-segment ideal with the same Hilbert
function as I. Then minimal graded resolutions (without the maps) for R/I
and R/L, respectively, can be given by
0 −→ R(−4) −→ R(−2)2 −→ R −→ R/I −→ 0,
respectively,
0 −→ R(−3) ⊕ R(−4) −→ R(−2)2 ⊕ R(−3) −→ R −→ R/L −→ 0.
One can then find Betti numbers basically by looking at the step of the
resolution and the number of copies of each shifted copy of R. In particular,
for the first resolution, we have (suppressing the R/I) β0,0 = 1, β1,2 =
2, β3,4 = 1, and βi,j = 0 for all other i, j.
80
BRIAN JOHNSON
If one is using Macaulay to find resolutions, it outputs tables of Betti
numbers as follow. We find βi,j in column i and row j − i, where the rows
and columns are numbered starting from 0. For R/I and R/L, we get
total
0
1
2
R/I
1
1
-
2
2
-
1
1
total
0
1
2
R/L
1
1
-
3
2
1
2
- .
1
1
Note that βi,j (R/I) ≤ βi,j (R/L) for all i, j, which is the Bigatti-HulettPardue Theorem (4.22).
7. Group Presentations
7.1. Lifting monomial ideals: A. Croll, C. Gibbons, and B. Johnson. References for this material are [15] and [3].
Definition 7.1. Let R be a ring and let u1 , . . . , ut ∈ R be an R-regular
sequence. Let S = R/(u1 , . . . , ut ), and suppose B is an S-module and A an
R-module. We say A is a t-lifting of B to R if u1 , . . . , ut is an A-regular
sequence and A/(u1 , . . . , ut )A ∼
= B. If t = 1, we may suppress the t and just
call A a lifting of B.
Grothendieck asked the question: when does one have a lifting? He was
interested in this problem as it related to a conjecture of Serre on multiplicities. With rings and modules as above and C an S-module, the useful fact
is that if A is a lifting of B to R, then
S
∼
TorR
i (A, C) = Tori (B, C).
For more, see [3].
Lemma 7.2 (2.2 from [15]). Let T be a ring, x ∈ T , and set Q = T /(x).
Let B be a Q-module, and let
φ2
φ1
F2 −→ F1 −→ F0
be an exact sequence of Q-modules, where coker φ1 ∼
= B. Suppose
ψ2
ψ1
G2 −→ G1 −→ G0
is a complex of T -modules such that
(1) x is a nonzerodivisor on each Gi
(2) Gi ⊗T Q = Fi
(3) ψi ⊗T Q = φi .
Then coker ψ1 is a lifting of B to T .
POWER OF MONOMIAL IDEALS
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81
7.1.1. Background.
Definition 7.3. Let M be an R-module. A resolution of M is a complex
of R-modules
d2
d1
· · · −→ P2 −→
P1 −→
P0 −→ 0
together with a map ε : P0 −→ M such that the augmented complex
d
d
ε
2
1
· · · −→ P2 −→
P1 −→
P0 −→ M −→ 0
is exact. If each Pi is projective, we call this a projective resolution.
For the remainder of this presentation, k is an algebraically closed field
of characteristic 0 and x = x1 , . . . , xn is a sequence of elements of S. Recall
Definition 3.37, which we repeat here:
Definition 7.4. Let R = k[x0 , . . . , xn ] and S = k[x1 , . . . , xn ], where k is an
algebraically closed field of characteristic 0. Let K ⊆ S be a homogeneous
ideal. We say that K lifts to an ideal I ⊆ R if
√
(1) I = I,
(2) x0 + I is a nonzerodivisor on R/I, and
(3) under the canonical isomorphism R/(x0 ) −→ S, we have (I, x0 )/(x0 )
mapped isomorphically to K.
We generalize this definition as follows:
Definition 7.5. Let S = k[x] and R = k[x, u], where u = u1 , . . . , ut is an
R-regular sequence. Let J ⊆ S and I ⊆ R be homogeneous ideals. We say
I is a t-lifting of J provided:
(1) u is an R/I-regular sequence and
(2) (I, u)/(u) ∼
= J.
If, in addition, we have
√
(3) I = I,
we say I is a reduced lifting of J.
7.1.2. The lifting method. Fix S = k[x] and R
(called the lifting matrix)

L1,1 L1,2 · · ·
 ..
A= .
Ln,1 Ln,2 · · ·
= k[x, u]. Define a matrix



such that for each xj , 1 ≤ j ≤ n, we have infinitely many linear forms
Lj,i ∈ k[xj , u] with nonzero coefficient on xj . For every monomial m =
Qn
aj
j=1 xj ∈ S, define
!
aj
n
Y
Y
m=
Lj,i ∈ R.
j=1
i=1
Suppose J = (m1 , . . . , mr ) is a monomial ideal. Define I = (m1 , . . . , mr ).
Then one can show I is a t-lifting of J. The strategy goes something like:
82
BRIAN JOHNSON
• Use Lemma 7.2 inductively with T = k[x, u], Q = k[x, u1 , . . . , ut−1 ],
and x = ut .
• Come up with an exact sequence with an appropriate cokernel isomorphic to J and a complex with all the properties in 7.2.
Proposition 7.6 (2.6 in [15]). Let J = (m1 , . . . , mr ) ⊆ S be a monomial
ideal. Let I = (m1 , . . . , mr ) ⊆ R as above (we’ve fixed a lifting matrix A).
Consider a free S-resolution of J:
F. : 0 −→ S βn −→ S βn−1 −→ · · · −→ S β0 −→ 0.
Then there exists a free R-resolution of I:
F . : 0 −→ Rβn −→ Rβn−1 −→ · · · −→ Rβ0 −→ 0,
where the maps S βi −→ S βi−1 “lift” to the maps Rβi −→ Rβi−1 via the
matrix A and the linear forms in A in such a way that the resulting complex
is exact (see Example 2.7 in [15]).
Remark 7.7.
(1) coker(S β1 −→ S β0 ) = J,
(2) coker(Rβ1 −→ Rβ0 ) = I,
(3) S β2 −→ S β1 −→ S β0 is exact,
(4) Rβ2 −→ Rβ1 −→ Rβ0 is a complex with the desired properties from
Lemma 7.2, and
(5) if F. is a minimal free resolution, F . is as well.
The proof of the previous proposition uses the Taylor resolution, which
can be found in [8].
7.1.3. Homological results.
Definition 7.8. If (R, m) is a local Noetherian ring and M is a finitely
generated R-module, we define
depthR M = max{i | there exists an M -regular sequence of length i in m}.
It’s worth noting here (as the astute reader may have already noticed)
that the situation we have introduced isn’t necessarily that of a local ring.
However, in our graded situation, with the proper hypotheses (which we
have), the notions of depth, minimal free resolutions, projective dimension,
etc. are all still well-defined and make sense. Henceforth, we’ll ignore such
concerns.
Theorem 7.9 (Auslander-Buchsbaum formula). If M 6= 0 is a finitely generated R-module with proj dimR M < ∞, then
depthR M + proj dimR M = depth R.
Corollary 7.10 (2.8 in [15]). With notation as in 7.6, we have
depthR (R/I) = depthS (S/J) + t.
In particular, depthR (R/I) ≥ t.
POWER OF MONOMIAL IDEALS
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83
Proof. To use the A-B formula, we need to show the projective dimensions of
both R/I and S/J are finite. Considering the resolutions from 7.6, one can
construct a free resolution for S/J (or R/I) by the following construction:
···
/ S β0 ε / J
@@
@@ ε
@@
@@
S
/0
/ S/J
/0
An additional consequence is that the projective dimensions are actually
equal, because 7.6 implies that if F. is a minimal resolution, F . is as well.
Thus,
depthR (R/I) = depth R − proj dimR (R/I)
= depth R − proj dimS (S/I)
= depth R − (depth S − depthS (S/J).
The variables x1 , . . . , xn are a maximal S-regular sequence. Also x, u is a
maximal R-regular sequence. Therefore,
depth R − depth S = n + t − n = t.
Corollary 7.11 (2.18 in [15]). Let J ⊆ S be a monomial ideal and fix
A (a lifting matrix). let I be the lift of J via A. If J has the primary
decomposition J = Q1 ∩ · · · ∩ Qr , then
I = Q1 ∩ · · · ∩ Qr ,
where Qi is generated by the lifts of the generators of Qi (this is not necessarily a primary decomposition).
The proof follows immediately from:
Lemma 7.12 (2.17 in [15]). Let J1 , J2 ⊆ S be monomial ideals and A a
lifting matrix. Let “ ” denote lifting. Then
(1) J1 ⊆ J2 if and only if J1 ⊆ J2 .
(2) J1 ∩ J2 = J1 ∩ J2
7.2. Representations of monomial orders: R. Brase, A. Denkert,
and M. Janssen. Material in this presentation comes from [8][§15.2, §15.8],
[17], and [19]. Recall that any monomial order on k[x1 , . . . , xn ] can be
realized as a total order on Nn (where N = N0 ).
Definition 7.13. A linear group order, or l.g.o., on a Q-vector space V is
a total order satisfying
(1) x > y if and only if x + z > y + z, and
(2) x > y if and only if αx > αy
for all x, y, z ∈ Q and α ∈ Q+ .
84
BRIAN JOHNSON
Remark 7.14. To completely determine a l.g.o. it suffices to specify which
elements are positive. Indeed, x > y if and only if x − y > 0.
Definition 7.15. A l.g.o. on Qn is admissible if α > 0 for α ∈ Nn r
{(0, . . . , 0)}.
Proposition 7.16. An ordering on Nn corresponding to a monomial order
is admissible, and conversely, any admissible order gives a monomial order.
Proof. Apparently boring and not enlightening.
An aside that will later be useful is the concept of an ordered polynomial
ring: Let F be an ordered field and t and indeterminate over F . Then F (t)
is an ordered field if we specify t > F (in F (t)). Then
1 < t < t2 < · · · .
Also, it then follows that an tn + · · · + a0 > 0 if and only if an > 0.
Example 7.17. Let F = R(t1 , . . . , ts ) and suppose ti > R(t1 , . . . , ti−1 ) for
each i. If
N
X
f=
ai tαi ,
i=1
where tα1 >revlex tα2 >revlex · · · , then f > 0 if and only if a1 > 0.
Theorem 7.18. Let > be a l.g.o. on a Q-vector space V ⊆ Qn . Then there
exists an ordered field F = R(t1 , . . . , ts ) and elements a1 , . . . , an ∈ F such
that if x = x1 e1 + · · · + xn en ∈ V , we have
x >V 0 ⇐⇒ a1 x1 + · · · + an xn >F 0.
Lemma 7.19. Let s = dimQ V . Then we may assume V = Qs and that
0 < e1 < · · · < es .
Proof. Let 0 < b1 < · · · < bs be a basis for V . Let φ : V −→ Qs via bi 7→ ei .
Let M be the matrix representing φ. I.e.,
φ(x)t = M xt
for x = (x1 , . . . , xt ) ∈ V . Define an order on Qs by setting y >Qs 0 if and
only if φ−1 (y) >V 0. Let a1 , . . . , as be as in the theorem for Qs (supposing
the theorem is true). Then for all v ∈ V , we have
x >V 0
⇐⇒
⇐⇒
φ(x) >Qs
a · φ(x) >F 0
⇐⇒
aφ(x)t >F 0
⇐⇒
a(M xt ) >F 0
⇐⇒
⇐⇒
(aM )xt >F 0
(aM ) · x >F 0.
So aM is a vector in F n which satisfies the theorem.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
85
Definition 7.20. A l.g.o. on Qn is restricted, or r.g.o., if 0 < e1 < · · · < en .
Lemma 7.21. Given a r.g.o. < on Qn , there exist {ai (t)}ni=1 ⊆ R(t) such
that
1 ≤R(t) a1 (t) ≤R(t) · · · ≤R(t) an (t),
P
P
and for x = (x1 , . . . , xn ) ∈ Qn ,
ai (t)xi >R(t) 0 implies
xi ei >Qn 0.
Remark 7.22. For 1 ≤R(t) a1 (t) ≤R(t) · · · ≤R(t) an (t),
(1) deg(ai (t)) ≥ deg(ai−1 (t)), and
(2) lc(ai (t)) > 0.
Proof of Lemma 7.21. We proceed by induction on n. If n = 1, then <Qn is
the usual order on Q, so take a1 = 1, and we’re done.
Suppose n > 1. Then <Qn induces a r.g.o. on Qn−1 ,→ Qn , where
(q1 , . . . , qn−1 ) 7→ (q1 , . . . , qn−1 , 0). Inductively, we have a1 (t), . . . , an−1 (t),
and we know
n−1
n−1
X
X
ai (t)xi >R(t) 0 implies
xi ei >Qn−1 0.
i=1
i=1
Define
A = {q ∈ Q | qen−1 >Qn en },
and set
(
inf A
b=
t
if A 6= ∅
if A = ∅
Note that en > en−1 , so any q ∈ A has q > 1, so b ≥ 1, whether or not A is
empty. Finally, set an (t) = ban−1 (t).
Consider x1 e1 + · · · + xn en ∈ Qn , and suppose
n
X
ai (t)xi >R(t) 0.
i=1
If xn = 0, we’re done by induction. Suppose xn 6= 0.
Case 1: A = ∅. Then b = t, so an = tan−1 has strictly larger degree than
a1 , . . . , an−1 . Thus
n
X
> 0 ⇐⇒ xn lc(an ) > 0
i=1
⇐⇒
xn > 0
(because lc(an ) > 0).
Since en > qen−1 for all q ∈ Q, we have en > qei for all q ∈ Q. Thus
n
X
xi ei > 0 if and only if xn > 0,
i=1
completing case 1.
Case 2: A 6= ∅. Then b = inf A. Let k ≥ 1 be such that
deg an = deg an−1 = · · · = deg an−k > deg an−k−1 .
86
BRIAN JOHNSON
Then, we have
Pn
i=1
>R(t) 0 if and only if
lc(an−k )xn−k + · · · + lc(an−1 )xn−1 + lc(an )xn > 0,
and this is if and only if
(11)
lc(an−k )xn−k + · · · + lc(an−1 )xn−1 + lc(an−1 b)xn > 0.
Choose q ∈ Q sufficiently close to b such that (11) holds with b replaced by
q. Choose q > b (resp., q < b) if xn < 0 (resp., xn > 0). If q > b and xn < 0,
then q ∈ A, so qen−1 > en , and thus
qxn en−1 < xn en .
In the other case, we also have qxn en−1 < xn en . Then
lc(an−k )xn−k + · · · + lc(an−1 )(xn−1 + qxn ) > 0
⇐⇒ a1 (t)x1 + · · · + an−1 (xn−1 + qxn ) >R(t)
=⇒
=⇒
x1 e1 + · · · + (xn−1 + qxn )en−1 >Qn 0
X
xi ei >Qn 0.
Proof of Theorem 7.18. We again use induction on n. If n = 1, we set
a1 = 1, and we’re done.
Suppose n > 1. We may assume V = Qn and 0 < e1 < · · · < en . Let
a = (a1 , . . . , an ) be the vector of elements from the lemma. Define
E = {x ∈ Qn | a · x = 0}.
Note that E is a proper linear subspace of Qn and dim E = s < n. By
induction, we have b = (b1 , . . . , bn ) such that for all x ∈ E, b · x > 0 if and
only if x > 0.
Case 1: a · x < 0 or a · x > 0. Then x < 0 or x > 0 by the lemma.
Case 2: a · x = 0. Then x > 0 if and only if b · x > 0. Define c = at + b
(t is a new indeterminate), where t is transcendental and t is greater than
everything else. So, if c · x > 0, then either a · x > 0, or a · x = 0. In the
latter case, x ∈ E implies b · x > 0, so x > 0.
Proposition 7.23. With notation as above, a1 , . . . , an define an admissible
order if and only if ai > 0 for each i.
Finally, we look at the relation of the previous material to monomial
ideals. Fix notation now with S = k[x1 , . . . , xn ].
Definition 7.24. Suppose >1 , >2 , . . . are partial orders on S. The product
order or lex product order, >, is the partial order where xα > xβ if xα >i xβ
for the first i where the two monomials are comparable.
If λ : Nn −→ N is a linear function, we define a weight order >λ , where
xα >λ xβ if λ(α) > λ(β). We say that > and >λ are compatible if xα >λ xβ
implies xα > xβ .
POWER OF MONOMIAL IDEALS
SUSAN COOPER
87
Robbiano showed that every monomial order > is the lex product of n
weight orders (here, n is still the number of variables).
Example 7.25. Let λi : Nn −→ N be the projection onto the ith coordinate.
(1) Then >lex is the lex product of >λ1 , . . . , >λn .
(2) Then >grevlex is the lex product of
>λτ , >−λn , . . . , >−λ2 ,
where τ is the total degree function (and there is some debate about
whether the minus signs are necessary and what exactly they mean,
but the point is that since it’s grevlex, we need to reverse the order
somehow).
7.3. Associated primes of monomial ideals and odd holes: D. Boeckner and D. Stolee. The main reference for this material is [9]
Definition 7.26. A simple graph is a graph G = (V, E), where E ⊆ V2
(this notation means that there are no multi-edges or loops).
Definition 7.27. A complete graph has all possible edges, the empty graph
has no edges. Given a graph G, its complement is
V (G)
G = V (G),
r E(G) .
2
Definition 7.28. Given a graph G and S ⊆ V (G), the induced subgraph
G[S] is
S
G[S] = S,
∩ E(G) .
2
A clique is a complete induced subgraph. We set
W (G) = the maximum order (# of vertices) of a clique.
An independent set is an induced empty graph. A coloring is a labeling in
which no adjacent vertices have the same label. We set
χ(G) = minimal # of colors allowing a proper coloring.
Proposition 7.29. W (G) ≤ χ(G).
Proof. A complete graph requires W (G) colors.
Furthermore, there exists G such that W (G) = 2 and χ(G) is arbitrarily
large.
Definition 7.30. A graph G is perfect if for every S ⊆ V (G),
W (G[S]) = χ(G[S]).
Theorem 7.31 (Perfect Graph Theorem–Lovász, 1972). A graph G is perfect if and only if G is perfect.
Theorem 7.32 (Strong Perfect Graph Theorem–CRST, 2002). A graph G
is perfect if and only if G has no odd holes and no odd antiholes.
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BRIAN JOHNSON
Definition 7.33. An odd hole is an induced odd cycle of length at least 5.
An odd antihole is the complement of an odd hole.
For the remainder, fix a graph G.
Definition 7.34. A vertex cover is a subset W ⊆ V (G) such that for all
ij ∈ E(G), either i ∈ W or j ∈ W . A k-cover is a vector α ∈ Nn (where
n = |V (G)|) such that for ij ∈ E(G)
αi + αj ≥ k.
A k-cover α is reducible if there exist an i-cover β and a j-cover γ such that
k =i+j
and
α = β + γ.
Otherwise α is irreducible.
Let S = k[x1 , . . . , xn ], n = |V (G)|. The edge ideal of G is
I(G) = (xi xj | ij ∈ E(G)).
This is a squarefree, quadratic monomial ideal.
Definition 7.35. Let I be a squarefree monomial ideal. The Alexander
dual of I is I ∨ . If I = (xα1 , . . . , xαr ), then
∨
I =
r
\
(xj | αi,j > 0).
i=1
Thus,
I(G)∨ =
\
(xi , xj ).
ij∈E(G)
Definition 7.36. Let I = P1 ∩ · · · ∩ Pr . The jth symbolic power of I is
I (j) = P1j ∩ · · · ∩ Prj .
Thus,
(I(G)∨ )(2) =
\
(x1 , xj )2 .
ij∈E
Then one can verify that xα ∈ (I(G)∨ )(2) if and only if α is a 2-cover.
Now, I(G)∨ is the set of 1-covers. If g1 , g2 ∈ I(G)∨ are two monomials,
then g1 g2 is a monomial whose exponents form a 2-cover which is the sum
of two 1-covers. So (I(G)∨ )2 is generated by all 2-covers formed by adding
1-covers. Set J = I(G)∨ . Recall that Ass I := Ass(R/I). Then we have
that J (2) ⊇ J 2 implies Ass(J 2 ) ⊇ Ass(J (2) ). What is Ass J?
\
\
J=
(xi , xj ) and J (2) =
(xi , xj )2 .
ij∈E
ij∈E
POWER OF MONOMIAL IDEALS
SUSAN COOPER
89
Definition 7.37 (from Ch.5, §2 of [16]). An irreducible monomial ideal in
n variables is one of the form
αi
α
(xi1 1 , . . . , xitit ),
denoted by mα , where m = (x1 , . . . , xn ) and α ∈ Nn . If αi = 0, omit xi
(rather than including 1 ∈ mα ).
Lemma 7.38 (5.18 in [16]). Every monomial ideal can be decomposed as
a finite intersection of irreducible monomial deals. A decomposition is irredundant if we can’t eliminate any of the ideals.
Theorem 7.39 (from [9]).
\
J2 =
[(x2i , xj ) ∩ (xi , x2j )] ∩
ij∈E
\
(x2i1 , . . . , x2is ).
{xi1 ,...,xis }
induces an
odd cycle
Lemma 7.40 (Dupont, Villarreal, 2007).
(1) If G is bipartite, then G has no irreducible 2-cover.
(2) If G is not bipartite, and c is a 2-cover that cannot be written as
the sum of two 1-covers ( not necessarily irreducible), then up to a
rearrangement of the vertices, we can write
c = (0, . . . , 0, b1 , b2 , . . . , b|B| , 1, . . . , 1),
| {z } |
{z
} | {z }
A
B
C
and
(1) bj ≥ 2,
(2) B is not a vertex cover of G, and A is an independent set, and
(3) the subgraph G[C] is not bipartite.
Also, B ⊇ N (A) (the neighborhod of A: stuff adjacent to something in A).
Proof of Theorem 7.39. Let L denote the right side of the expression in the
statement of the theorem. We’ll first show J 2 ⊆ L. Take a minimal generator g ∈ J 2 . Then g represents a 2-cover we can write as a sum of two
1-covers. So g ∈ (xi , xj )2 for all ij ∈ E, and g ∈ (xi , xj )2 ⊆ (x2i , xj ). Therefore it suffices to show g is in the right-most intersection in the expression
for L. Let {xij }sj=1 be the vertices induced from an odd cycle. We want to
show x2ij | g for some ij . Since g1 g2 = g and g1 , g2 are both 1-covers, at least
s+1
2 of the xij have an exponent of 1 or more in g1 , g2 . By the pigeonhole
principle, one of the xij divides both gi and thus divides g twice. Therefore,
g ∈ L.
To see L ⊆ J 2 , let h be a minimal generator of L. Since h ∈ [(x2i , xj ) ∩
(xi , x2j )] for all ij ∈ E, hi +hj ≥ 2 implies h is a 2-cover (hi , hj are exponents
of xi , xj I believe). So it suffices to show that h is the sum of two 1-covers.
Assume h is not the sum of two 1-covers. Then we can write
h = (0, . . . , 0, b1 , . . . , b|B| , 1, . . . , 1)
90
BRIAN JOHNSON
from the lemma. Thus
h=
Y
i∈B
xbi i
Y
xj .
j∈C
We also have G[C] is not bipartite, so G[C] contains an odd cycle. One can
then find an induced odd cycle (kinda like trapping a lion in the desert).
But because of the expression for L, we then know that h ∈ (x2i1 , . . . , x2is )
for some collection of vertices {xi1 , . . . , xis }. This implies the ci must be
at least 2, contradicting that they are 1. Therefore, h is the sum of two
1-covers, so h ∈ J 2 .
Therefore,
Ass(J 2 ) = {(xi , xj )}ij∈E
∪ {(xi1 , . . . , xis ) | {xij }sj=1 induces an odd cycle}.
Corollary 7.41. G is perfect if and only if neither Ass(S/(I(G)∨ )2 ) nor
Ass(S/(I(G)∨ )2 ) has a prime of height greater than 3.
Remark 7.42. Let G be a graph on n vertices, and let t denote the length
of the largest induced odd cycle. Then
(1) depth(S/J 2 ) ≤ n − t
(2) proj dim(S/J 2 ) ≥ t
Proof. By the Auslander-Buchsbaum formula, we need only prove (1). By
[2, 1.2.13], we get
depth(R/I) ≤ dim(R/p)
for all p ∈ Ass(R/I), and dim(R/p) ≤ n − t.
7.4. Resolutions and Betti numbers: J. DeVries and X. Yu. References for this material are [13] and [14]. Given a monomial ideal I, we can
construct a (graded) free resolution
M
M
· · · −→
R(−j)βi,j (I) −→ · · · −→
R(−j)β0,j (I) −→ I −→ 0.
j
j
Also, recall the edge ideal defined in the previous section:
I(G) = ({xi xj | {xi , xj } ∈ E(G)}).
Definition 7.43 (Eliahou, Kervaire). A monomial ideal I is splittable if
I = J + K such that
(1) if g(I) is the set of minimal generators of I, then we have that g(I)
is the disjoint union of g(J) and g(K),
(2) there exists a splitting function
g(J ∩ K) −→ g(J) × g(K)
w 7→ (f (w), h(w))
where
(a) if w ∈ g(J ∩ K), then w = lcm(f (w), h(w)), and
POWER OF MONOMIAL IDEALS
SUSAN COOPER
91
(b) if S ⊂ g(J ∩ K), then lcm(f (S)) and lcm(h(S)) strictly divide
lcm(S).
Theorem 7.44 (Eliahou-Kervaire, 1990; Fatabbi, 2001). Suppose I is a
splittable monomial ideal with splitting I = J + K. Then for all i, j ≥ 0,
βi,j (I) = βi,j (J) + βi,j (K) + βi−1,j (J ∩ K).
Definition 7.45. Let G be a simple graph with edge ideal I(G). Suppose
e = uv ∈ E(G). Set J = (uv) and K = I(G r e). If I = J + K is a splitting
of I, then we call e a splitting edge.
So now we want to characterize splitting edges (since it would be easier
than determining when some arbitrary ideal is splittable).
Theorem 7.46 (Hà, Adan). An edge e = uv is a splitting edge of G if and
only if N (u) ⊆ (N (v) ∪ {v}) or N (V ) ⊆ (N (u) ∪ {u}).
Lemma 7.47. With notation as above, suppose
A := N (u) r {v} = {u1 , . . . , un }
and
B := N (v) r {u} = {v1 , . . . , vm }.
If H = G r (N (u) ∪ N (v)), then
J ∩ K = uv((A, B) + I(H)).
Corollary 7.48.
g(J ∩ K) = {uvui | ui ∈ A r B} ∪ {uvvi | vi ∈ B r A}
∪{uvzi | zi ∈ A ∩ B} ∪ {uvm | m ∈ I(H)}.
Sketch of proof. ⇐=: Without loss of generality, we have (1) of 7.43. Write
g(J ∩ K) = {uvvi | vi ∈ N (v) r {u}} ∪ {uvm | m ∈ I(H)}.
We want to show e is a splitting edge. Then show
g(J ∩ K) −→ g(J) × g(K)
u
7→
(
(uv, vvi ) if w = uvvi
(f (w), h(w)) =
.
(uv, m) if w = uvm
=⇒: By contradiction.
Theorem 7.49. Let e = uv be a splitting edge of G, and set
n = |N (u) ∪ N (v)| − 2
and H = G r (N (u) ∪ N (v)). Then for all i ≥ 1, j ≥ 0, we have
i X
n
βi,j (I(G)) = βi,j (I(G r e)) +
βi−1−`,j−2−` (I(H)),
`
`=0
where β−1,0 (I(H)) = 1 and β−1, (I(H)) = 0 for j > 0.
92
BRIAN JOHNSON
Example 7.50. Consider the complete bipartite graph K1,d , or the “star.”
We claim

d

 i+1 j = i + 2, i ≥ 0
βi,j (I(K1,d )) = 1
j = 0, i = −1 .


0
otherwise
To see this is the case, note that any edge is a splitting edge. By the theorem,
for an edge e,
i X
d−1
βi,j (I(K1,d )) = βi,j (I(G
r e})) +
βi−1−`,j−2−` (I(H)),
| {z
`
K1,d−1
`=0
but I(H) = 0 because all vertices are adjacent to the center (thinking of
K1,d as the star). Certainly
(
1 i − 1 − ` = −1, j − 2 − ` = 0(⇔ i = `, j = ` + 2)
βi−1−`,j−2−` (0) =
0 otherwise
(from the base case of the theorem). Now, use induction on d. Suppose
j = i + 2, i ≥ 0. Then
d−1
βi,i+2 (0)
βi,i+2 (I(K1,d )) = βi,i+2 (I(K1,d−1 )) +
i
| {z }
=1
d−1
d−1
=
+
i+1
i
d
=
.
i+1
Finishing the other cases is an exercise.
Unfortunately, splitting edges don’t always exist, even in very small cases.
For example, if G = C4 , the cycle of length 4, there is no splitting edge (one
should verify this). This leads to the notion of splitting vertices. Introduce
the following notation. Suppose v ∈ V (G) with N (v) = {v1 , . . . , vd }. For
each i = 1, . . . , d, set Gi = G r (N (v) ∪ N (vi )), and finally set
G(v) = G{v1 ,...,vd }
∪ {v(e) | e ∈ E(G), e is incident to one of the vi , but not v}
An example of this notation is given in Figure 7.
Definition 7.51. A vertex v is a splitting vertex if deg v > 0 and
E(G r {v}) 6= ∅.
Theorem 7.52. Let v be a splitting vertex. Then I(G) has a splitting
J(vv1 , . . . , vvd ), K = I(G r {v}).
POWER OF MONOMIAL IDEALS
G
v1 s
v2 s
G1 = G2
v1 s
s
@ v
@s
SUSAN COOPER
s v3
@
@s
93
G3
∅
v2 s
G(v)
v1 s
v2 s
s
s v3
@
@s
Figure 7. Computing G(v) , etc. for a graph.
Theorem 7.53. Let v be a splitting vertex. Then
βi,j (I(G)) = βi,j (K1,d ) + βi,j (I(G r {v})) + βi−1,j (L),
where L = vI(G(v) ) + vv1 I(G1 ) + · · · + vvd I(Gd ).
Corollary 7.54. With notation above
βi,i+2 (I(G)) = βi,i+2 (I(K1,d )) + βi,i+2 (I(G r {v})) + βi−1,i+2 (I(G(v) )).
Betti numbers of the form βi,i+2 (I) are called the linear strand of I (when
I has a quadratic generating set).
Definition 7.55. A minimal cycle of G is a cycle of length greater than
equal to 4 with no chord. A graph is chordal if it has no minimal cycles. An
ideal I has N2,p if I is generated by quadratics and its resolution is linear
up to stage p (i.e., βi,j = 0 for 0 ≤ i ≤ p − 1 and j > i + 2).
Theorem 7.56. I(G) has N2,p for p > 1 if and only if every minimal cycle
of G has length greater than or equal to p + 3.
Corollary 7.57. If G is chordal, then I(G) has linear resolution (all nonzero
Betti numbers have the form βi,i+2 (I)).
Example 7.58. Consider the following graph and its complement. Since G
is chordal (doesn’t even have any cycles, let alone minimal ones) we can use
Corollary 7.54 to find all the Betti numbers. This is useful since this graph
does not have a splitting edge but does have a splitting vertex.
G
r
r
f
r
r
splitting vertex
G
r
Q
r
Q
QQ
Qr
r
Figure 8. A chordal graph and its complement.
94
BRIAN JOHNSON
References
1. Anna Bigatti, Anthony V. Geramita, and Juan C. Migliore, Geometric consequences
of extremal behavior in a theorem of Macaulay, Trans. Amer. Math. Soc. 346 (1994),
no. 1, 203–235.
2. Winfried Bruns and H. Jrgen Herzog, Cohen-macaulay rings (cambridge studies in
advanced mathematics), Cambridge University Press, 1998.
3. David A. Buchsbaum and David Eisenbud, Lifting modules and a theorem on finite
free resolutions, Ring theory (Proc. Conf., Park City, Utah, 1971), Academic Press,
New York, 1972, pp. 63–74.
4. G. F. Clements and B. Lindström, A generalization of a combinatorial theorem of
Macaulay, J. Combinatorial Theory 7 (1969), 230–238.
5. Susan M. Cooper and Leslie G. Roberts, Algebraic interpretation of a theorem of
Clements and Lindström, J. Commut. Algebra 1 (2009), no. 3, 361–380.
6. David Cox, John Little, and Donal O’Shea, Using algebraic geometry, second ed.,
Graduate Texts in Mathematics, vol. 185, Springer, New York, 2005.
7.
, Ideals, varieties, and algorithms, third ed., Undergraduate Texts in Mathematics, Springer, New York, 2007, An introduction to computational algebraic geometry and commutative algebra.
8. David Eisenbud, Commutative algebra, Graduate Texts in Mathematics, vol. 150,
Springer-Verlag, New York, 1995, With a view toward algebraic geometry.
9. Christopher A. Francisco, Huy Tai Ha, and Adam Van Tuyl, Associated primes of
monomial ideals and odd holes in graphs, 2008.
10. Christopher A. Francisco and Benjamin P. Richert, Lex-plus-powers ideals, Syzygies
and Hilbert functions, Lect. Notes Pure Appl. Math., vol. 254, Chapman & Hall/CRC,
Boca Raton, FL, 2007, pp. 113–144.
11. A. V. Geramita, D. Gregory, and L. Roberts, Monomial ideals and points in projective
space, J. Pure Appl. Algebra 40 (1986), no. 1, 33–62.
12. Curtis Greene and Daniel J. Kleitman, Proof techniques in the theory of finite sets,
Studies in combinatorics, MAA Stud. Math., vol. 17, Math. Assoc. America, Washington, D.C., 1978, pp. 22–79.
13. Huy Tài Hà and Adam Van Tuyl, Resolutions of square-free monomial ideals via facet
ideals: a survey, Algebra, geometry and their interactions, Contemp. Math., vol. 448,
Amer. Math. Soc., Providence, RI, 2007, pp. 91–117.
, Splittable ideals and the resolutions of monomial ideals, J. Algebra 309 (2007),
14.
no. 1, 405–425.
15. Juan C. Migliore and Uwe Nagel, Lifting monomial ideals, 1999.
16. Ezra Miller and Bernd Sturmfels, Combinatorial commutative algebra, Graduate Texts
in Mathematics, vol. 227, Springer-Verlag, New York, 2005.
17. Lorenzo Robbiano, Term orderings on the polynomial ring, EUROCAL ’85, Vol. 2
(Linz, 1985), Lecture Notes in Comput. Sci., vol. 204, Springer, Berlin, 1985, pp. 513–
517.
18. Richard P. Stanley, Hilbert functions of graded algebras, Advances in Math. 28 (1978),
no. 1, 57–83.
19. Volker Weispfenning, Admissible orders and linear forms, SIGSAM Bull. 21 (1987),
no. 2, 16–18.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
95
Appendix A. Problem Set 1
Due: Tuesday, February 16
(1) Let S = k[x1 , . . . , xn ] where k is a field. Fix a monomial order >σ
on Zn≥0 .
(a) Show that multideg(f g) = multideg(f )+ multideg(g) for nonzero polynomials f, g ∈ S.
(b) A special case of a weight order is constructed as follows. Fix
u ∈ Zn≥0 . Then, for α , β in Zn≥0 , define α >u,σ β if and only if
u · α > u · β,
(2)
(3)
(4)
(5)
or
u·α = u·β
and
α >σ β ,
where · denotes the usual dot product of vectors. Verify that
>u,σ is a monomial order.
(c) A particular example of a weight order is the elimination order
which was introduced by Bayer and Stillman. Fix an integer
1 ≤ i ≤ n and let u = (1, . . . , 1, 0, . . . , 0), where there are i 1’s
and n − i 0’s. Then the ith elimination order >i is the weight
order >u,grevlex . Prove that >i has the following property: if xα
is a monomial in which one of x1 , . . . , xi appears, then xα >i xβ
for any monomial xβ involving only xi+1 , . . . , xn . Does this
property hold for the graded reverse lexicographic order?
Let I be a non-zero ideal in k[x1 , . . . , xn ]. Let G = {g1 , . . . , gt } and
F = {f1 , . . . , fr } be two minimal Gröbner bases for I with respect
to some fixed monomial order. Show that {LT (g1 ), . . . , LT (gt )} =
{LT (f1 ), . . . , LT (fr )}.
Suppose that I = (g1 , . . . , gt ) is a non-zero ideal of k[x1 , . . . , xn ] and
fix a monomial order on Zn≥0 . Suppose that for all f in I we obtain
a zero remainder upon dividing f by G = {g1 , . . . , gt } using the
Division Algorithm. Prove that G is a Gröbner basis for I. (We
showed the converse of this statement in class.)
Consider the ideal I = (xy + z − xz, x2 − z) ⊂ k[x, y, z]. For what
follows, use the graded reverse lexicographic order with x > y > z.
You are not permitted to use a computer algebra system for this
exercise. Be sure to show all of your work.
(a) Apply Buchberger’s Algorithm to find a Gröbner basis for I. Is
the result a reduced Gröbner basis for I?
(b) Use your answer from part (a) to determine if f = xy 3 z−z 3 +xy
is in I.
Consider the affine variety V = V(x2 +y 2 +z 2 −4, x2 +2y 2 −5, xz −1)
in C3 . Use a computer algebra system and Gröbner bases to find all
the points of V .
96
BRIAN JOHNSON
Appendix B. Problem Set 2
Due: Thursday, March 25
(1) Fix H := (1, 4, 6, 9, 10, 13, 13, . . .) and let S := k[x1 , x2 , x3 , x4 ] where
k is a field. Does there exist a homogeneous ideal I ⊂ S such that
H(S/I) = H? Provide two reasons for your answer: one using an
O-sequence approach and one using an order ideal of monomials
approach.
(2) For this exercise we use the same notation that was set up in our discussion of lifting monomial ideals. Let f = xα ∈ S = k[x1 , . . . , xn ].
Prove the following two facts:
β ) = 0 if and only if α 6≤ β ;
(a) f (β
α) (except for α itself).
(b) f (γγ ) = 0 for all γ with deg(γγ ) ≤ deg(α
(3) In this exercise we further explore Hilbert functions of distinct points
in projective 2-space. Let S = k[x1 , x2 ], where k is an algebraically
closed field of characteristic
zero. Further, let J ⊂ S be a homoge√
neous ideal such that J = (x1 , x2 ). We set α(J) to be the least
degree of a non-zero homogeneous polynomial in J.
(a) Set B = S/J. Prove that
t+1
for t < α(J)
H(B, t) =
≤ α(J) for t ≥ α(J).
(b) Let V ⊂ St be a subspace of St . Denote by S1 V the subspace
of St+1 generated by {Lv | L ∈ S1 and v ∈ V }. Prove that
dimk (S1 V ) ≥ (dimk V ) + 1.
(c) Let X = {P1 , . . . , Pt } be a set of distinct points in P2 . We set
α = α(X) to be the least degree of a non-zero homogeneous
polynomial in I(X). Show that ∆H(X) has the form
∆H(X) = {1, 2, 3, . . . , α − 1, α, ∆H(X, α), ∆H(X, α + 1), . . .}
where α ≥ ∆H(X, α) ≥ ∆H(X, α + 1) ≥ ∆H(X, α + 2) ≥ · · · .
(4) Find all possible Hilbert functions for 9 distinct points in P2 . Pick
one of the Hilbert functions H and find a set X ⊂ P2 of 9 distinct
points in P2 such that H(X) = H. How do you know that the
constructed set of points has the selected Hilbert function?
(5) Suppose that I is a homogeneous ideal in the ring R = k[x0 , . . . , xn ]
where k is an algebraically closed field of characteristic 0. Suppose
that Id 6= 0 and that H(R/I) has maximal growth in degree d. Prove
that Id and Id+1 have a greatest common divisor of positive degree
in the following two cases:
(a) n = 1 and H(R/I, d) ≥ 1;
(b) n = 2 and H(R/I, d) ≥ d + 1.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
97
Appendix C. Problem Set 3
Due: Thursday, April 15
This problem set involves choices! Submit solutions to 2 exercises from Part
I and 1 exercise from Part II.
Part I - Exercises Related to Hilbert Functions & Regular
Sequences
(1) For parts (b) - (d) of this exercise use reverse-lexicographic order
with x1 >revlex > x2 >revlex · · · .
(a) Find a (3, 4, 5)-lex-plus-powers ideal L ⊂ S = k[x1 , x2 , x3 ] such
that H(S/L, 3) = 9 and H(S/L, 6) = 5.
(b) Fix m to be a monomial of degree d in
S = k[x1 , x2 , x3 , x4 ]/(x51 , x42 , x43 , x34 ).
Recall that L(m) denotes the set of all degree d monomials in S
which are greater than or equal to m. Decompose
|L(x31 x32 x24 )|
e1 ,...,ej . Give an algebraic
in terms of integers of the form
l
description of each term in the decomposition.
(c) Assume I ⊂ S = k[x1 , x2 , x3 , x4 ] is a homogeneous ideal containing {x51 , x42 , x43 , x34 }. If H(S/I, 8) = 17, then what is the
largest value possible for H(S/I, 9)?
(d) Assume that the EGH Conjecture is true. Can there be a
homogeneous (3, 4, 4, 5)-ideal I ⊂ S = k[x1 , x2 , x3 , x4 ] with
H(S/I) = (1, 4, 10, 18, 24, 29, . . .)?
(2) EGH Points Conjecture in P2 : Fix integers 2 ≤ d1 ≤ d2 . Let ∆H =
{ht }t≥0 be the first difference Hilbert function of some finite set of
distinct points in P2 such that ht ≤ H(k[x1 , x2 ]/(xd11 , xd22 ), t) for all
t ≥ 0. Prove that there exist finite sets of distinct points X ⊆ Y ⊂ P2
where Y is a complete intersection of type {d1 , d2 } and ∆H(X) =
(t)
∆H if and only if ht+1 ≤ ht for all t ≥ 1.
(3) Classical Cayley-Bacharach Theorem: Let X = {P1 , . . . , P9 } be the
complete intersection of two cubics in P2 . Use the Cayley-Bacharach
Theorem to show that any cubic passing through 8 of the 9 points
of X must also pass through the remaining 9th point.
Part II - Exercises From Group Presentations
(1) From Croll-Gibbons-Johnson: Our exercise outlines a proof of the
following lemma due to Buchsbaum and Eisenbud:
Lemma C.1. Let R be a ring, x ∈ R, and S = R/(x). Let B be an
S-module, and let
F:
F2
φ2
/ F1
φ1
/ F0
98
BRIAN JOHNSON
be an exact sequence of S-modules with coker(φ1 ) ∼
= B. Suppose that
G:
ψ2
G2
/ G1
ψ1
/ G0
is a complex of R-modules such that
(i) x is a non-zero divisor on each Gi ,
(ii) Gi ⊗R S ∼
= Fi , and
(iii) ψi ⊗R S = φi .
Then A = coker(ψ1 ) is a lifting of B to R.
(a) With the conditions of the lemma and i ∈ {0, 1, 2}, prove that
the sequence
/ Gi
0
·x
q
/ Gi
/ Gi /xGi
/0
is exact, where ·x is the map given by multiplication by x and q
is the canonical quotient map.
(b) In the diagram below, show that each square of the diagram
commutes.
0
···
0
/ G2
/0
ψ2
·x
···
/0
/ G2
···
/0
/ F2
ψ2
φ2
0
/ G1
ψ1
/ G0
/0
/ ···
/ G0
/0
/ ···
/ F0
/0
/ ···
·x
/ G1
/ F1
ψ1
φ1
0
·x
0
0
Conclude that
0
/G
·x
/G
/F
/0
is an exact sequence of complexes (briefly explain why each column is exact).
(c) Given any exact sequence of complexes
0
/ D.
·x
/ D.
/ C.
/0,
POWER OF MONOMIAL IDEALS
SUSAN COOPER
99
there is a corresponding long exact sequence in homology given
by
···
/ H2 (C. )
v
·x
/ H1 (D. )
/ H1 (C. )
v
·x
/ H0 (D. )
/ H0 (C. )
H1 (D. )
H0 (D. )
/ H2 (D. )
/ 0.
Use the long exact sequence in homology with the exact sequence
of complexes to determine that A/xA ∼
= B and x is a non-zero
divisor on A. Conclude that A is a lifting of B to R.
(2) From Brase-Denkert-Janssen: Accept that any monomial ordering
> on k[x1 , . . . , xn ] can be obtained by taking pairwise orthogonal
vectors v1 , . . . , vr ∈ k n where v1 has only non-negative entries and
α = vi ·β
β
where xα > xβ if and only if there exists t ≤ r such that vi ·α
for all i ≤ t − 1 and vt · α > vt · β .
(a) Let r = n and vi = ei for all i where ei is the ith standard basis
vector for k n . Show that > is the lexicographic order.
(b) Let r = n and define vectors as follows:
v1 = (1, . . . , 1)
vi = (1, 1, . . . , 1, i − (n + 1), 0, 0, . . . , 0)
where the entry i − (n + 1) is in the (n + 2 − i)th position for
i ∈ {2, . . . , n}. Show that > is the graded reverse-lexicographic
order.
100
BRIAN JOHNSON
Appendix D. Problem Set 4
Due: Thursday, April 22
This problem set involves choices! Submit solutions to 2 exercises from Part
I and 1 exercise from Part II.
Part I - Exercises Related to Borel-Fixed and Generic Initial
Ideals
The following exercises are taken from the Chapter 2 Exercises of “Combinatorial Commutative Algebra” by E. Miller and B. Sturmfels.
(1) [Exercise 2.2] Can you find a general formula for the number B(r, d)
of Borel-fixed ideals generated by r monomials of degree d in three
unknowns {x1 , x2 , x3 }?
(2) [Exercise 2.4] Is the class of Borel-fixed ideals closed under the idealtheoretic operations of taking intersections, sums, and products?
Either prove your claims or give counter-examples.
(3) [Modified Exercise 2.11] Let I = (x1 x2 , x2 x3 , x1 x3 ) be an ideal of
S = k[x1 , x2 , x3 ]. Compute the generic initial ideal gin< (I) for the
lexicographic and reverse lexicographic monomial orders. Also, compute the lex-segment ideal L ⊆ S with H(S/I) = H(S/L). (Note:
Although you can use a computer algebra program to support your
solution, you should avoid finding the generic initial ideals by using
the pre-defined function.)
Part II - Exercises From Group Presentations
(1) From Boeckner-Stolee: Recall the definition of a perfect graph is a
graph for which every induced subgraph, we have the chromatic
number equal to the clique number.
It is well known that the Petersen graph, described as follows and
shown below, is not perfect.
The Petersen graph is the graph on 10 vertices, given by subsets
of size 2 from a set of 5 elements. The edges are formed if the two
vertices (as subsets) are disjoint.
POWER OF MONOMIAL IDEALS
SUSAN COOPER
101
{1, 2} u
ZZ
Z
Z
{3, 5} u
C
Z
Z
Z
Z
Z
5}
ZZ{4,
u
H
LL H
H
5}
C
HH{2,
L
u
u
C
L
Q
{1, 3}
C
Q
L
Q C L
Q
Q C
L
Q
L
Q C
C
Q
L
QCu{1, 4}
{2, 4} u
L
A
L
A
L
A
L
A L A {2, 3}
Au
{1, 5} Lu
{3, 4} u
H
Z
C
C
C
(a) Show that the chromatic number of the Petersen graph is 3, but
the clique number is 2.
(b) Find an odd hole.
(c) Let J := I(G)∨ , where G is the Petersen graph. Give an associated prime of height > 3 in Ass(J 2 ).
(2) From DeVries-Yu: Let Kn,d be the complete bipartite graph on n
and d vertices (i.e. let L be a set of n vertices and R a set of d vertices
with L ∩ R = ∅. Then the vertex set of Kn,d is L ∪ R, and the edge
set of Kn,d is the set of all pairs with one element from L and one
element from R). Let I(Kn,d ) denote the edge ideal of Kn,d . Write
a recursive formula for βi,j (I(Kn,d )) in terms of the Betti numbers
of I(Km,d ) for m < n. Use your formula to compute β1,j (I(Kn,d ))
for all j.

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