Notes 10: Monomial Ideals
Let f = xy 2 − x, f1 = xy + 1, f2 = y 2 − 1. When we divide f by f1 , f2 we obtain
xy 2 − x = y(xy + 1) + 0(y 2 − 1) + (−x − y).
When we divide f byf2 , f1 , we obtain
xy 2 − x = y(xy = 1) + 0.
From the second computation we conclude that f ∈ I =< f1 , f2 >, and from this we see
that the remainder from the first calculation, namely (−x − y), is in I also. We want to
make a guess as to why the first calculation failed, and from that guess see how to improve
the division algorithm. The difficulty is that neither of the leading terms of f1 , f2 divides
either of the terms in the remainder −x − y. Our plan is to adjoined to the list f1 , f2 more
terms in the ideal f1 , f2 so that we have sufficiently many leading terms. For example if
we adjoin the term −x − y (which is in I) to the list f1 , f2 , then we would not have run
into this difficulty.
Example 1. We use the lexicographic order with x > y in k[x, y]. Let I =< f1 = x3y 2 +
y, f2 = xy 3 − x > . The element g = yf1 − xf2 = x2 − xy is an element of I, but neither
of its leading terms is divisible by either of the leading terms of f1 or f2 . If we applied the
division algorithm to g we would get g = 0f1 + 0f2 + x2 − xy. The division algorithm fails
to detect that g ∈ I. Again the problem is the paucity of leading terms appearing in our
list of generators.
Our plan is to add more terms to the set of generators of a given ideal, so that we
have more leading terms. The first step in carrying out this program is to look at ’leading
terms’ more carefully.
Definition 1. An ideal I that can be written
X
I={
aα xα |α ∈ A, aα ∈ k}
for some A ⊆ Zn≥0 is a monomial ideal.
Example 2. Let I =< x2 > . Then I is a monomial ideal and A = {n ∈ |Z, n ≥ 2}.
Example 3. Let I =< xy 3 , x4 y > . We have the notation:
(a, b) + Z2≥0 = {(a, b) + (x, y)|(x, y) ∈ Z2 such that x ≥ 0, y ≥ 0}.
Let
A = (1, 3) + Z2≥0 ∪ (4, 1) + Z2≥0 .
Then I is the monomial ideal < xα , α ∈ A > .
It is often easier to use a figure to see exactly what the set A looks like.
P
Note. Let I be a monomial ideal. If f = α hα xα , h ∈ k is an element of I, then each of
the terms hα xα is also in I.
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Note. Let B ⊆ Zn≥0 . The ideal J =< xα , α ∈ B > is a monomial ideal. Indeed if we set
A=
[
α + Zn≥0 ,
α∈B
then J = {
P
α∈A
hα xα }.
Corollary 1. Two monomial ideals are the same iff they contain the same monomials.
Theorem 1. Dickson’s Lemma Let I =< xα , α ∈ A > be a monomial idea. Then there
exists a finite subset B ⊆ A so that I =< xα , α ∈ B > .
Proof. We prove the theorem by induction on the number of variables. If there is a single
variable x = x1 , then I =< xα , α ∈ A ⊂ Z≥0 > . Let β be the smallest element in this list.
Then for any α ∈ A, xα = xβ xγ for some γ ∈ Z≥0 . This says that I is generated by the
single element xβ .
We do the induction step. We assume that any monomial ideal in n − 1 variables
n−1
generated by a subset A ⊆ Z≥0
• is generated by a finite set of monomials, and
• we may take these monomials from the set A.
We write xn = y so k[x1 , · · · , xn−1 , y] = k[x1 , x2 , · · · , xn ]. We can write any monon−1
mial in the form xα y m for some α ∈ Z≥0
. Let J ⊆ k[x1 , x2 , · · · , xn−1 ] be the ideal of
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k[x1 , x2 , · · · , xn−1 ] generated by all the xα so that xα y m ∈ I for some m ≥ 0. Here is where
we use the induction hypothesis. By the induction hypothesis J is generated by a finite
∗
set of α ∈ Zn−1
≥0 . Call this set B . This is just one part of the induction hyothesis. The
second part tells us that we may take these xα so that there is a mα such that
xα y mα ∈ I.
Let m be the maximum among the integers mα so that xα xmα ∈ B ∗ . Now let B =
{x y |α ∈ B ∗ }.
For each k = 0, 1, · · · < m, let Jk denote the ideal of k[x1 , x2 , · · · xn−1 ] generated by xα so
that xα y k ∈ I. By induction this ideal is finitely generated by elements xαk (i) , i = 1, · · · , sk .
Let Bk = {xαk (i) y k }.
We claim that I is generated by
α m
B ∪ B0 ∪ B1 ∪ · · · ∪ Bm−1 .
We prove the claim: Let xβ y t ∈ I. If t ≥ m, the there is an α ∈ B ∗ so that xα divides
xβ , furthermore, since t ≥ m we have xα y mα divides xβ y t . One the other hand, suppose
that t < m. In this case there is an element of Bt that divides xβ y t . We have shown that
the ideal I has a finite set of generators. Call this set of generators C.
We show that we can find a finite set of generators in the set xα , α ∈ A. We have implicitly changed notation here. The notation xα now refers to a monomial in the complete
set of variables x1 , x2 , · · · , xn and not just the first n − 1 variables. For each element xα
in C there is some element xβ in A so that xβ divides xα . The set of such xβ generates I
and is a subset of xβ , β ∈ A.
The above proof has a puzzling point. Can we proceed as follows: Instead of taking
generators of the form xα xm , α ∈ B ∗ , m = max{mα }, why not take as generators xα y mα .
Does this allow us to avoid dealing with the ideals Jk ?
The answer is no. We look at an example. Let I =< xy 3 , x4 y > .
• The ideal J is generated by a single element x.
• This comes from the monomial xy 3 .
• The monomial xy 3 does not generate I.
• We have J0 = (0), J1 = x4 , J2 = x4 .
• Hence we add to the generator xy 3 coming from J, the generator x4 y.
Corollary 2. Consider an total order on the monomials in x1 , x2 , · · · , xn . Assume it
satisfies the property:
xα > xβ ⇐⇒ xα+γ > xβ+γ .
Then this is a well ordering if and only if xα > x0 for all non-zero α.
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Proof. We have already done the proof in the direction =⇒ . We use the theorem above
to prove the implication ⇐= .
Let A ⊆ Zn≥0 and let I be the monomial ideal generated by A. By the theorem above
the ideal has a finite set of generators, {xα1 , xα2 , · · · xαt }. Relabeling if necessary, we may
assume that xα1 < xα2 < · · · < xαt . It is easy to see that xα1 is the smallest element in the
set A..
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