Koszul cycles and Golod rings

0.15 MB·

by  Jürgen Herzog

KOSZUL CYCLES AND GOLOD RINGS
JU¨RGEN HERZOGAND RASOULAHANGARIMALEKI
7
1
0
Abstract. Let S be the power series ring or the polynomial ring over a field K in
2
the variables x1,...,xn, and let R = S/I, where I is proper ideal which we assume to
n be graded if S is the polynomial ring. We give an explicit description of the cycles of
a the Koszul complex whose homology classes generate the Koszul homology of R= S/I
J
with respect to x1,...,xn. The description is given in terms of the data of the free
4 S-resolution of R. The result is used to determine classes of Golod ideals, among them
2 proper ordinary powers and proper symbolic powers of monomial ideals. Our theory is

also applied to stretched local rings.
]
C
A
.
h Introduction
t
a
Let S be the power series ring or the polynomial ring over a field K in the variables
m
x ,...,x , and let I ⊂ S be a proper ideal of S, which we assume to be a graded ideal, if
[ 1 n
S is the polynomial ring. Consider a finitely generated R = S/I-module M. The formal

1 power series ring PR(t) = dim TorR(K,M)ti is called the Poincar´e series of M.
v M i≥0 K i
8 Since M is also an S-moduleP, we may as well consider the Poincar´e series PMS(t), similarly
3 defined. Note that PS(t) is a polynomial, since S is regular. By a result of Serre [11],
M
7
there is a coefficientwise inequality
6
0 (1+t)n
1. PKR(t) ≤ 1−t(PS(t)−1).
R
0
7 The ring R (or I itself) is called Golod, if equality holds. In this case the Poincar´e series
1 PR(t) is a rational function, which in general for the Poincar´e series of the residue field
: K
v K is not always the case, see [2]. Interestingly, over Golod rings we have rationality not
Xi only for PKR(t) but also for Poincar´e series of all finitely generated R-modules.
In Section 1 of this note we give a canonical and explicit description of the cycles of the
r
a Koszul complex whose homology classes generate the Koszul homology of R with respect
to x ,...,x , see Theorem 1.3. The description is given in terms of the data of the free
1 n
S-resolution of R, and allows us to identify interesting classes of Golod ideals.
The same strategy, namely to give a nice description of Koszul cycles, has been applied
by the first author and Huneke [8] to show, among other results, that in the polynomial
case, the powers Ik of I are Golod for all k ≥ 2, provided the characteristic of K is zero.
In that result, the annoying assumption that the characteristic of K should be zero, arises
from the fact that if the authors use the result from [7] which says that char(K)= 0, then
the desired Koszul cycles can be described in terms of Jacobians derived from the maps in
the graded minimal free S-resolution of R. To avoid this drawback, we choose a different
description of the Koszul cycles which can be given for any characteristic of the base field,
and only depends on the given order of the variables.
Apart from an explicit description of Koszul cycles, our approach to prove Golodness
for certain classes of ideals and rings is based on the Golod criterion [1, Proposition 1.3],
due to the second author of this paper. He showed that if I ⊂ J ⊂ S are ideals with
1
J2 ⊂ I and such that the natural maps TorS(K,S/I) → TorS(K,S/J) are zero for all
i i
i ≥ 1, then I is Golod. As an easy consequence of our description of the Koszul cycles it
is shown in Section 2 that TorS(K,S/I) → TorS(K,S/d(I)) is the zero map for all i ≥ 1,
i i
where d(I) is the ideal generated by the elements di(f ) for i = 1,...,n and f ,...,f a
j 1 m
system of generators of I. The operator di is defined by
di(f)= (f(0,...,0,x ,...,x )−f(0,...,0,x ,...,x ))/x forf ∈ n.
i n i+1 n i
In combination with the above Golod criterion we then obtain that I is a Golod ideal, if
d(I)2 ⊂ I. If this is the case we say that I is d-Golod. The operators di depend on the
order of the variables. If I happens to be d-Golod after a permutation σ of the variables,
then we say that I is d -Golod, and we call I strongly d-Golod, if it is d -Golod for any
σ σ
permutation σ.
As one of the main applications of this approach we obtains (that for a monomial ideal
I, all proper ordinary powers, saturated powers or symbolic powers of I are Golod. The
same holds true for the integral closures Ik of the powers Ik for k ≥ 2, see Theorem 3.1
and Proposition 3.2. The same results can be found in [8] for graded ideals, but under
the additional assumption that char(K) = 0. Here we have no assumptions on the the
characteristic but we prove these results only for monomial ideal. However, with our
methods, a new class of non-monomial Golod ideals in the formal power series ring is
detected, see Proposition 3.4.
As a last application of the techniques presented in this paper we have a result of more
general nature. In Theorem 3.5 it is shown that if R is a stretched local ring or a standard
graded stretched K-algebra, then R is Golod, if one of the following conditions is satisfied:
(i) R is standard graded, (ii) R is not Artinian, (iii) R is Artinian and the socle dimension
of R coincides with its embedding dimension.
The result that R is Golod, if (iii) is satisfied, has been shown in the recent paper
[4]. Our proof of this case can be deduced without any big efforts from the case that
R is standard graded. The latter case is accessible to our theory, since after a suitable
extension of the base field, the defining ideal of a standard graded stretched K-algebra R
turns out to be d -Golod for a suitable permutation of the variables.
σ
1. A description of the Koszul cycles
LetK beafield, andlet S stand for power series ringK[[x ,...,x ]] or thepolynomial
r r n
ring K[x ,...,x ] over K. For S we simply write S. Since S is naturally embedded into
r n 1 r
S we may view any element in S also as an element in S. We denote by n = (x ,...,x )
r 1 n
the maximal (resp. the graded) maximal ideal of S.
Let f ∈ n. Then
f = α xr1···xrn,
r1,...,rn 1 n
(r1,..X.,rn)∈Nn0
where the coefficients α belong to K.
r1,...,rn
For f ∈ n and r = 1,...,n, we set
f(0,...,0,x ,...,x )−f(0,...,0,x ,...,x )
(1) dr(f)= r n r+1 n .
x
r
Then the following rules hold:
(i) f = d1(f)x +···+dn(f)x , and
1 n
(ii) dr(f)∈ S for 1,...,n.
r
2
Note that the operators dr are uniquely determined by (i) and (ii).
The following example demonstrates this definition: let S = K[[x ,x ,x ,x ]] and
1 2 3 4
f = x2x +x x3+x2x3+x2x . Then d1(f)= x x +x3, d2(f)= x x3,d3(f)= x x and
1 3 1 2 2 3 3 4 1 3 2 2 3 3 4
d4(f)= 0.
Of course the definition of the dr(f) depend on the order of the variables. Like partial
derivatives, theoperators d :S → S areK-linear maps, andthereis aproductrulewhich
i i
is however less simple than that for partial derivatives. Indeed one has
Lemma 1.1. Let f,g ∈ n and r be an integer with 1 ≤ r ≤ n. Then
(i) dr is a K-linear map and so dr(f +g) = dr(f)+dr(g);
(ii) dr(fg) = dr(f)dr(g)x + (dr(f)di(g)+dr(g)di(f))x .
r i>r i
Proof. (i) is obvious. (ii) followsPeasily from (1) and the equations
n
(fg)(0,...,0,x ,...,x ) = di(f)dj(g)x x ,
s n i j
i,j=s
X
for s = r and s = r+1. (cid:3)
Let Ω be the free S-module of rank n with basis dx ,...,dx . We denote by Ω the
1 1 n
exterior algebra Ω of Ω . Note that Ω is a graded S-algebra with graded components
1 1
i
Ω = Ω . In particular, Ω = S and Ω is a free S-module with basis
i 1 0 i
V
dx ∧...∧dx , 1 ≤ r < r < ... < r ≤ n.
V r1 ri 1 2 i
Let ∂ :Ω → Ω be the S-linear map given by
i i i−1
i
∂ (dx ∧...∧dx ) = (−1)k+1x dx ∧...∧dx ∧dx ∧...∧dx
i r1 ri rk r1 rk−1 rk+1 i
k=1
X
To simplify our notation we shall write dx ...dx for dx ∧...∧dx .
r1 ri r1 ri
ThecomplexΩisnothingbuttheKoszulcomplexwithrespecttothesequencex ,...,x ,
1 n
and so is a minimal free S-resolution of the residue field K of S.
Now let I be a proper ideal of S and (F,δ) a minimal free S-resolution of R = S/I.
Then for each i = 0,...,n we have the following isomorphism
ψ : K ⊗F = H (K ⊗F)∼= H (Ω⊗F)∼= H (Ω⊗S/I).
i i i i i
Tracing through this isomorphism one obtains i-cycles in Ω⊗S/I whose homology classes
form a K-basis of H (Ω⊗S/I).
i
i
We recall that an element z = (z ...,z ) ∈ (Ω⊗F) = Ω ⊗F is a cycle in
0 i i j=0 j i−j
(Ω⊗F) if and only if
L
(2) (id⊗δ )(z ) = (−1)j(∂ ⊗id)(z ) for all j.
i−j j j+1 j+1
Now the isomorphism ψ can be describe as follows: let 1⊗f ∈ K ⊗F and choose a
i i
cycle z = (z ...,z ) ∈ (Ω⊗F) such that z maps to 1⊗f under canonical epimorphism
0 i i 0
Ω ⊗F → K ⊗F . Then
0 i i
ψ (1⊗f)= [z¯],
i i
where z¯ denotes the image of z in Ω ⊗S/I and [z¯] its homology class in H (Ω⊗S/I).
i i i i i
In order to make this description of ψ more explicit one has to choose suitable cycles
i
z = (z ...,z ) ∈ (Ω ⊗ F) . There are of course many choices for such cycles. In [7]
0 i i
3
the partial derivatives were used to describe these cycles. Here we replace the partial
derivatives by our dr-operators.
Let b be the rank of F . For each i we choose a basis e ,...,e , and let
i i i1 ibi
bi
(i)
δ (e ) = α e
i ij kj i−1k
k=1
X
for all i,j.
The following result is the crucial technical statement of this note.
Proposition 1.2. Consider the element (z ,...,z )∈ (Ω⊗F) with z = 1⊗e and
0 i i 0 ij
z =
i−k
bk bk+1 bi−1
··· drk+1(α(k+1) )...dri(α(i) )dx ...dx ⊗e
jkjk+1 ji−1j rk+1 ri kjk
jXk=1(cid:18)1≤rk+1X<...<ri≤njkX+1=1 jiX−1=1 (cid:19)
for all 0 ≤ k <i.
Then
(id⊗δ )(z ) = (∂ ⊗id)(z )
k+1 i−k−1 i−k i−k
for all k.
Proof. Since α(i) = n dr(α(i) )x it is obvious that (id⊗δ )(z ) = (∂ ⊗ id)(z ).
ji−1j r=1 ji−1j r i 0 1 1
Thus the assertion is true for i−k = 1. Now let i−k > 1. Then
P
bk−1
(id⊗δ )(z )= h ⊗e ,
k i−k jk−1 k−1jk−1
jkX−1=1
where
bk bi−1
h = ... α(k) drk+1(α(k+1) )···dri(α(i) )dx ···dx .
jk−1 jk−1jk jkjk+1 ji−1j rk+1 ri
1≤rk+1X<...<ri≤njXk=1 jiX−1=1
In order to prove the assertion it suffices to show that for all j = 1,...,b we have
k−1 k−1
(3) h = ∂ (g ),
jk−1 i−k+1 jk−1
where
bk bi−1
g = ... drk(α(k) )···dri(α(i) )dx ···dx .
jk−1 jk−1jk ji−1j rk ri
1≤rk<X...<ri≤njXk=1 jiX−1=1
By applying rule (i), we see that
h = a ,
jk−1 rk+1,...,ri
1≤rk+1X<...<ri≤n
where
n bk bi−1
a = ... dr(α(k) )drk+1(α(k+1) )···dri(α(i) )x dx ···dx .
rk+1,...,ri jk−1jk jkjk+1 ji−1j r rk+1 ri
Xr=1jXk=1 jiX−1=1
4
Since α(k)α(k+1) = 0, Lemma 1.1 implies that
bk
(4) drk+1(α(k) )drk+1(α(k+1) )x =
jk−1jk jkjk+1 rk+1
jXk=1
bk bk
− dr(α(k) )drk+1(α(k+1) )x − drk+1(α(k) )dr(α(k+1) )x .
jk−1jk jkjk+1 r jk−1jk jkjk+1 r
rk+X1<r≤njXk=1 rk+X1<r≤njXk=1
By using this identity, we obtain for a the expression
rk+1,...,ri
a =
rk+1,...,ri
bk bi−1
... dr(α(k) )drk+1(α(k+1) )···dri(α(i) )x dx ···dx
jk−1jk jkjk+1 ji−1j r rk+1 ri
1≤rX<rk+1jXk=1 jiX−1=1
n bk bi−1
− ... drk+1(α(k) )dr(α(k+1) )dkr+2(α(k+2) )···
jk−1jk jkjk+1 jk+1jk+2
rk+X1<r≤njXk=1 jiX−1=1
···dri(α(i) )x dx ···dx .
ji−1j r rk+1 ri
Next we use the analogue to formula (4) corresponding to the fact that α(k+1)α(k+2) = 0.
Then the sum in the bottom row of the previous expression for a can be rewritten
rk+1,...,ri
as
bk bi−1
... drk+1(α(k) )dr(α(k+1) )···dri(α(i) )x dx ···dx
jk−1jk jkjk+1 ji−1j r rk+1 ri
rk+1<Xr<rk+2jXk=1 jiX−1=1
n bk bi−1
− ... drk+1(α(k) )drk+2(α(k+1) )dr(α(k+2) )drk+3(α(k+3) )···
jk−1jk jkjk+1 jk+1jk+2 jk+2jk+3
rk+X2<r≤njXk=1 jiX−1=1
···dri(α(i) )x dx ···dx .
ji−1j r rk+1 ri
Proceeding this way, we obtain that
i
(5) h = (−1)l−kcl ,
jk−1 rk+1,...,ri
1≤rk+1X<...<ri≤nXl=k
where
bk bi
cl = ... drk+1(α(k) )···drl(α(l−1) )dr(α(l) )·
rk+1,...,ri jk−1,jk jl−2,jl−1 jl−1,jl
rl<Xr<rl+1jXk=1 jiX−1=1
drl+1(α(l+1) )···dri(α(i) )x dx ···dx .
jl,jl+1 ji−1,j r rk+1 i
Here, by definition, r = 1 and r = n.
k i+1
5
On the other hand,
bk bi−1 ′ ′
∂i−k+1(gjk−1) = ∂i−k+1( ... drk(α(jkk−)1jk)...dri(α(jii)−1j)dxrk′ ...dxri′)
1≤rk′<X...<ri′≤njXk=1 jiX−1=1
bk bi−1 s=i−k+1 ′
= ... (−1)s+1drk(α(k) )...
jk−1jk
1≤rk′<X...<ri′≤njXk=1 jiX−1=1 Xs=1
′
...dri(α(i) )x ′ dx ′ ...dx ′ dx ′ dx ′
ji−1j rk+1−s rk rk−s rk−s+2 ri
i−k+1
= (−1)s+1ck+1−s .
r′,...,r′ ,r′ ,...,r′
k k−s k−s+2 i
1≤rk′<X...<ri′≤n Xs=1
Comparing this with (5) the desired equality (3) follows. (cid:3)
As a consequence of Proposition 1.2 we obtain
Theorem 1.3. For i= 1,...,n and j = 1,...,b let
i
b1 bi−1
z = ... dr1(α(1) )...dri(α(i) )dx ...dx ,
ij j0j1 ji−1j r1 ri
1≤r1<X...<ri≤njX1=1 jiX−1=1
and denote its image in Ω ⊗S/I by z¯ . Then for all i and j, the element z¯ are cycles of
i ij ij
Ω⊗S/I, and the homology classes [z¯ ] with j = 1,...,b form a K-basis of H (Ω⊗S/I).
ij i i
Proof. The elements 1⊗e with 1 ≤ j ≤ b form a K-basis of K ⊗F . Proposition 1.2
ij i i
together with (2) implies that ψ (1⊗e ) = ±[z¯ ], and this yields the assertion. (cid:3)
i ij ij
2. A Golod criterion
Let S be as before and R = S/I with I ⊂ n. As before we assume that I is a graded
ideal, if S is the polynomial ring.
We will use the following Golod criterion given in [1] by the second author.
Theorem 2.1. Let I ⊂ J ⊂ S be ideals with J2 ⊂ I and such that the natural maps
TorS(K,S/I) → TorS(K,S/J)
i i
are zero for all i≥ 1. Then I is Golod.
Let I = (f ,...,f ). We define d(I) to be the ideal generated by the elements di(f )
1 m j
with i = 1,...,n and j = 1,...,m. Note that I ⊂ d(I) and that d(I) does not depend
on the particular choice of the generators, but of course on the order of the variables.
If x ,...,x is a relabeling the variables given by the permutation σ, then for i =
σ(1) σ(n)
1,...,n we set
diσ(f)= σ(di(f(xσ−1(1),...,xσ−1(n))),
and let d (I) be ideal generated by the elements di(f ).
σ σ j
The important conclusion that arises from our description of the Koszul cycles is the
following
6
Corollary 2.2. Let I be a proper ideal of S and σ be any permutation of the integers
1,...,n. Then the natural map
TorS(K,S/I) → TorS(K,S/d (I))
i i σ
induced by the surjection S/I → S/d (I) is zero for all i≥ 1.
σ
Proof. We first notice that for any S-module M the Koszul homology H (Ω;M) is func-
i
torially isomorphic to TorS(K,M). Thus it suffice to show that the map H (Ω;S/I) →
i i
H (Ω;S/d (I)) is the zero map for all i ≥ 1. It is enough to show this for the basis [z ]
i σ ij
elements of H (Ω;S/I) as given in Theorem 1.3. These cycles have coefficients in d (I)
i σ
and hence their image, already in Ω⊗S/d (I), is zero. (cid:3)
σ
Now combining Theorem 2.1 with Corollary 2.2 we obtain
Theorem 2.3. Let I ⊂ S be a proper ideal such that d (I)2 ⊂ I for some permutation σ.
σ
Then I is a Golod ideal.
We say that I is d -Golod, if I satisfies the condition of the theorem, and simply say
σ
that I d-Golod, if I is d -Golod for σ = id. Finally we say that I is strongly d-Golod, if I
σ
is d -Golod for all permutations σ of the set [n]= {1,2,...,n}.
σ
For monomial ideals d(I) can be easily computed. If u is a monomial and r is the
smallest integer such that x divides u, then dr(u) = u/x and di(u) = 0 for all i 6= r.
r r
Thus, for example, if I = (x x ,x2), then d(I) = (x ), and d (I) = (x ,x ) for σ the
1 2 2 2 σ 1 2
permutation σ with σ(1) = 2 and σ(2) = 1.
Obviously, one has the following implications:
I is strongly d-Golod ⇒ I is d -Golod ⇒ I is Golod.
σ
In the above example, I is d-Golod, but not d -Golod. In particular, d-Golod does not
σ
imply strongly d-Golod. Also, a Golod ideal is in general not d -Golod for any σ. For
σ
example, the ideal I =(x x ) is Golod, but not d -Golod.
1 2 σ
On the other hand, if char(K)= 0 and I is a monomial ideal, then I is strongly Golod
inthesenseof [8]if andonly ifitis strongly d-Golod. Thisfollows fromtheremarksatthe
begin of Section 3 in [8], where it is observed that a monomial ideal I is strongly Golod
if and only if for all monomial generators u,v ∈ I and all integers i and j with x |u and
i
x |v it follows that uv/x x ∈ I. Indeed, it is obvious that strongly Golod implies strongly
j i j
d-Golod. Conversely, let u,v ∈ I be monomials with x | u and x | v. If x | u or x | v,
i j j i
then clearly uv/x x ∈ I. Suppose now that x ∤ u and x ∤ v. Then i 6= j, and we may
i j j i
assume that i < j. Choose any permutation σ of [n] such that σ(1) = i and σ(2) = j.
Then d (u) = u/x and d (v) = v/x , and since I is d -Golod we get that uv/x x ∈ I.
σ i σ j σ i j
3. Applications
Let I and J be ideals of S. The ideal I: Jt is called the saturation of I with
t≥1
respect to J. For J = n, this saturation is denoted by I. The kth symbolic power of I,
S
denoted by I(k), is the saturation of Ik with respect to the ideal which is the intersection
of all associated, non-minimal prime ideals of Ik. e
As a first application we prove a result analogue to [8, Theorem 2.3], whose proof
follows very much the line of arguments given there. The new and important fact is that
no assumptions on the characteristic of the base field are made.
7
Theorem 3.1. Let I,J ⊂ S be ideals. Assume that σ is a permutation of the integers [n].
Then the following holds:
(a) If I and J are d -Golod, then I ∩J and IJ are d -Golod.
σ σ
(b) If I and J are d -Golod and d (I)d (J) ⊂ I +J, then I +J is d -Golod.
σ σ σ σ
(c) If I is a strongly d-Golod monomial ideal and J is an arbitrary monomial ideal
such that I :J = I : J2, then I : J is strongly d-Golod.
(d) If I is a monomial ideal, then Ik, I(k) and Ik are strongly d-Golod for all k ≥2.
(e) If I ⊂ J are monomial ideals and I is d -Golod, then IJ is d -Golod.
σ σ
e
Proof. For the proofs of (a) and (b) we may assume that σ = id. The proof for a general
permutation σ is the same.
(a) By assumption, d(I)2 ⊂ I and d(J)2 ⊂ J. Hence, since d(I ∩J) ⊂ d(I)∩d(J), it
follows that d(I ∩J)2 ⊂d(I)2 ∩d(J)2 ⊂ I ∩J, and this shows that I ∩J is d-Golod.
Now let f ∈ I and g ∈ J. Then Lemma 1.1(ii) implies that dr(fg) ∈ d(I)d(J).
Therefore, d(IJ) ⊂ d(I)d(J), and hence, d(IJ)2 ⊂ d(I)2d(J)2 ⊂ IJ, as desired.
(b) Since by assumption, d(I)2 ⊂ I, d(J)2 ⊂ J and d(I)d(J) ⊂ I +J, it follows that
d(I +J)2 = (d(I)+d(J))2 = d(I)2+d(I)d(J)+d(J)2 ⊂ I+J, which shows that I +J is
d-Golod.
(c) Let w ,w ∈ I :J be two monomials and i and j be integers with x |w and x |w .
1 2 i 1 j 2
We must show that w w /x x ∈I :J.
1 2 i j
Assume that u,v ∈ J are arbitrary. Therefore w u∈ I and w v ∈ I. Since I is strongly
1 2
d-Golod, it follows that (w w /x x )uv = (w u/x )(w v/x ) ∈ I. Hence, w w /x x ∈ I :
1 2 i j 1 i 2 j 1 2 i j
J2 = I : J.
(d) We first show that Ik is strongly d-Golod for all k ≥ 2. For this we have to show
that Ik is d -Golod for all k ≥ 2 and all permutations σ of [n]. We prove this for σ = id.
σ
The proof for general σ is the same. So now let w = w ···w be a monomial generator
1 k
of Ik with w ∈ I for j = 1,...k, and let r be the smallest integer which divides w. We
j
may assume that x divides w . Then di(w) = 0 for i 6= r and dr(w) ∈ Ik−1. It follows
r 1
that d(Ik)⊂ Ik−1, and hence d(Ik)2 ⊂ I2(k−1) ⊂ Ik.
The remaining statements of (d) now result from the following more general fact: let I
be a strongly d-Golod ideal and J an arbitrary monomial ideal. Then the saturation of I
with respect to J is strongly d-Golod.
For the proof of this we observe that, due to the fact that S is Noetherian, there
exists an integer t such that I : Jt = I : Js for s ≥ t . Thus if L = Jt0. Then
0 t≥0 0
I : Jt = I : L = I :L2, and the claim follows from (c).
t≥0 S
(e) Let u ,u ∈ I and v ,v ∈ J be monomials. Assume that i is the smallest integer
1 2 1 2
S
such that x |u v and j is the smallest integer such that x |u v . We need to show
σ(i) 1 1 σ(j) 2 2
that u v u v /x x ∈IJ.
1 1 2 2 σ(i) σ(j)
If x |u and x |u , then (u u /x x )v v ∈ IJ since I is d -Golod. If x |v
σ(i) 1 σ(j) 2 1 2 σ(i) σ(j) 1 2 σ σ(i) 1
and x |v , then u u (v v /x x ) ∈ IJ since I ⊂ J. If x |u and x (j)|v , then
σ(j) 2 1 2 1 2 σ(i) σ(j) σ(i) 1 σ 2
u /x (v u v /x ) ∈ IJ since I is d -Golod. The case x |v and x (j)|u is similar.
1 σ(i) 1 2 2 σ(j) σ σ(i) 1 σ 2
(cid:3)
Let I be a monomial ideal. We denote by I¯ the integral closure of I. The following
result and its proofare completely analogue to that of [8, Proposition 3.1], wherea similar
result is shown for strongly Golod monomial ideals.
Proposition 3.2. Let I be a monomial ideal which is d -Golod. Then I¯is d -Golod. In
σ σ
particular, I¯k is strongly d-Golod for all k ≥ 2.
8
A monomial ideal I of polynomial ring S = K[x ,...,x ] is called stable if for all
1 n
monomial u ∈ I one has x u/x for all i ≤ m(u), where m(u) is the largest integer j
i m(u)
such that x divides u.
j
We use Theorem 3.1 to reprove and generalize a result of Aramova and [3] who showed
that stable monomial ideals are Golod.
Corollary 3.3. Let I be a stable monomial ideal. Then IJ is a Golod ideal for any J
with I ⊂ J. In particular, I is Golod.
Proof. Let σ be the permutation reversing the order of the variables. Then d (u) =
σ
u/x . Hence for all monomials u,v ∈ I one has d (u)d (v) ∈ I, since I is stable. It
m(u) σ σ
followsthatI isd -Golod. ThereforethedesiredresultfollowsfromTheorem3.1(e). (cid:3)
σ
In the following proposition we present new family of d-Golod ideals which are not
necessarily monomial ideals.
Proposition 3.4. Let J ⊂ K[[x ,...,x ]] be an ideal for i = 1,...,n, and J ⊂ S =
i i n
K[|x ,...,x |] be the ideal generated by n x J . Then Jk is d-Golod for all k ≥ 2.
1 n i=1 i i
Proof. Note that for any k ≥ 2 we have P
Jk = xa1···xanJa1···Jan
1 n 1 n
a1+·X··+an=k
where the a are non-negative integers. By convention, xaiJai = S if a = 0. Thus we see
i i i i
that Jk is generated by elements of the form
xa1···xanf ···f
1 n 1 n
with integers a ≥ 0 such that a +...+a = k and with f ∈ Jai for i= 1,...,n,
i 1 n i i
One has
0, if there exists j < i with a > 0 or a = 0,
di(f)= j i .
(xiai−1···xannfi···fn, otherwise.
Hence, we see that di(f)dj(g) ∈ Jk for all f,g ∈ Jk and all i,j with 1≤ i,j ≤ n, and this
shows that Jk is d-Golod for all k ≥ 2. (cid:3)
Thelast application we have in mind is of more general nature and deals with stretched
local rings. Let (R,m,K) be a Noetherian local ring with maximal ideal m and residue
field K or a standard graded K-algebra with graded maximal ideal m. The ring R is said
to be stretched if m2 is a principal ideal.
We set n = dim m/m2 and τ = dim Soc(R), where Soc(R) = (0 : m) is the socle of
K K R
R. Moreover, if R is Artinian, we let s be the largest integer such that ms 6= 0. Note the
s+1 is the Loewy length of R.
Stretched local rings have been introduced by Sally [10]. She showed that the Poincar´e
series of K is a rational function. Indeed, she showed that
1/(1−nt), if τ = n
(6) PR(t) =
K (1/(1−nt+t2), if τ 6= n
Very recently in [4] it was shown that all finitely generated modules over a stretched
Artinian local ring R have a rational Poincar´e series with a common denominator by
studying the algebra structure of the Koszul homology of R. Among other results they
9
proved in [4, Theorem 5.4] that R is Golod, if τ = h. By using our methods we give an
alternative proof of the result and show
Theorem 3.5. Let (R,m,K) be a stretched local ring or a stretched standard graded K-
algebra. Then R is Golod if one of the following conditions is satisfied:
(i) R is standard graded, (ii) R is not Artinian, or (iii) R is Artinian and τ = n.
The following lemma will be needed for the proof of the theorem.
Lemma 3.6. Let R be a stretched standard graded K-algebra with n ≥ 2 and s ≥ 3 in
Artinian case. Then the following holds:
(i) τ = n, if R is Artinian.
(ii) τ = n−1, if R is not Artinian.
Proof. Since R is standard graded, R R = R 6= 0 for 2 ≤ i < s if R is Artinian, and
1 i i+1
R R = R 6= 0 for all i ≥ 2 if R is not Artinian. Since dim R = 1 for all i ≥ 2 with
1 i i+1 K i
R 6= 0,itfollowsthatR Soc(R) R ⊕R ifRisArtinian,andSoc(R) R ifRisnot
i s 1 s 1
Artinian. Thus in order to prove (i) and (ii) we must show that dimSoc(R)∩R = h−1.
1
After an extension of the base field we may assume that K is algebraically closed.
Indeed, a base field extnesion does not change the Hilbert function, not does it change the
socle dimension of the R.
We proceed by induction on n. We first assume that n= 2. Since dim R > dim R
K 1 K 2
and since K is algebraically closed, [9, Lemma 2.8] implies that there exists a non-zero
linear form x such that x2 = 0. Assume that x R 6= 0. Then R = x R , and R =
1 1 1 1 2 1 1 3
R R = x R2 = x2R = 0, contradicting the assumption that s ≥ 3. Thus x R = 0,
1 2 1 1 1 1 1 1
and hence x ∈ Soc(R), and the assertion follows for n = 2. Now let n > 2, and let R′
1 1
be a K-linear subspace of R such that R′ ⊕Kx = R . Then it follows that (R′)2 =
1 1 1 1 1
R2. Therefore the standard graded K-algebra R′ = K[R′] is a stretched K-algebra of
1 1
embedding dimension n−1. By induction hypothesis, dim Soc(R′)∩R′ = n−2. Let
K 1
x ,...,x span the K-vector space Soc(R′) ∩ R′. These elements x are also socle
2 n−1 1 i
elements of R and together with x , they span a vector space of dimension n − 1, as
1
desired. (cid:3)
Remark 3.7. Suppose that R = S/I is a stretched standard graded K-algebra, where
S = K[x ,...,x ] is the polynomial ring with K is an algebraically closed field, and where
1 n
I ⊂ n2. The proof of Lemma 3.6 shows that after a suitable linear change of coordinates
one has that
(i) I = (x ,...,x )2+x (x ,...,x )+(xs+1), if R is Artinian;
1 n−1 n 1 n−1 n
(ii) I = (x ,...,x )2+x (x ,...,x ), if R is not Artinian.
1 n−1 n 1 n−1
Proof of Theorem 3.5. Let us first assume that R is a standard graded K-algebra. Af-
ter a suitable base field extension, which does not affect the Golod property, we may
assume that I is generated as described in Remark 3.7. We order the variables as
follows: x ,x ,...,x , and let σ be the corresponding permutation of [n]. Then
n 1 n−1
d (I) = (x ,...,x ,xs) in the Artinian case, and and d (I) = (x ,...,x ) in the
σ 1 n−1 n σ 1 n−1
non-Artinian case. Clearly d (I)2 ⊂ I in both cases. Thus in any case R is a Golod ring.
σ
This proves case (i).
In order to prove Golodness of R in the case (ii), we consider the associated graded ring
G(R) of R, which, as can be seen from its Hilbert function, is a standard graded stretched
K-algebra. We claim that G(R) is a Koszul algebra. Koszulness of a standard graded
K-algebra is characterized by the property that PR(t) = 1/H (−t), where H (t) denotes
K R R
10