Koszul cycles and Golod rings

TL;DR

This paper provides an explicit description of Koszul cycles for quotients of polynomial or power series rings, enabling the identification of Golod ideals and applications to various classes of rings.

Contribution

It introduces a new explicit description of Koszul cycles in terms of free resolutions, advancing the understanding of Golod rings and their ideals.

Findings

Explicit description of Koszul cycles in terms of free resolutions

Identification of classes of Golod ideals including powers of monomial ideals

Application to stretched local rings

Abstract

Let $S$ be the power series ring or the polynomial ring over a field $K$ in the variables $x_1,\ldots,x_n$, and let $R=S/I$, where $I$ is proper ideal which we assume to be graded if $S$ is the polynomial ring. We give an explicit description of the cycles of the Koszul complex whose homology classes generate the Koszul homology of $R=S/I$ with respect to $x_1,\ldots,x_n$. The description is given in terms of the data of the free $S$-resolution of $R$. The result is used to determine classes of Golod ideals, among them proper ordinary powers and proper symbolic powers of monomial ideals. Our theory is also applied to stretched local rings.

Equations

$$P_K^R(t)\le \frac{(1+t{)}^n}{1-t(P_R^S(t)-1)}.$$

$$d^i(f)=(f(0,\dots ,0,x_i,\dots ,x_n)-f(0,\dots ,0,x_{i+1},\dots ,x_n))/x_i\text{for}f\in \mathfrak{n}.$$

$$f=\sum _{(r_1,\dots ,r_n)\in {\mathbb{N}}_0^n}{\alpha }_{r_1,\dots ,r_n}{x}_1^{r_1}\cdots x_n^{r_n},$$

$$d^r(f)=\frac{f(0,\dots ,0,x_r,\dots ,x_n)-f(0,\dots ,0,x_{r+1},\dots ,x_n)}{x_r}.$$

$$(fg)(0,\dots ,0,x_s,\dots ,x_n)=\sum _{i,j=s}^n{d}^i(f)d^j(g)x_i{x}_j,$$

$$d{x}_{r_1}\wedge \dots \wedge d{x}_{r_i},1\le r_1<r_2<\dots <r_i\le n.$$

$${\partial }_i(d{x}_{r_1}\wedge \dots \wedge d{x}_{r_i})=\sum _{k=1}^i(-1{)}^{k+1}{x}_{r_k}d{x}_{r_1}\wedge \dots \wedge d{x}_{r_{k-1}}\wedge d{x}_{r_{k+1}}\wedge \dots \wedge d{x}_i$$

$${\psi }_i:K\otimes F_i=H_i(K\otimes \mathbb{F})\cong H_i(\Omega \otimes \mathbb{F})\cong H_i(\Omega \otimes S/I).$$

$$(\mathrm{id}\otimes {\delta }_{i-j})(z_j)=(-1{)}^j({\partial }_{j+1}\otimes \mathrm{id})(z_{j+1})\text{for all}j.$$

$${\psi }_i(1\otimes f)=[\overline{z_i}],$$

$${\delta }_i(e_{ij})=\sum _{k=1}^{b_i}{\alpha }_{kj}^{(i)}{e}_{i-1k}$$

$$z_{i-k}=$$

$$\displaystyle\sum_{j_{k}=1}^{b_{k}}\bigg{(}\sum_{1\leq r_{k+1}<\ldots<r_{i}\leq n}\sum_{j_{k+1}=1}^{b_{k+1}}\cdots\sum_{j_{i-1}=1}^{b_{i-1}}d^{r_{k+1}}(\alpha^{(k+1)}_{j_{k}j_{k+1}})\ldots d^{r_{i}}(\alpha^{(i)}_{j_{i-1}j})dx_{r_{k+1}}\ldots dx_{r_{i}}\bigg{)}\otimes e_{kj_{k}}$$

$$(\mathrm{id}\otimes {\delta }_{k+1})(z_{i-k-1})=({\partial }_{i-k}\otimes \mathrm{id})(z_{i-k})$$

$$(\mathrm{id}\otimes {\delta }_k)(z_{i-k})=\sum _{j_{k-1}=1}^{b_{k-1}}{h}_{j_{k-1}}\otimes e_{k-1j_{k-1}},$$

$$h_{j_{k-1}}=\sum _{1\le r_{k+1}<\dots <r_i\le n}\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}{\alpha }_{j_{k-1}{j}_k}^{(k)}{d}^{r_{k+1}}({\alpha }_{j_k{j}_{k+1}}^{(k+1)})\cdots d^{r_i}({\alpha }_{j_{i-1}j}^{(i)})d{x}_{r_{k+1}}\cdots d{x}_{r_i}.$$

$$h_{j_{k-1}}={\partial }_{i-k+1}(g_{j_{k-1}}),$$

$$g_{j_{k-1}}=\sum _{1\le r_k<\dots <r_i\le n}\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}{d}^{r_k}({\alpha }_{j_{k-1}{j}_k}^{(k)})\cdots d^{r_i}({\alpha }_{j_{i-1}j}^{(i)})d{x}_{r_k}\cdots d{x}_{r_i}.$$

$$h_{j_{k-1}}=\sum _{1\le r_{k+1}<\dots <r_i\le n}{a}_{r_{k+1},\dots ,r_i},$$

$$a_{r_{k+1},\dots ,r_i}=\sum _{r=1}^n\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}{d}^r({\alpha }_{j_{k-1}{j}_k}^{(k)})d^{r_{k+1}}({\alpha }_{j_k{j}_{k+1}}^{(k+1)})\cdots d^{r_i}({\alpha }_{j_{i-1}j}^{(i)})x_{r}d{x}_{r_{k+1}}\cdots d{x}_{r_i}.$$

$$\sum _{j_k=1}^{b_k}{d}^{r_{k+1}}({\alpha }_{j_{k-1}{j}_k}^{(k)})d^{r_{k+1}}({\alpha }_{j_k{j}_{k+1}}^{(k+1)})x_{r_{k+1}}=$$

$$-\sum _{r_{k+1}<r\le n}\sum _{j_k=1}^{b_k}{d}^r({\alpha }_{j_{k-1}{j}_k}^{(k)})d^{r_{k+1}}({\alpha }_{j_k{j}_{k+1}}^{(k+1)})x_r-\sum _{r_{k+1}<r\le n}\sum _{j_k=1}^{b_k}{d}^{r_{k+1}}({\alpha }_{j_{k-1}{j}_k}^{(k)})d^r({\alpha }_{j_k{j}_{k+1}}^{(k+1)})x_r.$$

$$a_{r_{k+1},\dots ,r_i}=$$

$$\sum _{1\le r<r_{k+1}}\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}{d}^r({\alpha }_{j_{k-1}{j}_k}^{(k)})d^{r_{k+1}}({\alpha }_{j_k{j}_{k+1}}^{(k+1)})\cdots d^{r_i}({\alpha }_{j_{i-1}j}^{(i)})x_{r}d{x}_{r_{k+1}}\cdots d{x}_{r_i}$$

$$-\sum _{r_{k+1}<r\le n}^n\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}{d}^{r_{k+1}}({\alpha }_{j_{k-1}{j}_k}^{(k)})d^r({\alpha }_{j_k{j}_{k+1}}^{(k+1)})d^{k_{r+2}}({\alpha }_{j_{k+1}{j}_{k+2}}^{(k+2)})\cdots$$

$$\cdots d^{r_i}({\alpha }_{j_{i-1}j}^{(i)})x_{r}d{x}_{r_{k+1}}\cdots d{x}_{r_i}.$$

$$\sum _{r_{k+1}<r<r_{k+2}}\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}{d}^{r_{k+1}}({\alpha }_{j_{k-1}{j}_k}^{(k)})d^r({\alpha }_{j_k{j}_{k+1}}^{(k+1)})\cdots d^{r_i}({\alpha }_{j_{i-1}j}^{(i)})x_{r}d{x}_{r_{k+1}}\cdots d{x}_{r_i}$$

$$-\sum _{r_{k+2}<r\le n}^n\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}{d}^{r_{k+1}}({\alpha }_{j_{k-1}{j}_k}^{(k)})d^{r_{k+2}}({\alpha }_{j_k{j}_{k+1}}^{(k+1)})d^r({\alpha }_{j_{k+1}{j}_{k+2}}^{(k+2)})d^{r_{k+3}}({\alpha }_{j_{k+2}{j}_{k+3}}^{(k+3)})\cdots$$

$$\cdots d^{r_i}({\alpha }_{j_{i-1}j}^{(i)})x_{r}d{x}_{r_{k+1}}\cdots d{x}_{r_i}.$$

$$h_{j_{k-1}}=\sum _{1\le r_{k+1}<\dots <r_i\le n}\sum _{l=k}^i(-1{)}^{l-k}{c}_{r_{k+1},\dots ,r_i}^l,$$

$$c_{r_{k+1},\dots ,r_i}^l=\sum _{r_l<r<r_{l+1}}\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_i}{d}^{r_{k+1}}({\alpha }_{j_{k-1},j_k}^{(k)})\cdots d^{r_l}({\alpha }_{j_{l-2},j_{l-1}}^{(l-1)})d^r({\alpha }_{j_{l-1},j_l}^{(l)})\cdot$$

$$d^{r_{l+1}}({\alpha }_{j_l,j_{l+1}}^{(l+1)})\cdots d^{r_i}({\alpha }_{j_{i-1},j}^{(i)})x_{r}d{x}_{r_{k+1}}\cdots d{x}_i.$$

$${\partial }_{i-k+1}(g_{j_{k-1}})={\partial }_{i-k+1}(\sum _{1\le r_k^{{}^{\prime }}<\dots <r_i^{{}^{\prime }}\le n}\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}{d}^{r_k^{{}^{\prime }}}({\alpha }_{j_{k-1}{j}_k}^{(k)})\dots d^{r_i^{{}^{\prime }}}({\alpha }_{j_{i-1}j}^{(i)})d{x}_{r_k^{{}^{\prime }}}\dots d{x}_{r_i^{{}^{\prime }}})$$

$$=\sum _{1\le r_k^{{}^{\prime }}<\dots <r_i^{{}^{\prime }}\le n}\sum _{j_k=1}^{b_k}\dots \sum _{j_{i-1}=1}^{b_{i-1}}\sum _{s=1}^{s=i-k+1}(-1{)}^{s+1}{d}^{r_k^{{}^{\prime }}}({\alpha }_{j_{k-1}{j}_k}^{(k)})\dots$$

$$\dots d^{r_i^{{}^{\prime }}}({\alpha }_{j_{i-1}j}^{(i)})x_{r_{k+1-s}^{{}^{\prime }}}d{x}_{r_k^{{}^{\prime }}}\dots d{x}_{r_{k-s}^{{}^{\prime }}}d{x}_{r_{k-s+2}^{{}^{\prime }}}d{x}_{r_i^{{}^{\prime }}}$$

Full text

Introduction

Let $S$ be the power series ring or the polynomial ring over a field $K$ in the variables $x_{1},\ldots,x_{n}$, and let $I\subset S$ be a proper ideal of $S$, which we assume to be a graded ideal, if $S$ is the polynomial ring. Consider a finitely generated $R=S/I$-module $M$. The formal power series ring $P^{R}_{M}(t)=\sum_{i\geq 0}\dim_{K}\operatorname{Tor}_{i}^{R}(K,M)t^{i}$ is called the Poincaré series of $M$. Since $M$ is also an $S$-module, we may as well consider the Poincaré series $P^{S}_{M}(t)$, similarly defined. Note that $P^{S}_{M}(t)$ is a polynomial, since $S$ is regular. By a result of Serre [11], there is a coefficientwise inequality

[TABLE]

The ring $R$ (or $I$ itself) is called Golod, if equality holds. In this case the Poincaré series $P^{R}_{K}(t)$ is a rational function, which in general for the Poincaré series of the residue field $K$ is not always the case, see [2]. Interestingly, over Golod rings we have rationality not only for $P^{R}_{K}(t)$ but also for Poincaré 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 Koszul complex whose homology classes generate the Koszul homology of $R$ with respect to $x_{1},\ldots,x_{n}$, see Theorem 1.3. The description is given in terms of the data of the free $S$-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 $I^{k}$ of $I$ are Golod for all $k\geq 2$, 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 $\operatorname{char}(K)=0$, 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 $I\subset J\subset S$ are ideals with $J^{2}\subset I$ and such that the natural maps $\operatorname{Tor}_{i}^{S}(K,S/I)\rightarrow\operatorname{Tor}_{i}^{S}(K,S/J)$ are zero for all $i\geq 1$, then $I$ is Golod. As an easy consequence of our description of the Koszul cycles it is shown in Section 2 that $\operatorname{Tor}_{i}^{S}(K,S/I)\rightarrow\operatorname{Tor}_{i}^{S}(K,S/d(I))$ is the zero map for all $i\geq 1$, where $d(I)$ is the ideal generated by the elements $d^{i}(f_{j})$ for $i=1,\ldots,n$ and $f_{1},\ldots,f_{m}$ a system of generators of $I$. The operator $d^{i}$ is defined by

[TABLE]

In combination with the above Golod criterion we then obtain that $I$ is a Golod ideal, if $d(I)^{2}\subset I$. If this is the case we say that $I$ is $d$-Golod. The operators $d^{i}$ depend on the order of the variables. If $I$ happens to be $d$-Golod after a permutation $\sigma$ of the variables, then we say that $I$ is $d_{\sigma}$-Golod, and we call $I$ strongly $d$-Golod, if it is $d_{\sigma}$-Golod for any permutation $\sigma$.

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 $\overline{I^{k}}$ of the powers $I^{k}$ for $k\geq 2$, 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 $\operatorname{char}(K)=0$. 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 $d_{\sigma}$-Golod for a suitable permutation of the variables.

1. A description of the Koszul cycles

Let $K$ be a field, and let $S_{r}$ stand for power series ring $K[[x_{r},\ldots,x_{n}]]$ or the polynomial ring $K[x_{r},\ldots,x_{n}]$ over $K$. For $S_{1}$ we simply write $S$. Since $S_{r}$ is naturally embedded into $S$ we may view any element in $S_{r}$ also as an element in $S$. We denote by ${\mathfrak{n}}=(x_{1},\ldots,x_{n})$ the maximal (resp. the graded) maximal ideal of $S$.

Let $f\in{\mathfrak{n}}$. Then

[TABLE]

where the coefficients $\alpha_{r_{1},\ldots,r_{n}}$ belong to $K$.

For $f\in{\mathfrak{n}}$ and $r=1,\ldots,n$, we set

[TABLE]

Then the following rules hold:

1. (i)

$f=d^{1}(f)x_{1}+\cdots+d^{n}(f)x_{n}$, and 2. (ii)

$d^{r}(f)\in S_{r}$ for $1,\ldots,n$.

Note that the operators $d^{r}$ are uniquely determined by (i) and (ii).

The following example demonstrates this definition: let $S=K[[x_{1},x_{2},x_{3},x_{4}]]$ and $f=x_{1}^{2}x_{3}+x_{1}x_{2}^{3}+x_{2}^{2}x_{3}^{3}+x_{3}^{2}x_{4}$. Then $d^{1}(f)=x_{1}x_{3}+x_{2}^{3},$ $d^{2}(f)=x_{2}x_{3}^{3},d^{3}(f)=x_{3}x_{4}$ and $d^{4}(f)=0$.

Of course the definition of the $d^{r}(f)$ depend on the order of the variables. Like partial derivatives, the operators $d_{i}:S\rightarrow S_{i}$ are $K$-linear maps, and there is a product rule which is however less simple than that for partial derivatives. Indeed one has

Lemma 1.1**.**

Let $f,g\in{\mathfrak{n}}$ and $r$ be an integer with $1\leq r\leq n$. Then

1. (i)

$d^{r}$* is a $K$-linear map and so $d^{r}(f+g)=d^{r}(f)+d^{r}(g)$;* 2. (ii)

$d^{r}(fg)=d^{r}(f)d^{r}(g)x_{r}+\sum_{i>r}(d^{r}(f)d^{i}(g)+d^{r}(g)d^{i}(f))x_{i}$.

Proof.

(i) is obvious. (ii) follows easily from (1) and the equations

[TABLE]

for $s=r$ and $s=r+1$.

Let $\Omega_{1}$ be the free $S$-module of rank $n$ with basis $dx_{1},\ldots,dx_{n}$. We denote by $\Omega$ the exterior algebra $\bigwedge\Omega_{1}$ of $\Omega_{1}$. Note that $\Omega$ is a graded $S$-algebra with graded components $\Omega_{i}=\bigwedge^{i}\Omega_{1}$. In particular, $\Omega_{0}=S$ and $\Omega_{i}$ is a free $S$-module with basis

[TABLE]

Let $\partial_{i}:\Omega_{i}\rightarrow\Omega_{i-1}$ be the $S$-linear map given by

[TABLE]

To simplify our notation we shall write $dx_{r_{1}}\ldots dx_{r_{i}}$ for $dx_{r_{1}}\wedge\ldots\wedge dx_{r_{i}}$.

The complex $\Omega$ is nothing but the Koszul complex with respect to the sequence $x_{1},\ldots,x_{n}$, 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 $({\mathbb{F}},\delta)$ a minimal free $S$-resolution of $R=S/I$. Then for each $i=0,\ldots,n$ we have the following isomorphism

[TABLE]

Tracing through this isomorphism one obtains $i$-cycles in $\Omega\otimes S/I$ whose homology classes form a $K$-basis of $H_{i}(\Omega\otimes S/I)$.

We recall that an element $z=(z_{0}\ldots,z_{i})\in(\Omega\otimes{\mathbb{F}})_{i}=\bigoplus_{j=0}^{i}\Omega_{j}\otimes F_{i-j}$ is a cycle in $(\Omega\otimes{\mathbb{F}})$ if and only if

[TABLE]

Now the isomorphism $\psi_{i}$ can be describe as follows: let $1\otimes f\in K\otimes F_{i}$ and choose a cycle $z=(z_{0}\ldots,z_{i})\in(\Omega\otimes{\mathbb{F}})_{i}$ such that $z_{0}$ maps to $1\otimes f$ under canonical epimorphism $\Omega_{0}\otimes F_{i}\rightarrow K\otimes F_{i}$. Then

[TABLE]

where $\bar{z_{i}}$ denotes the image of $z_{i}$ in $\Omega_{i}\otimes S/I$ and $[\bar{z_{i}}]$ its homology class in $H_{i}(\Omega\otimes S/I)$.

In order to make this description of $\psi_{i}$ more explicit one has to choose suitable cycles $z=(z_{0}\ldots,z_{i})\in(\Omega\otimes{\mathbb{F}})_{i}$. There are of course many choices for such cycles. In [7] the partial derivatives were used to describe these cycles. Here we replace the partial derivatives by our $d^{r}$-operators.

Let $b_{i}$ be the rank of $F_{i}$. For each $i$ we choose a basis $e_{i1},\ldots,e_{ib_{i}}$, and let

[TABLE]

for all $i,j$.

The following result is the crucial technical statement of this note.

Proposition 1.2**.**

Consider the element $(z_{0},\ldots,z_{i})\in(\Omega\otimes{\mathbb{F}})_{i}$ with $z_{0}=1\otimes e_{ij}$ and

[TABLE]

for all $0\leq k<i$.

Then

[TABLE]

for all $k$.

Proof.

Since $\alpha^{(i)}_{j_{i-1}j}=\sum_{r=1}^{n}d^{r}(\alpha^{(i)}_{j_{i-1}j})x_{r}$ it is obvious that $(\operatorname{id}\otimes\delta_{i})(z_{0})=(\partial_{1}\otimes\operatorname{id})(z_{1})$. Thus the assertion is true for $i-k=1$. Now let $i-k>1$. Then

[TABLE]

where

[TABLE]

In order to prove the assertion it suffices to show that for all $j_{k-1}=1,\ldots,b_{k-1}$ we have

[TABLE]

where

[TABLE]

By applying rule (i), we see that

[TABLE]

where

[TABLE]

Since $\alpha^{(k)}\alpha^{(k+1)}=0$, Lemma 1.1 implies that

[TABLE]

By using this identity, we obtain for $a_{r_{k+1},\ldots,r_{i}}$ the expression

[TABLE]

Next we use the analogue to formula (4) corresponding to the fact that $\alpha^{(k+1)}\alpha^{(k+2)}=0$. Then the sum in the bottom row of the previous expression for $a_{r_{k+1},\ldots,r_{i}}$ can be rewritten as

[TABLE]

Proceeding this way, we obtain that

[TABLE]

where

[TABLE]

Here, by definition, $r_{k}=1$ and $r_{i+1}=n$.

On the other hand,

[TABLE]

Comparing this with (5) the desired equality (3) follows.

As a consequence of Proposition 1.2 we obtain

Theorem 1.3**.**

For $i=1,\ldots,n$ and $j=1,\ldots,b_{i}$ let

[TABLE]

and denote its image in $\Omega_{i}\otimes S/I$ by $\bar{z}_{ij}$. Then for all $i$ and $j$, the element $\bar{z}_{ij}$ are cycles of $\Omega\otimes S/I$, and the homology classes $[\bar{z}_{ij}]$ with $j=1,\ldots,b_{i}$ form a $K$-basis of $H_{i}(\Omega\otimes S/I)$.

Proof.

The elements $1\otimes e_{ij}$ with $1\leq j\leq b_{i}$ form a $K$-basis of $K\otimes F_{i}$. Proposition 1.2 together with (2) implies that $\psi_{i}(1\otimes e_{ij})=\pm[\bar{z}_{ij}]$, and this yields the assertion.

2. A Golod criterion

Let $S$ be as before and $R=S/I$ with $I\subset{\mathfrak{n}}$. 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 $I\subset J\subset S$ be ideals with $J^{2}\subset I$ and such that the natural maps

[TABLE]

are zero for all $i\geq 1$. Then $I$ is Golod.

Let $I=(f_{1},\ldots,f_{m})$. We define $d(I)$ to be the ideal generated by the elements $d^{i}(f_{j})$ with $i=1,\ldots,n$ and $j=1,\ldots,m$. Note that $I\subset d(I)$ and that $d(I)$ does not depend on the particular choice of the generators, but of course on the order of the variables. If $x_{\sigma(1)},\ldots,x_{\sigma(n)}$ is a relabeling the variables given by the permutation $\sigma$, then for $i=1,\ldots,n$ we set

[TABLE]

and let $d_{\sigma}(I)$ be ideal generated by the elements $d_{\sigma}^{i}(f_{j})$.

The important conclusion that arises from our description of the Koszul cycles is the following

Corollary 2.2**.**

Let $I$ be a proper ideal of $S$ and $\sigma$ be any permutation of the integers $1,\ldots,n$. Then the natural map

[TABLE]

induced by the surjection $S/I\rightarrow S/d_{\sigma}(I)$ is zero for all $i\geq 1$.

Proof.

We first notice that for any $S$-module $M$ the Koszul homology $H_{i}(\Omega;M)$ is functorially isomorphic to $\operatorname{Tor}^{S}_{i}(K,M)$. Thus it suffice to show that the map $H_{i}(\Omega;S/I)\rightarrow H_{i}(\Omega;S/d_{\sigma}(I))$ is the zero map for all $i\geq 1$. It is enough to show this for the basis $[z_{ij}]$ elements of $H_{i}(\Omega;S/I)$ as given in Theorem 1.3. These cycles have coefficients in $d_{\sigma}(I)$ and hence their image, already in $\Omega\otimes S/d_{\sigma}(I)$, is zero.

Now combining Theorem 2.1 with Corollary 2.2 we obtain

Theorem 2.3**.**

Let $I\subset S$ be a proper ideal such that $d_{\sigma}(I)^{2}\subset I$ for some permutation $\sigma$. Then $I$ is a Golod ideal.

We say that $I$ is $d_{\sigma}$-Golod, if $I$ satisfies the condition of the theorem, and simply say that $I$ $d$-Golod, if $I$ is $d_{\sigma}$-Golod for $\sigma=\operatorname{id}$. Finally we say that $I$ is strongly $d$-Golod, if $I$ is $d_{\sigma}$-Golod for all permutations $\sigma$ of the set $[n]=\{1,2,\ldots,n\}$.

For monomial ideals $d(I)$ can be easily computed. If $u$ is a monomial and $r$ is the smallest integer such that $x_{r}$ divides $u$, then $d^{r}(u)=u/x_{r}$ and $d^{i}(u)=0$ for all $i\neq r$. Thus, for example, if $I=(x_{1}x_{2},x_{2}^{2})$, then $d(I)=(x_{2})$, and $d_{\sigma}(I)=(x_{1},x_{2})$ for $\sigma$ the permutation $\sigma$ with $\sigma(1)=2$ and $\sigma(2)=1$.

Obviously, one has the following implications:

[TABLE]

In the above example, $I$ is $d$-Golod, but not $d_{\sigma}$-Golod. In particular, $d$-Golod does not imply strongly $d$-Golod. Also, a Golod ideal is in general not $d_{\sigma}$-Golod for any $\sigma$. For example, the ideal $I=(x_{1}x_{2})$ is Golod, but not $d_{\sigma}$-Golod.

On the other hand, if $\operatorname{char}(K)=0$ and $I$ is a monomial ideal, then $I$ is strongly Golod in the sense of [8] if and only if it is strongly $d$-Golod. This follows from the remarks at the 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 $u,v\in I$ and all integers $i$ and $j$ with $x_{i}|u$ and $x_{j}|v$ it follows that $uv/x_{i}x_{j}\in I$. Indeed, it is obvious that strongly Golod implies strongly $d$-Golod. Conversely, let $u,v\in I$ be monomials with $x_{i}\mid u$ and $x_{j}\mid v$. If $x_{j}\mid u$ or $x_{i}\mid v$, then clearly $uv/x_{i}x_{j}\in I$. Suppose now that $x_{j}\nmid u$ and $x_{i}\nmid v$. Then $i\neq j$, and we may assume that $i<j$. Choose any permutation $\sigma$ of $[n]$ such that $\sigma(1)=i$ and $\sigma(2)=j$. Then $d_{\sigma}(u)=u/x_{i}$ and $d_{\sigma}(v)=v/x_{j}$, and since $I$ is $d_{\sigma}$-Golod we get that $uv/x_{i}x_{j}\in I$.

3. Applications

Let $I$ and $J$ be ideals of $S$. The ideal $\bigcup_{t\geq 1}I\colon J^{t}$ is called the saturation of $I$ with respect to $J$. For $J={\mathfrak{n}}$, this saturation is denoted by $\widetilde{I}$. The $k$th symbolic power of $I$, denoted by $I^{(k)}$, is the saturation of $I^{k}$ with respect to the ideal which is the intersection of all associated, non-minimal prime ideals of $I^{k}$.

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.

Theorem 3.1**.**

Let $I,J\subset S$ be ideals. Assume that $\sigma$ is a permutation of the integers $[n]$. Then the following holds:

1. (a)

If $I$ and $J$ are $d_{\sigma}$-Golod, then $I\cap J$ and $IJ$ are $d_{\sigma}$-Golod. 2. (b)

If $I$ and $J$ are $d_{\sigma}$-Golod and $d_{\sigma}(I)d_{\sigma}(J)\subset I+J$, then $I+J$ is $d_{\sigma}$-Golod. 3. (c)

If $I$ is a strongly $d$-Golod monomial ideal and $J$ is an arbitrary monomial ideal such that $I:J=I:J^{2}$, then $I:J$ is strongly $d$-Golod. 4. (d)

If $I$ is a monomial ideal, then $I^{k}$, $I^{(k)}$ and $\widetilde{I^{k}}$ are strongly $d$-Golod for all $k\geq 2$. 5. (e)

If $I\subset J$ are monomial ideals and $I$ is $d_{\sigma}$-Golod, then $IJ$ is $d_{\sigma}$-Golod.

Proof.

For the proofs of (a) and (b) we may assume that $\sigma=\operatorname{id}$. The proof for a general permutation $\sigma$ is the same.

(a) By assumption, $d(I)^{2}\subset I$ and $d(J)^{2}\subset J$. Hence, since $d(I\cap J)\subset d(I)\cap d(J)$, it follows that $d(I\cap J)^{2}\subset d(I)^{2}\cap d(J)^{2}\subset I\cap J$, and this shows that $I\cap J$ is $d$-Golod.

Now let $f\in I$ and $g\in J$. Then Lemma 1.1(ii) implies that $d^{r}(fg)\in d(I)d(J)$. Therefore, $d(IJ)\subset d(I)d(J)$, and hence, $d(IJ)^{2}\subset d(I)^{2}d(J)^{2}\subset IJ$, as desired.

(b) Since by assumption, $d(I)^{2}\subset I$, $d(J)^{2}\subset J$ and $d(I)d(J)\subset I+J$, it follows that $d(I+J)^{2}=(d(I)+d(J))^{2}=d(I)^{2}+d(I)d(J)+d(J)^{2}\subset I+J$, which shows that $I+J$ is $d$-Golod.

(c) Let $w_{1},w_{2}\in I:J$ be two monomials and $i$ and $j$ be integers with $x_{i}|w_{1}$ and $x_{j}|w_{2}$. We must show that $w_{1}w_{2}/x_{i}x_{j}\in I:J$.

Assume that $u,v\in J$ are arbitrary. Therefore $w_{1}u\in I$ and $w_{2}v\in I$. Since $I$ is strongly $d$-Golod, it follows that $(w_{1}w_{2}/x_{i}x_{j})uv=(w_{1}u/x_{i})(w_{2}v/x_{j})\in I$. Hence, $w_{1}w_{2}/x_{i}x_{j}\in I:J^{2}=I:J$.

(d) We first show that $I^{k}$ is strongly $d$-Golod for all $k\geq 2$. For this we have to show that $I^{k}$ is $d_{\sigma}$-Golod for all $k\geq 2$ and all permutations $\sigma$ of $[n]$. We prove this for $\sigma=\operatorname{id}$. The proof for general $\sigma$ is the same. So now let $w=w_{1}\cdots w_{k}$ be a monomial generator of $I^{k}$ with $w_{j}\in I$ for $j=1,\ldots k$, and let $r$ be the smallest integer which divides $w$. We may assume that $x_{r}$ divides $w_{1}$. Then $d^{i}(w)=0$ for $i\neq r$ and $d^{r}(w)\in I^{k-1}$. It follows that $d(I^{k})\subset I^{k-1}$, and hence $d(I^{k})^{2}\subset I^{2(k-1)}\subset I^{k}$.

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 $t_{0}$ such that $\bigcup_{t\geq 0}I:J^{t}=I:J^{s}$ for $s\geq t_{0}$. Thus if $L=J^{t_{0}}$. Then $\bigcup_{t\geq 0}I:J^{t}=I:L=I:L^{2}$, and the claim follows from (c).

(e) Let $u_{1},u_{2}\in I$ and $v_{1},v_{2}\in J$ be monomials. Assume that $i$ is the smallest integer such that $x_{\sigma(i)}\mid u_{1}v_{1}$ and $j$ is the smallest integer such that $x_{\sigma(j)}\mid u_{2}v_{2}$. We need to show that $u_{1}v_{1}u_{2}v_{2}/x_{\sigma(i)}x_{\sigma(j)}\in IJ.$

If $x_{\sigma(i)}|u_{1}$ and $x_{\sigma(j)}|u_{2}$, then $(u_{1}u_{2}/x_{\sigma(i)}x_{\sigma(j)})v_{1}v_{2}\in IJ$ since $I$ is $d_{\sigma}$-Golod. If $x_{\sigma(i)}|v_{1}$ and $x_{\sigma(j)}|v_{2}$, then $u_{1}u_{2}(v_{1}v_{2}/x_{\sigma(i)}x_{\sigma(j)})\in IJ$ since $I\subset J$. If $x_{\sigma(i)}|u_{1}$ and $x_{\sigma}(j)|v_{2}$, then $u_{1}/x_{\sigma(i)}(v_{1}u_{2}v_{2}/x_{\sigma(j)})\in IJ$ since $I$ is $d_{\sigma}$-Golod. The case $x_{\sigma(i)}|v_{1}$ and $x_{\sigma}(j)|u_{2}$ is similar.

Let $I$ be a monomial ideal. We denote by $\bar{I}$ the integral closure of $I$. The following result and its proof are completely analogue to that of [8, Proposition 3.1], where a similar result is shown for strongly Golod monomial ideals.

Proposition 3.2**.**

Let $I$ be a monomial ideal which is $d_{\sigma}$-Golod. Then $\bar{I}$ is $d_{\sigma}$-Golod. In particular, $\bar{I^{k}}$ is strongly $d$-Golod for all $k\geq 2$.

A monomial ideal $I$ of polynomial ring $S=K[x_{1},\ldots,x_{n}]$ is called stable if for all monomial $u\in I$ one has $x_{i}u/x_{m(u)}$ for all $i\leq m(u)$, where $m(u)$ is the largest integer $j$ such that $x_{j}$ divides $u$.

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 $I\subset J$. In particular, $I$ is Golod.

Proof.

Let $\sigma$ be the permutation reversing the order of the variables. Then $d_{\sigma}(u)=u/x_{m(u)}$. Hence for all monomials $u,v\in I$ one has $d_{\sigma}(u)d_{\sigma}(v)\in I$, since $I$ is stable. It follows that $I$ is $d_{\sigma}$-Golod. Therefore the desired result follows from Theorem 3.1(e).

In the following proposition we present new family of $d$-Golod ideals which are not necessarily monomial ideals.

Proposition 3.4**.**

Let $J_{i}\subset K[[x_{i},\ldots,x_{n}]]$ be an ideal for $i=1,\ldots,n$, and $J\subset S=K[|x_{1},\ldots,x_{n}|]$ be the ideal generated by $\sum_{i=1}^{n}x_{i}J_{i}$. Then $J^{k}$ is $d$-Golod for all $k\geq 2$.

Proof.

Note that for any $k\geq 2$ we have

[TABLE]

where the $a_{i}$ are non-negative integers. By convention, $x_{i}^{a_{i}}J_{i}^{a_{i}}=S$ if $a_{i}=0$. Thus we see that $J^{k}$ is generated by elements of the form

[TABLE]

with integers $a_{i}\geq 0$ such that $a_{1}+\ldots+a_{n}=k$ and with $f_{i}\in J_{i}^{a_{i}}$ for $i=1,\ldots,n$,

One has

[TABLE]

Hence, we see that $d^{i}(f)d^{j}(g)\in J^{k}$ for all $f,g\in J^{k}$ and all $i,j$ with $1\leq i,j\leq n$, and this shows that $J^{k}$ is $d$-Golod for all $k\geq 2$.

The last application we have in mind is of more general nature and deals with stretched local rings. Let $(R,{\mathfrak{m}},K)$ be a Noetherian local ring with maximal ideal ${\mathfrak{m}}$ and residue field $K$ or a standard graded $K$-algebra with graded maximal ideal ${\mathfrak{m}}$. The ring $R$ is said to be stretched if ${\mathfrak{m}}^{2}$ is a principal ideal.

We set $n=\dim_{K}{\mathfrak{m}}/{\mathfrak{m}}^{2}$ and $\tau=\dim_{K}\operatorname{Soc}(R)$, where $\operatorname{Soc}(R)=(0:_{R}{\mathfrak{m}})$ is the socle of $R$. Moreover, if $R$ is Artinian, we let $s$ be the largest integer such that ${\mathfrak{m}}^{s}\neq 0$. Note the $s+1$ is the Loewy length of $R$.

Stretched local rings have been introduced by Sally [10]. She showed that the Poincaré series of $K$ is a rational function. Indeed, she showed that

[TABLE]

Very recently in [4] it was shown that all finitely generated modules over a stretched Artinian local ring $R$ have a rational Poincaré series with a common denominator by studying the algebra structure of the Koszul homology of $R$. Among other results they proved in [4, Theorem 5.4] that $R$ is Golod, if $\tau=h$. By using our methods we give an alternative proof of the result and show

Theorem 3.5**.**

Let $(R,{\mathfrak{m}},K)$ 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 $\tau=n$.*

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 $n\geq 2$ and $s\geq 3$ in Artinian case. Then the following holds:

1. (i)

$\tau=n$, if $R$ is Artinian. 2. (ii)

$\tau=n-1$, if $R$ is not Artinian.

Proof.

Since $R$ is standard graded, $R_{1}R_{i}=R_{i+1}\neq 0$ for $2\leq i<s$ if $R$ is Artinian, and $R_{1}R_{i}=R_{i+1}\neq 0$ for all $i\geq 2$ if $R$ is not Artinian. Since $\dim_{K}R_{i}=1$ for all $i\geq 2$ with $R_{i}\neq 0$, it follows that $R_{s}\varsubsetneq\operatorname{Soc}(R)\varsubsetneq R_{1}\oplus R_{s}$ if $R$ is Artinian, and $\operatorname{Soc}(R)\varsubsetneq R_{1}$ if $R$ is not Artinian. Thus in order to prove (i) and (ii) we must show that $\dim\operatorname{Soc}(R)\cap R_{1}=h-1$.

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 $\dim_{K}R_{1}>\dim_{K}R_{2}$ and since $K$ is algebraically closed, [9, Lemma 2.8] implies that there exists a non-zero linear form $x_{1}$ such that $x_{1}^{2}=0$. Assume that $x_{1}R_{1}\neq 0$. Then $R_{2}=x_{1}R_{1}$, and $R_{3}=R_{1}R_{2}=x_{1}R_{1}^{2}=x_{1}^{2}R_{1}=0$, contradicting the assumption that $s\geq 3$. Thus $x_{1}R_{1}=0$, and hence $x_{1}\in\operatorname{Soc}(R)$, and the assertion follows for $n=2$. Now let $n>2$, and let $R_{1}^{\prime}$ be a $K$-linear subspace of $R_{1}$ such that $R_{1}^{\prime}\oplus Kx_{1}=R_{1}$. Then it follows that $(R_{1}^{\prime})^{2}=R_{1}^{2}$. Therefore the standard graded $K$-algebra $R^{\prime}=K[R_{1}^{\prime}]$ is a stretched $K$-algebra of embedding dimension $n-1$. By induction hypothesis, $\dim_{K}\operatorname{Soc}(R^{\prime})\cap R_{1}^{\prime}=n-2$. Let $x_{2},\ldots,x_{n-1}$ span the $K$-vector space $\operatorname{Soc}(R^{\prime})\cap R_{1}^{\prime}$. These elements $x_{i}$ are also socle elements of $R$ and together with $x_{1}$, they span a vector space of dimension $n-1$, as desired.

Remark 3.7**.**

Suppose that $R=S/I$ is a stretched standard graded $K$-algebra, where $S=K[x_{1},\ldots,x_{n}]$ is the polynomial ring with $K$ is an algebraically closed field, and where $I\subset{\mathfrak{n}}^{2}$. The proof of Lemma 3.6 shows that after a suitable linear change of coordinates one has that

1. (i)

$I=(x_{1},\ldots,x_{n-1})^{2}+x_{n}(x_{1},\ldots,x_{n-1})+(x_{n}^{s+1})$, if $R$ is Artinian; 2. (ii)

$I=(x_{1},\ldots,x_{n-1})^{2}+x_{n}(x_{1},\ldots,x_{n-1})$, if $R$ is not Artinian.

Proof of Theorem 3.5.

Let us first assume that $R$ is a standard graded $K$-algebra. After 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: $x_{n},x_{1},\ldots,x_{n-1}$, and let $\sigma$ be the corresponding permutation of $[n]$. Then $d_{\sigma}(I)=(x_{1},\ldots,x_{n-1},x_{n}^{s})$ in the Artinian case, and and $d_{\sigma}(I)=(x_{1},\ldots,x_{n-1})$ in the non-Artinian case. Clearly $d_{\sigma}(I)^{2}\subset I$ in both cases. Thus in any case $R$ is a Golod ring. 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 $P^{R}_{K}(t)=1/H_{R}(-t)$, where $H_{R}(t)$ denotes the Hilbert series for $R$. Therefore, it is enough to prove the claim for a suitable base extension, because a base extension does not change the Poincaré series of a $K$-algebra, nor does it change its Hilbert series, as already noticed before. Hence after this base field extension we may assume that $I$ is generated by quadratic monomials as described in Remark 3.7. By a result of Fröberg [6], this implies that $G(R)$ is Koszul. Now by another result of Fröberg [5] it follows that $P^{R}_{K}(t)=P^{G(R)}_{K}(t)$. By case (i), $G(R)$ is Golod, and hence

[TABLE]

The coefficientwise inequality in these formulas follows from the well-known fact that there is the coefficientwise inequality $P^{S}_{R}(t)\leq P^{S}_{G}(R)(t)$. Since the opposite inequality $P^{S}_{K}(t)/(1-t(P^{S}_{R}(t)-1)\geq P^{R}_{K}(t)$ always holds, we have equality and $R$ is Golod.

Finally suppose that (iii) is satisfied. Then $G(R)$ is a stretched Artinian $K$-algebra, and hence by Lemma 3.6 we have $\tau=n$ for $G(R)$. By our assumption, $\tau=n$ also for $R$. Thus (6) implies that $P^{R}_{k}(t)=P^{G(R)}_{k}(t)$. As in (ii) it follows from this equation that $R$ is Golod, since $G(R)$ is Golod.