Seminar Series in Mathematics: Algebra
2003, 1–16
MONOMIAL SUBRINGS OF COMPLETE
GRAPH ∗
Introduction
The complete graph, denoted Kn , has every pair of its n vertices adjacent. We
note that
µ
¶
n
(i) the number of the edges of Kn is
;
2
(ii) Kn is not bipartite;
(iii) α0 (Kn ) = n − 1, so ht(I(Kn )) = n − 1, by Corollary 6.1.18 of [V1].
Now, let Kn be the complete graph on the vertex set X = {x1 , . . . , xn } and let
R = K [x1 , . . . , xn ] be the polynomial ring over the field K. According to the
definition of Kn , the edge set is E(Kn ) = { {xi ; xj }| 1 ≤ i < j ≤ n}. Hence,
the monomial subring or edge subring of the graph Kn is the K-subalgebra
K [Kn ] = K [ {xi xj | 1 ≤ i < j ≤ n}] .
Of course, the complete graph is a special kind of graph on the vertex set
{x1 , . . . , xn } and we expect some special properties. Our main purpose is to
show that K [Kn ] is a CM ring. We also find the Hilbert series and the multiplicity e (K [Kn ]) by a presentation of K [Kn ].
The internal logic of our lecture is strongly related to the following facts.
(1) The first step is to identify a presentation of K [Kn ], that is a polynomial
ring An and an ideal Pn so that K [Kn ] ' An /Pn . It was proved (see [G], [M])
that An /in(Pn ) is CM implies An /Pn is CM, where in(Pn ) is the initial ideal
of Pn (in the lexicographic order). Thus, to show that K [Kn ] is the CM it is
∗ This lecture was held by Eduard Asadurian
University of Pitesti, Pitesti, Romania
e-mail: [email protected]
1
2
Monomial subrings of complete graph
sufficient to prove that An /in(Pn ) is CM .
(2) In the second step a graph will be built, denoted Gσn , so that in(Pn ) =
I (Gσn ).
(3) The third step is to show that I (Gσn ) is a CM ideal.
(4) The fourth step is to write the Hilbert series Fn (t) of the presentation
An /Pn of K [Kn ]. Here, we recall that An /Pn and An /in(Pn ) have the same
Hilbert function (see [V1], Corollary 2.4.13). After all, the Hilbert series Fn (t)
of An /Pn satisfies
µ
¶
µ
¶
µ
¶
n
n(n − 3)
n ¤
n
n
n
2
3
£
(1 − t) Fn (t) = 1 +
t+
t +
t +···+
t[ 2 ] .
n
4
6
2
2
2
Consequently, e (K [Kn ]) = 2n−1 − n.
1
A presentation of monomial subring K [Kn ]
We set fij , 1 ≤ i < j ≤ n, for the generators of K [Kn ]. Accordingly, let Vn
be the set of indeterminates
Vn = { Tij | (i, j) ∈ σn } ,
µ
where σn = {(i, j)|1 ≤ i < j ≤ n}. Obviously, |Vn | =
n
2
¶
. Consider the
homomorphism ψ : An = K[Vn ] → K [Kn ], induced by Tij 7→ fij . The ideal
Pn := Kerψ is the toric ideal or presentation ideal of K [Kn ].
Observe that if |{i, j, k, l}| = 4, then all primitive binomials [V1, Definition
8.1.3] Tij Tkl − Tik Tjl , Til Tjk − Tik Tjl etc. are in Pn . A question is if these
binomials really generate the toric ideal Pn .
Proposition 1 ([V2]). Let R = K [x1 , . . . , xn ] be a polynomial ring over the
field K and Pn be the presentation ideal of K [Kn ]. Set
B = {Tij Tkl − Til Tjk , Tik Tjl − Til Tjk |1 ≤ i < j < k < l ≤ n} .
If the terms in An = K [ {Tij | 1 ≤ i < j ≤ n}] are ordered lexicographically by
T12 < T13 < · · · < T1n < T23 < · · · < T2n < · · · < T(n−1)n ,
then
(i) B is a minimal generating set for Pn ;
(1)
Monomial subrings of complete graph
3
(i) B is a reduced Gröbner basis for Pn .
Proof. (i) It is a consequence of the property that says that the toric ideal of
the subring of k-products K[Vk ] ⊂ R, where
Vk = {xi1 . . . xik | 1 ≤ i1 < · · · < ik ≤ n} ,
is generated by homogenous binomials of degree two ([V1, Corollary 7.1.7] and
[FH]).
(ii) It suffices to verify that (f, g) →B 0 (the S-polynomial of f and g reduced
to zero with respect to B), for all f, g ∈ B. There are two possibilities for f
and g:
f = Tij Tkl − Til Tjk
or
f = Tik Tjl − Til Tjk , where i < j < k < l
and
g = Tpq Trs − Tps Tqr
or f = Tpr Tqs − Tps Tqr , where p < q < r < s.
First assume f = Tij Tkl − Til Tjk and g = Tpq Trs − Tps Tqr . Now, there are two
cases.
(ii.1) The leading terms of f and g have no common factors, that is,
(lt(f ), lt(g)) = 1. In this case S(f, g) →F 0, where F = {f, g} ([V1, Lemma
2.4.14]).
(ii.2) The leading terms of f and g have common factors.
(ii.2.1) Assume Tij = Tpq , that is , i = p and j = q. The S-polynomial of f
and g is
S(f, g) = Tkl Tps Tqr − Trs Tpl Tqk .
If {k, l} ∩ {r, s} 6= ∅, then S(f, g) →B 0. Indeed, in this assumption,

Tqr (Tlr Tps − Trs Tpl ) ,
if k = r,



Tps (Tqr Tsl − Trs Tql ) + Trs (Tps Tql − Tpl Tqs ) ,
if k = s,
S(f, g) =
if l = r,
 Tkr (Tps Tqr − Tpr Tqs ) − Tpr (Trs Tqk − Tqs Tkr ) ,


Tps (Tsk Tqr − Trs Tqk ) ,
if l = r.
If {k, l} ∩ {r, s} = ∅, then we distinguish again four cases. So,
- if k < r and l > s, then
S(f, g) = Tps (Tkl Tqr − Tql Tkr )+Tkr (Tps Tql − Tpl Tqs )−Tpl (Trs Tqk − Tkr Tqs ) ;
- if k > r and l > s, then
S(f, g) = Tps (Tkl Tqr − Tql Trk )+Trk (Tps Tql − Tpl Tqs )−Tpl (Trs Tqk − Trk Tqs ) ;
4
Monomial subrings of complete graph
- if k > r and l < s, then
S(f, g) = Tps (Tqr Tkl − Tql Trk )−Tpl (Tqk Trs − Tqs Trk )−Trk (Tps Tql − Tpl Tqs ) ;
- if k < r and l < s, then
S(f, g) = Tps (Tkl Tqr − Tkr Tql )−Tpl (Trs Tqk − Trk Tqs )−Tkr (Tpl Tqs − Tps Tql ) .
(ii.2.2) Assume Tij = Trs .Then i = r and j = s, hence p < q < r < s < k < l.
Here,
S(f, g) = Tkl Tps Tqr − Tpq Trl Tsk , respectively we write
S(f, g) = Tqr (Tps Tkl − Tpl Tsk ) + Tsk (Tqr Tpl − Tpq Trl ) .
(ii.2.3) The case Tkl = Tpq is symmetric with the previous one.
(ii.2.4) Finally, the last case is Tkl = Trs with the S-polynomial
S(f, g) = Tij Tps Tqr − Tpq Tis Tjr .
If i ≥ q, then
S(f, g) = Tij (Tps Tqr − Tqs Tpr )+Tpr (Tij Tqs − Tis Tqj )−Tis (Tpq Tjr − Tpr Tqj ) .
If i < q, then
S(f, g) = Tps (Tij Tqr − Tir Tqj )+Tqj (Tps Tir − Tis Tpr )−Tis (Tpq Tjr − Tpr Tqj ) .
In all the cases S(f, g) →B 0. ¤
Corollary 2. (a rule to describe B). If Pn is the toric ideal of the subring
K [Kn ] and n ≥ 4, then Pn is the ideal I2 (Y ) generated by the 2-minors of
symmetric array
•
T12
T13
Y = ...
T1(n−2)
T1(n−1)
T1n
T12
•
T23
..
.
T13
T23
•
..
.
...
...
...
T1(n−2)
T2(n−2)
T3(n−2)
..
.
T1(n−1)
T2(n−1)
T3(n−1)
..
.
T1n
T2n
T3n
..
.
T2(n−2)
T2(n−1)
T2n
T3(n−2)
T3(n−1)
T3n
...
...
...
•
T(n−2)(n−1)
T(n−2)n)
T(n−2)(n−1)
•
T(n−1)n
T(n−2)n
T(n−1)n
•
and the 2-minors of Y form a Gröbner basis for I2 (Y ) with respect to the lex
ordering (1).
Remark 3. If it is used a total ordering induced by a different order from the
one given by (1), then the corresponding reduced Gröbner basis for the toric
ideal Pn may be different from B. In [V1, Example 9.2.2] there is such an
example for the toric ideal of K[K5 ] with respect to the ordering
T45 < T34 < T35 < T23 < T24 < T25 < T12 < T13 < T14 < T15 .
Monomial subrings of complete graph
2
5
The initial ideal in(Pn ) seen like an edge ideal
A virtue of Proposition 1 is that it discovers a generating set for the ideal
in(Pn ). Thus, the initial ideal in(Pn ) is generated by the leading terms of the
generators of B, that is
in(Pn ) = ({Tij Tkl , Tik Tjl , | 1 ≤ i < j < k < l ≤ n}) .
Observe that
in(Pn ) = ({ Tij Tkl | i < k, j < l and {i, j} ∩ {k, l} = ∅}) .
Starting from the last equality we construct a suitable graph so that in(Pn )
is exactly an edge ideal. As above , let Vn = {Tij | (i, j) ∈ σn }. Now, let Gσn
be the graph on the vertex set Vn with the edge set
E = {{Tij Tkl }| (i, j), (k, l) ∈ σn , i < k, j < l and {i, j} ∩ {k, l} = ∅} .
If V ⊂ Vn , then we denote by GV the maximal subgraph of Gσn on the vertex
set V . Finally, let G be a graph on vertex set Vn . The edge ideal of G is
I(G) = ( {Tij Tkl | {Tij , Tkl } ∈ E(G)}) ≤ An ,
that is, the ideal of An generated by the square-free monomials Tij Tkl so that
Tij is adjacent to Tkl in G.
Of course, I(Gσn ) = in(Pn ).
3
The CM property of Gσn and ht (I(Gσn ))
We begin with the following lemma.
Lemma 4. Let n ≥ 4 be a fixed integer and r be a positive integer so that
1 ≤ r ≤ n − 1. For 1 ≤ k ≤ r − 1 set
Dr0 = {Tij ∈ Vn | r ≤ i ≤ n − 2, r + 1 ≤ j ≤ n − 1} ,
00
Dkr
= {Tir ∈ Vn | k ≤ i ≤ r − 1} ∪ {Tin ∈ Vn | k ≤ i ≤ r − 1}
and
00
Dkr = Dr0 ∪ Dkr
.
for all i < n, then I (GDkr ) is a CM
If I (Gσi ) is a CM ideal of height i(i−3)
2
ideal of height
µ
¶
n−r−1
+ r − k − 1,
2
for all 1 ≤ k ≤ r.
6
Monomial subrings of complete graph
Proof. First, we specify that
ª
©
Dr0 = Tr(r+1) , Tr(r+2) , . . . , T(n−2)(n−1) ,
ª ©
ª
©
00
Dkr
= Tkr , T(k+1)r , . . . , T(r−1)r ∪ Tkn , . . . , T(r−1)n ,
and there is a isomorphism Gσn−r → GDr0 given by the translation Tij 7→
T(i+r−1)(j+r−1) . Now, we proceed by descending induction on k ≤ r:
00
(i)¡ If k =¢ r, then Drr
= ∅ and thus GDrr ' Gσn−r . Therefore, I (GDrr ) ∼
=
I Gσn−r of height
µ
¶
(n − r)(n − r − 3)
n−r−1
=
+ r − r − 1.
2
2
(ii) Assume that the statement is true for k + 1 ≤ r and prove it for k.
By definition,
Dkr = D(k+1)r ∪ {Tkr , Tkn }
and Tkn is an isolated vertex of GDkr . Then,
¡
¢
E (GDkr ) = E GD(k+1)r ∪ { {Tij , Tkr }| Tij ∈ Dkr is adjacent to Tkr }
Let
N = {Tij ∈ Dkr | Tij is adjacent to Tkr } , hence
¡
¢
E (GDkr ) = E GD(k+1)r ∪ { {Tij , Tkr }| Tij ∈ N } .
N is a minimal vertex cover for GDkr with
µ
¶
n−r−1
|N | =
+ r − k − 1.
2
Let’s calculate ht (I (GDkr )). On the one hand, Tkn being an isolated vertex
of GDkr we have α0 (GDkr \ Tkn ) = α0 (GDkr ) .
On the other hand, N can not be contained in a minimal vertex cover for
GDkr \ Tkr because T(k+1)n ∈ N is an isolated vertex in GDkr \ Tkr . So, a
minimal vertex cover for GDkr \ Tkr contains < |N | vertices, hence
α0 (GDkr \ Tkr ) < α0 (GDkr ) .
Using [V1, Proposition 6.1.20] we obtain
But α0 (GDkr
α0 (GDkr ) = α0 (GDkr \ Tkr ) + 1.
¡
¢
\ Tkr ) = α0 GD(k+1)r implies
¡
¢
¡
¢
α0 GD(k+1)r = α0 GD(k+1)r + 1,
7
Monomial subrings of complete graph
and
¡
¢
ht (I (GDkr )) = ht GD(k+1)r + 1,
so
¡
¢
ht I(GD(k+1)r ) + 1
¶
·µ
¸
n−r−1
+ r − (k + 1) − 1 + 1
2
µ
¶
n−r−1
+ r − k − 1.
2
ht (I(GDkr )) =
=
=
By the definition of N and because N is a minimal vertex cover for GDkr , we
have in the ring An = K [{Tij |1 ≤ i < j ≤ n}] the following equalities:
¡ ¡
¢
¢
(I (GDkr ) , N ) = (N ) and (I (GDkr ) , Tkr ) = I GD(k+1)r , Tkr .
In order to show that I (GDkr ) is CM, consider the exact sequence
T
kr
0 → An / (I (GDkr ) , N ) →
An /I (GDkr ) → An / (I (GDkr ) , Tkr ) → 0,
(2)
in fact, the exact sequence
T
kr
0 → An / (N ) →
An /I (GDkr ) → An / (I (GDkr ) , Tkr ) → 0.
Here, An /(N ) is CM of dimension |Vn | − |N | since An /(N ) ' K[Vn \ N ] and
dim K[Vn \ N ] = |Vn \ N | = |Vn | − |N ].
Then, dim(An /(I(D(k+1)r ), Tkr )) = |Vn | − 1 − ht(I(GD(k+1)r ) = |Vn | − |N |.
Therefore, the ends of the exact sequence (2) are both CM rings of dimension
equal to dim(An /(I(GDkr ). Finally, we apply ”Depth Lemma” ([V1, Lemma
1.3.9]) for the exact sequence (2). Hence, An /I(GDkr ) is CM as required. ¤
Theorem 5. The edge ideal I(Gσn ) is CM of height equal to
n(n−3)
.
2
Proof. For n ≤ 4 the statement is verified directly. So, if n = 3 then we have
V3 = {T12 , T13 , T23 } and E(Gσ3 ) = ∅. Hence, I(Gσ3 ) = (0) is CM of height
equal to 3(3−3)
. If n = 4, then we have V4 = {T12 , T13 , T14 , T23 , T24 , T34 } and
2
E(Gσ4 ) = {{T12 , T34 } , {T13 T24 }}. The edge ideal is I(Gσ4 ) = (T12 T34 , T13 T24 ).
A minimal vertex cover for I(Gσ4 ) is {T12 , T13 } and α0 (Gσ4 ) = 2, therefore
. Let’s show that I(Gσ4 ) is CM. Observe that
ht(I(Gσ4 )) = 4(4−3)
2
An /I(Gσ4 ) '
k[x1 , . . . , x6 ]
k[x1 , x2 , x3 , x4 ]
'
[x5 , x6 ].
(x1 x2 , x3 x4 )
(x1 x2 , x3 x4 )
8
Monomial subrings of complete graph
k[x1 ,x2 ,x3 ,x4 ]
(x1 x2 ,x3 x4 ) is CM. Using [V1, Proposition 2.2.20] we
k[xi ,xj ]
k[x1 ,x2 ,x3 ,x4 ]
k[x1 ,x2 ]
3 ,x4 ]
have (x1 x2 ,x3 x4 ) ' (x1 x2 ) ⊗ k[x
(x3 x4 ) , and (xi xj ) (i 6= j) is CM of dimension
xj k[x ,x ]
k[x ,x ]
k[x ,x ]
equal to 1 (0 → (xi i ) j (−1) → (xiixj j) → (xij ) j → 0 is an exact sequence
³
´
k[x ,x ]
with the ends CM rings of dimensions equal to dim (xiixj j) ). For n ≥ 4 we
It suffices to show that
proceed by induction on n. We assume that the assertion is true for all i < n,
that is I(Gσi ) is CM and ht (I(Gσi )) = i(i−3)
2 . The idea is that starting from
the CM ideal I(Gσn−1 ), step by step, we can prove that
I(GC1 ), I(GC2 ), . . . , I(GCn−1 ) = I(Gσn )
are CM, where C1 = Vn−1 ∪ {T1n }, C2 = C1 ∪ {T2n }, . . . , Cn−1 = Cn−2 ∪
{T(n−1)n } = Vn . Actually set
Cr = Vn ∪ {T1n , . . . , Trn }, 1 ≤ r ≤ n − 1,
and claim that GCr is CM with ht (I(GCr )) = (r−1)+ (n−1)(n−4)
. We proceed
2
by induction on r. If r = 1, then C1 = Vn−1 ∪ {T1n } and T1n is an isolated
vertex of C1 . Hence I(GC1 ) = I(Gσn−1 ) and
ht (I(GC1 )) = (1 − 1) +
(n − 1)(n − 4)
.
2
Now, we assume GCi is CM with ht(I(GCi )) = (i − 1) +
i < r. Let
Ur = {Tij ∈ Cr |Tij is adjacent to Trn } .
(n−1)(n−4)
,
2
Observe that
Ur = Vr−1 ∪ {T1(r+1) , . . . , T1(n−1) , T2(r+1) , . . .
. . . , T2(n−1) , . . . , T(r−1)(r+1) , . . . , T(r−1)(n−1) }
and
|Ur | =
(r − 2)(r − 1)
+ (n − r − 1)(r − 1).
2
00
Let D1r be as in Lemma 4, that is, D1r = Dr0 ∪ D1r
, where
Dr0 = {Tij | r ≤ i ≤ n − 2, r + 1 ≤ j ≤ n − 1} ,
00
Dkr
= {Tir | 1 ≤ i ≤ r − 1} ∪ {Tin | 1 ≤ i ≤ r − 1} .
for all
Monomial subrings of complete graph
9
Put Lr = (I(GCr ), Ur ) . We have (I(GCr ), Ur ) = (I(GD1r ), Ur ). Using Lemma
4 it follows that Lr is CM and
ht(Lr ) =
=
=
ht((I(GD1r )) + |Ur |
¶
¶
µ
µ
n−r−1
r−1
+ (r − 2) + (r − 1)(n − r − 1) +
2
2
(r − 1) +
(n − 1)(n − 4)
.
2
¡
¢
Set Jr = (I(GCr ), Trn ). We note that (I(GCr ), Trn ) = I(GCr−1 ), Trn , so Jr
is CM with
¡
¢
(n − 1)(n − 4)
.
ht(Jr ) = ht I(GCr−1 ) + 1 = (r − 1) +
2
Therefore, we got that Lr and Jr are both CM of the same height; consequently, An /Jr and
More,
¡ An /Lr are ¢CM rings
¡ with the same
¢ dimension.
¡
¢
observe that Lr = I(GCr−1 ), Ur and ht I(GCr−1 ), Ur = ht I(GCr−1 ) + 1.
The last equation implies that
¡ Ur can¢not be a minimal vertex cover for GCr−1
and thus ht (I(GCr )) = ht I(GCr−1 ) + 1. The exact sequence
T
rn
0 → (An /Lr )(−1) →
An /I(GCr ) → An /Jr → 0
has its ends both CM rings of dimension equal to dimAn /I(GCr ). Hence,
An /I(GCr ) is CM. In particular, I(GCn−1 ) = I(Gσn ) is CM of height equal to
n(n−3)
. ¤
2
Corollary 6. The monomial subring K[Kn ] is CM.
Proof. Let An /Pn be the presentation of K[Kn ]. Because I(Gσn ) is CM and
I(Gσn ) = in(Pn ), we obtain that An /in(Pn ) is CM. Therefore An /Pn is CM,
as required. ¤
4
Hilbert series
First, we identify the Hilbert series of the ring An /I(Gσn ). Let
Fn (t) =
hn (t)
, hn (t) ∈ Z[t], hn (1) 6= 0,
(1 − t)d
be the Hilbert series of the ring An /I(Gσn ), where d =dim(An /I(Gσn )). Because
dim (An /I(Gσn )) = dimAn − ht (I(Gσn )) ,
10
Monomial subrings of complete graph
µ
by [V1, Corollary 2.1.7], we obtain that d =
Fn (t) =
n
2
¶
−
n(n−3)
2
= n. Therefore,
hn (t)
, hn (t) ∈ Z[t], hn (1) 6= 0.
(1 − t)n
Proposition 7 ([V1],[V2]). Let Fn (t) = hn (t)/(1−t)−n be the Hilbert series
of the ring An /I(Gσn ). Then the polynomial hn (t) satisfies the difference
equation
hn (t) = 2hn−1 (t) + (t − 1)hn−2 (t) + (n − 2)t2
for n ≥ 5 with boundary condition h3 (t) = 1 and h4 (t) = 1 + 2t + t2 .
Proof. For n = 3, recall that V (Gσ3 ) = V3 = {T12 , T13 , T23 }, E(Gσ3 ) = ∅.
1
Thus, I(Gσ3 ) = (0) and A3 /I(Gσ3 ) ' K[T12 , T13 , T12 ]. Hence, F3 (t) = (1−t)
3
([V1, Exercise 4.1.18]). For n = 4, V (Gσ4 ) = V4 = {T12 , T13 , T14 , T23 , T24 , T34 },
E(Gσ4 ) = {{T12 , T34 } , {T13 T24 }} and
A4 /I(Gσ4 ) = V4 = K [T12 , T13 , T14 , T23 , T24 , T34 ] / (T12 T34 , T13 T24 ) .
Note that f1 = T12 T34 , f2 = T13 T24 is a homogenous regular sequence in A4
with f1 , f2 , of degree 2.. Using [V1, Exercise 4.1.21] we obtain
F (A4 /I(Gσ4 ), t) =
(1 − t2 )2
(1 + t)2
=
,
6
(1 − t)
(1 − t)4
hence h4 (t) = 1 + 2t + t2 .
For n ≥ 4 we proceed by induction on n. Set
Cr = Vn−1 ∪ {T1n , T2n , . . . , Trn } .
00
,
Let D1r and Ur be as in Lemma 4 and Theorem 5, that is D1r = Dr0 ∪ D1r
where
Dr0 = {Tij | r ≤ i ≤ n − 2, r + 1 ≤ j ≤ n − 1} ,
©
ª ©
ª
00
D1r
= T1r , T2r , . . . , T(r−1)r ∪ T1n , . . . , T(r−1)n
and
Ur = {Tij ∈ Cr |Tij is adjacent to Trn } .
Once again observe that Cn−1 = Vn , so I(GCn−1 ) = I(Gσn ). We assert that
the sequences
0 → (An /Ln−1 ) (−1)
T(n−1)n
→
An /I(GCn−1 ) → An /Jn−1 → 0
(3)
Monomial subrings of complete graph
and
T
rn
0 → (An /Lr ) (−1) →
An /Jr+1 → An /Jr → 0
11
(4)
are exact, where
¡
¢
Lr = I(GD1r ), Ur , T(r+1)n , . . . , T(n−1)n
and
¡
¢
Jr+1 = I(GCr ), T(r+1)n , . . . , T(n−1)n .
Indeed,
¡
¢ ¡
¢
Ln−1 = I(GCn−1 ), Un−1 = I(GCn−1 ), Un−1 ,
¡
¢ ¡
¢
Jn−1 = I(GCn−2 ), T(n−1)n = I(GCn−1 ), T(n−1)n ,
and the exact sequence (3) is exactly the same exact sequence in the proof
of Theorem 5 for r = n − 1. Using (3) and the sequences (4) recursively, for
r = n − 2, n − 1, . . . , 2 yields
Fn (t) = t [F (An /Ln−1 , t) + · · · + F (An /L2 , t)] + F (An /J2 , t),
(5)
where
¡
¢ ¡
¢
J2 = I(GC1 ), T2n , . . . , T(n−1)n = I(Gσn−1 ), T2n , . . . , T(n−1)n .
Observe that
An
J2
=
=
K[Vn ]
¢
I(Gσn−1 ), T2n , . . . , T(n−1)n
µ
¶
µ
¶
K[Vn−1 ]
An−1
[T1n ] =
[T1n ],
I(Gσn−1 )
I(Gσn−1 )
¡
hence
¡
¢
(1 − t)n F (An /J2 , t) = (1 − t)n F An−1 /I(Gσn−1 ), t = hn−1 (t)
By induction on r we obtain
½
(r − 2)t + 1,
if r ∈ {n − 2, n − 1},
(1 − t)n F (An /Lr , t) =
. (6)
(r − 1)t + hn−r (t), if
2≤r ≤n−2
¡
¢
Indeed, if r = 2, then L2 = I(GD12 ), U2 , T3n , . . . , T(n−1)n , where
D12 = {Tij | 2 ≤ i ≤ n − 2, 3 ≤ j ≤ n − 1} ,
and
©
ª
U2 = T13 , T14 , . . . , T1(n−1) .
12
Monomial subrings of complete graph
Therefore,
¢
¡
L2 = I(GD12 ), T13 , T14 , . . . , T1(n−1) , T3n , . . . , T(n−1)n .
We have
µ
dim(An /L2 ) =
n
2
¶
− [ht (I(GD12 )) + 2(n − 3)] = n,
hence
F (An /L2 , t) =
h(t)
.
(1 − t)n
Recall that Gσn−2 ' GD20 and consequently I(GD20 ) ' I(Gσn−2 ). Note that
¢
¡
¢
¡
I(GD12 ) = I GD20 ∪{T12 ,T1n } = I GD20 ∪{T12 } = (I (D20 ) , {T12 Tij |Tij ∈ D30 }) .
Finally,
¢
¡
L2 = I(D20 ), {T12 Tij |Tij ∈ D30 } , T13 , . . . , T1(n−1) , T3n , . . . , T(n−1)n .
Now, we have
An
L2
=
'
=
where
©
ª¤
£
K D20 ∪ T12 , T13 , . . . , T1n , T2n , T3n , . . . , T(n−1)n
¡
¢
I(D20 ) ∪ {T12 , Tij , |Tij ∈ D30 } , T13 . . . , T1(n−1) , T3n , . . . , T(n−1)n
K [D20 ∪ {T12 , T1n , T2n }]
¢
I(GD20 ), {T12 , Tij , |Tij ∈ D30 }
Ã
!
K [D20 ∪ {T12 }]
¡
¢ [T12 , T2n ] ,
I(GD20 ), {T12 , Tij , |Tij ∈ D30 }
¡
K[D20 ]
I(GD0 )
2
0→
'
An−2
I(Gσn−2 ) .
The exact sequence
K[D20 ∪ {T12 }
K[D20 ∪ {T12 }]
K[D20 ∪ {T12 }]
T12
(−1)
→
→
→0
(D30 )
(I(GD20 ), {T12 Tij |Tij ∈ D30 })
(I(GD20 , T12 ))
is equivalent to the exact sequence
0
T
→
12
K[T23 , T24 , . . . , T2(n−1) , T12 ](−1) →
→
An−2
→ 0.
I(Gσn−2 )
K[D20 ∪ {T12 }]
(I(GD20 ), {T12 Tij |Tij ∈ D30 })
13
Monomial subrings of complete graph
Thus, we obtain
·
¸
¡
¢
1
1
Fn (An /L2 , t) =
t
+ F An−2 /I(Gσn−2 ), t ,
(1 − t)2 (1 − t)n−2
respectively,
Fn (An /L2 , t) =
t + hn−2 (t)
.
(1 − t)n
Using (5) and (6) we obtain
" n−1
#
n−1
X
X
hn (t) = t
(1 + (k − 2)t) +
((n − k)t + hk−1 (t)) + hn−1 (t)
k=n−2
k=4
and the correspondent
" n−2
#
n−2
X
X
hn−1 (t) = t
(1 + (k − 2)t) +
((n − k − 1)t + hk−1 (t)) + hn−2 .
k=n−3
k=4
Subtract the last two equalities and obtain
hn (t) = 2hn−1 (t) + (t − 1)hn−2 (t) + (n − 2)t2 ,
as required. ¤
Theorem 8 ([V2]). . Let Kn be the complete graph on n vertices and An /Pn
be the presentation of K[Kn ]. If n ≥ 3, then the Hilbert series Fn (t) of An /Pn
satisfies
¶
[n/2] µ
X
n(n − 3)
n
n
(1 − t) Fn (t) = 1 +
tm .
t+
2m
2
m=2
Proof. Recall that An /Pn and An /in(Pn ) = An /I(Gσn ) have the same
Hilbert series ([V1, Corollary 2.4.13]). Let
(1 − t)n Fn (t) = hn (t) = a0n + a1n t + · · · + aθ(n)n tθ(n) .
Writing hp (t) =
½
amn =
θ(m)
P
m=0
amp and using Proposition 7 we obtain
2am(n−1) + a(m−1)(n−1) − am(n−2) ,
2am(n−1) + a(m−1)(n−2) − am(n−2) + (n − 2),
Let
Gm (x) =
∞
X
n=3
amn xn
if n ≥ 5, m 6∈ {0, 2}
.
if n ≥ 5, m = 2
14
Monomial subrings of complete graph
be the generating function of the sequence (amn )n≥3 . Recall that h3 (t) = 1,
∞
P
h4 (t) = 1 + 2t + t2 and by Proposition 7 we have G0 (x) =
xn . Hence,
n=3
a0n = 1. For m ≥ 3 we obtain
Gm (x)
∞
X
=
amn xn
n=5
∞
X
=
2
n
am(n−1) x +
n=5
∞
X
n
a(m−1)(n−2) x −
n=5
∞
X
am(n−2) xn
n=5
2xGm (x) + x2 Gm−1 (x) − x2 Gm (x)
=
and consequently
Gm (x) =
x2
Gm−1 (x), for m ≥ 3.
(1 − x)2
Also, we obtain
∞
x4 (2 − x) X n(n − 3) n
n(n − 3)
=
x , hence a1n =
3
(1 − x)
2
2
n=3
G1 (x) =
4
x
and G2 (x) = (1−x)
5.
Finally, we can write
Gm (x) =
¶
∞ µ
X
xm
n
=
xm , for n ≥ 2.
2m
(1 − x)2m+1
n=2m
Therefore,
µ
amn =
n
2m
¶
for m ≥ 2 and n ≥ 3. ¤
Corollary 9. The multiplicity e (K[Kn ]) of the ringK[Kn ] is equal to 2n−1−n.
Proof. Indeed,
µ
¶ µ
¶
n(n − 3)
n
n
e (K[Kn ]) = 1 +
+
+
+ ...
4
6
2
·µ
¶
¸ µ
¶ µ
¶
n
n
n
= 1+
−n +
+
+ ...
2
4
6
=
2n−1 − n.¤
Monomial subrings of complete graph
15
References
[BH] W.Bruns, J.Herzog, Cohen-Macanlay rings, Cambridge Studies in Advanced Mathematics 39, Cambridge University Press, 1993.
[G] H.G.Gräbe, Streckungsringe, Dissertation B, Erfurt/Mühlhausen, 1998.
[M] M.Lejeune-Jalabert, Effectivité de calculs polynomiaux, Course de D.E.A., Institute
Fourier, Université de Grenoble I, 1985.
[Ma] H.Matsumura, Commutative ring theory, Cambridge Studies in Advanced Mathematics 8, Cambridge University Press, 1986.
[V1] R.Villarreal, Monomials algebras, Monographs and Textbooks in Pure and Applied
Mathematics 238, M.Dekker, Ink., New York, 2001.
[V2] R.Villarreal, Normality of subrings generated by square-free monomials, J. Pure &
Applied Algebra 113(1996), 91-106.
[FH] H.Flaschka and L.Haine, Torus orbits on G/P , Pacific J.Math. 149(1991), 251-292.
[SSV] A.Simis, W.V.Vasconcelos and R.Villarreal, On the ideal theory of graphs, J.Algebra,
167(1994), 389-416.
16
Monomial subrings of complete graph