Some criteria for detecting large homomorphisms of local rings

TL;DR

This paper investigates ideals in local rings that produce large homomorphisms upon quotienting, providing characterizations over specific classes like complete intersections, Koszul, and Golod rings.

Contribution

It offers new characterizations of ideals inducing large homomorphisms in various classes of local rings, expanding understanding of their structural properties.

Findings

Characterization over complete intersections

Characterization over Koszul rings

Characterization over Golod rings

Abstract

We study ideals in a local ring $R$ whose quotient rings induce large homomorphisms of local rings. We characterize such ideals over complete intersections, Koszul rings, and over some classes of Golod rings.

Equations

$$0\to R\to R/I\oplus R/J\to k\to 0.$$

$$E_{p,q}^2\cong {\mathrm{Tor}}_p^S(k,k)\otimes {\mathrm{Tor}}_q^R(S,k)⟹{\mathrm{Tor}}_{p+q}^R(k,k).$$

$$\begin{array}{ccccccccc}({\oplus }_{i<j\le n}R{e}_i\wedge e_j)\oplus ({\oplus }_{\ell =1}^{r}R{T}_{\ell })& \to & {\oplus }_{i=1}^{n}R{e}_i& \to & R& \to & k& \to & 0\\ \downarrow f_2& & \downarrow f_1& & \downarrow f_0& & \downarrow =& & \\ ({\oplus }_{i<j\le m}S{e}_i\wedge e_j)\oplus ({\oplus }_{\ell =1}^{s}S{\tau }_{\ell })& \to & {\oplus }_{i=1}^{m}S{e}_i& \to & S& \to & k& \to & 0,\end{array}$$

$$\begin{array}{ccccccccc}{F}_2=({\oplus }_{i<j\le l}R{e}_i\wedge e_j)\oplus ({\oplus }_{\ell =1}^{q}R{S}_{\ell })& \to & F_1={\oplus }_{i=1}^{l}R{e}_i& \to & R& \to & R/I& \to & 0\\ \downarrow {\varphi }_2& & \downarrow {\varphi }_1& & \downarrow =& & \downarrow \pi & & \\ G_2=({\oplus }_{i<j\le n}R{e}_i\wedge e_j)\oplus ({\oplus }_{\ell =1}^{r}R{T}_{\ell })& \to & G_1={\oplus }_{i=1}^{n}R{e}_i& \to & R& \to & k& \to & 0,\end{array}$$

$${\mathrm{ld}}_R(M)=\sup \{i|H_i({\mathrm{lin}}^R(F))\ne 0\}.$$

$$P_k^S(t)=\frac{P_k^R(t)}{1-t(P_S^R(t)-1)},$$

$$\begin{array}{ccccccccc}0& \to & \mathfrak{m}I& \to & I& \to & I/\mathfrak{m}I& \to & 0\\ \downarrow & & \downarrow & & \iota \downarrow & & & & \\ 0& \to & {\mathfrak{m}}^2& \to & \mathfrak{m}& \to & \mathfrak{m}/{\mathfrak{m}}^2& \to & 0,\end{array}$$

$$\begin{array}{ccccc}{\mathrm{Tor}}_i^R(\mathfrak{m}I,k)& \stackrel{f_i}{\to }& {\mathrm{Tor}}_i^R(I,k)& \stackrel{{\gamma }_i}{\to }& {\mathrm{Tor}}_i^R(I/\mathfrak{m}I,k)\\ \downarrow & & g_i\downarrow & & {\iota }_i\downarrow \\ {\mathrm{Tor}}_i^R({\mathfrak{m}}^2,k)& \to & {\mathrm{Tor}}_i^R(\mathfrak{m},k)& \to & {\mathrm{Tor}}_i^R(\mathfrak{m}/{\mathfrak{m}}^2,k)\end{array}$$

$$\begin{array}{ccccc}{\mathrm{Tor}}_i^R(R/\mathfrak{m}I,k)& \stackrel{h_i}{\to }& {\mathrm{Tor}}_i^R(k,k)& \to & {\mathrm{Tor}}_{i-1}^R(\mathfrak{m}/\mathfrak{m}I,k)\\ & & & & \\ {\psi }_i\downarrow & & \downarrow & & \\ & & & & \\ {\mathrm{Tor}}_i^{R/\mathfrak{m}I}(k,k)& \stackrel{\cong }{\to }& {\mathrm{Tor}}_{i-1}^{R/\mathfrak{m}I}(\mathfrak{m}/\mathfrak{m}I,k),& & \end{array}$$

$$\begin{array}{ccccccc}\dots & \to & F_n& \stackrel{{\partial }_n^F}{\to }& F_{n-1}& \to & \dots \\ f_n\downarrow & & f_{n-1}\downarrow & & & & \\ \dots & \to & G_n& \stackrel{{\partial }_n^G}{\to }& G_{n-1}& \to & \dots \end{array}$$

$$P_k^R(t)=\frac{(1+t{)}^{{\nu }_R(\mathfrak{m})}}{1-{\sum }_{i\ge 1}{\dim }_k(H_i(R))t^{i+1}},$$

$$\cdots \to H_i(J)\stackrel{{\overline{x}}_r}{⟶}{H}_i(J)⟶H_i(I)⟶H_{i-1}(J)\stackrel{{\overline{x}}_r}{⟶}\dots .$$

$$\begin{array}{ccccccccc}0& \to & {\mathfrak{m}}^{p-1}{K}_i(J)& \to & {\mathfrak{m}}^{p-1}{K}_i(I)& \to & {\mathfrak{m}}^{p-1}{K}_{i-1}(J)& \to & 0\\ f_i\downarrow & & g_i\downarrow & & f_{i-1}\downarrow & & & & \\ 0& \to & H_i(J)& \to & H_i(I)& \to & H_{i-1}(J)& \to & 0.\end{array}$$

$$\begin{array}{ccccccccc}0& \to & {\mathfrak{m}}^{p-1}{K}_1(J)& \to & {\mathfrak{m}}^{p-1}{K}_1(I)& \to & {\mathfrak{m}}^{p-1}& \to & 0\\ f_1\downarrow & & g_1\downarrow & & f_0\downarrow & & & & \\ 0& \to & H_1(J)& \to & H_1(I)& \to & (0{:}_{R/J}{\overline{x}}_r)& \to & 0.\end{array}$$

$$\begin{array}{ccc}{\mathfrak{m}}^{p-1}{K}_i(R)& \to & H_i(R)\\ \downarrow & & \downarrow \\ {\overline{\mathfrak{m}}}^{p-1}{K}_i(S)& \to & H_i(S)\end{array}$$

Full text

2010 Mathematics Subject Classification. 13C13, 13D40

Key words and phrases. Large homomorphism, small homomorphism, Golod homomorphism, Poincaré series, Koszul homology, Golod ring, Koszul module.

Ryo Takahashi was partly supported by JSPS Grant-in-Aid for Scientific Research 16K05098 and JSPS Fund for the Promotion of Joint International Research 16KK0099

1. Introduction

In 1978 Avramov [4] introduced and studied small homomorphisms of local rings, and a year after, Levin [15] introduced large homomorphisms of local rings as a dual notion of the small homomorphisms. A surjective local homomorphism $f:R\to S$ is called small if the induced homomorphism of graded algebras $f_{*}:\operatorname{Tor}^{R}_{*}(k,k)\to\operatorname{Tor}^{S}_{*}(k,k)$ is injective, and it is called large if $f_{*}$ is surjective. Levin proved that a local homomorphism $R\to S$ is large if and only if for every finitely generated $S$-module $M$ there is an equality of Poincaré series $\operatorname{P}^{R}_{M}(t)=\operatorname{P}^{R}_{S}(t)\cdot\operatorname{P}^{S}_{M}(t)$. This result makes large homomorphisms a very useful tool for understanding betti numbers and computing Poincaré series.

Small homomorphisms are closely related to another class of local homomorphisms namely Golod homomorphisms; see Definition 3.3. In fact, each Golod homomorphism is small, and these homomorphisms have been studied very well in several articles; see for example [4], [16], [17], and [20]. On the other hand, we are only aware of relatively few results about the large homomorphisms. The goal of this paper is not only to prove new results, but also to collect known facts and ideas about large homomorphisms that have been used in many articles without stating them. The organization of the paper is as follows.

In section 2, after some preliminaries and examples, we give various conditions under which a surjective homomorphism $R\to S$ is large, specifically when $R$ is a complete intersection; see Theorem 2.11.

In section 3, over a local ring $(R,\mathfrak{m})$ with an ideal $I$, we give a necessary and sufficient condition when $R\to R/I$ is large and $R\to R/\mathfrak{m}I$ is small simultaneously; see Theorem 3.4. As a result, we show that a sujective local homomorphism $R\to S$ where $S$ is a Koszul $R$-module is large; see Corollary 3.5. Then we provide a sufficient condition for large homomorphisms over Golod rings; see Proposition 3.12.

2. Large homomorphisms over complete intersections

Throughout $(R,\mathfrak{m},k)$ is a commutative Noetherian local ring with maximal ideal $\mathfrak{m}$ and residue field $k$. We start this section by recalling some definitions and preliminaries.

A differential graded algebra (DG algebra) $A$ is a complex $(A,\partial)$ equipped with an algebra structure such that $A_{0}$ is a commutative unitary ring, $A_{i}=0$ for $i<0$, and the product satisfies the Leibniz rule $\partial(ab)=\partial(a)b+(-1)^{|a|}a\partial(b)$, for $a,b\in A$.

** 2.1****.**

Acyclic Closure. Let $I$ be an ideal of $R$. An acyclic closure $R\langle X_{i}|\ {i\geq 1}\rangle$ of the augmentation $R\to R/I$ obtained from the Tate construction is a DG algebra resolution for $R/I$ over $R$, where $X_{i}$ is a set of exterior variables when $i$ is odd, and it is a set of divided variables if $i$ is even number. Note that $X_{i}$ are added in a given homological degree $i$ in such a way that their images under the differential minimally generate the $(i-1)$th homology. For $i\geq 1$, we set $\pi_{i}(R)$ to be the $k$-vector space with basis $X_{i}$. By using the notations of [3], we write $\varepsilon_{i}=\dim_{k}\pi_{i}(R)$.

An acyclic closure $R\langle X_{i}\rangle$ of $R/I$ is not a minimal resolution in general but if $I=\mathfrak{m}$, then it is a minimal resolution by Gulliksen’s theorem; see [12, Theorem 1.6.2].

Let $U$ be a DG algebra resolution of $k$ over $R$ described in 2.1, and let $\mathfrak{a}$ be an ideal of $R$. Then $A=U\otimes_{R}R/\mathfrak{a}$ is a DG algebra, and hence, $\operatorname{Tor}^{R}(k,R/\mathfrak{a})$ also has the graded algebra structure induced from $A$.

Let $M$ be a finitely generated $R$-module. The Poincaré series $\operatorname{P}^{R}_{M}(t)$ of $M$ over $R$ is defined to be the formal power series $\operatorname{P}^{R}_{M}(t)=\sum^{\infty}_{i=0}\dim_{k}\operatorname{Tor}^{R}_{i}(M,k)t^{i}$.

Definition 2.2**.**

[15, Theorem 1.1] Let $(R,\mathfrak{m},k)$ be a local ring, and let $f:R\rightarrow S$ be a surjective local homomorphism. Then $f$ is called large if one of the following equivalent conditions holds.

1. (1)

The induced map $f_{*}:\operatorname{Tor}^{R}_{*}(k,k)\rightarrow\operatorname{Tor}^{S}_{*}(k,k)$ of graded algebras is surjective. 2. (2)

The induced map $\varphi_{*}:\operatorname{Tor}^{R}_{*}(S,k)\rightarrow\operatorname{Tor}^{R}_{*}(k,k)$ of graded algebras is injective. 3. (3)

$\operatorname{P}^{R}_{k}(t)=\operatorname{P}^{R}_{S}(t)\cdot\operatorname{P}^{S}_{k}(t)$.

** 2.3****.**

Necessary Condition. Suppose $R\rightarrow R/I$ is a large homomorphism. Then by Definition 2.2(2) we have $\operatorname{Tor}^{R}_{1}(R/I,k)\to\operatorname{Tor}^{R}_{1}(k,k)$ is injective which specifically tells us that a minimal generating set of $I$ can be completed to a minimal generating set of $\mathfrak{m}$, equivalently $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. We will need this condition in most of our results.

In the following we list some well-known examples of large homomorphisms from the literature.

Example 2.4**.**

Let $(R,\mathfrak{m},k)$ be a local ring, and let $I$ be an ideal such that $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. In either of the following, the induced map $R\rightarrow R/I$ is a large homomorphism.

1. (1)

$\mathrm{pd}_{R}(I)<\infty$. 2. (2)

$(0:_{R}I)=\mathfrak{m}$. 3. (3)

The ring $R/I$ is complete intersection. 4. (4)

The map ${R\to R/}I$ is a quasi-complete intersection homomorphism. 5. (5)

The ring $R$ is Cohen-Macaulay non-Gorenstein, $\mathfrak{m}^{2}\subseteq I$, and $\operatorname{G-dim}_{R}(I)<\infty$. 6. (6)

$\mathfrak{m}=I\oplus J$.

Proof.

(1) We show that if $\mathrm{pd}_{R}(I)<\infty$ then $I$ is generated by a regular sequence. The claim is obvious if $I$ is a principal ideal. Hence we may assume $I=(x_{1},\dots,x_{n})$ such that $n\geq 2$ and $x_{i}\in\mathfrak{m}\setminus\mathfrak{m}^{2}$ for all $i$. Since $\mathrm{pd}_{R}(I)<\infty$, we have $\operatorname{grade}_{R}(I)>0$ by [9, Corollary 1.4.6]. By prime avoidance, one can choose a non zero-divisor $x\in I\setminus\mathfrak{m}^{2}$. It follows from [3, Theorem 2.2.3] that $\mathrm{pd}_{R/(x)}R/I<\infty$. Since $I/(x)$ is generated by $n-1$ elements, by induction it is generated by a regular sequence over $R/(x)$. Hence $I$ is generated by a regular sequence over $R$. Therefore $R\rightarrow R/I$ is a large homomorphism by [15, Theorem 2.2].

(2) also follows by [15, Theorem 2.2]. See [15, Theorem 2.4] for (3), [5, Theorem 6.2] for (4), and [11, Theorem 1.2] for (5).

To see (6), consider the exact sequence

[TABLE]

Applying $-\otimes_{R}k$, we see that the induced map $\operatorname{Tor}^{R}_{i}(R/I,k)\to\operatorname{Tor}^{R}_{i}(k,k)$ is injective for all $i\geq 0$. ∎

The following provides more examples of large homomorphisms.

Remark 2.5**.**

Let $R$ be a local ring and let $I$ be an ideal of $R$. If there exists a local ring $Q$ and a surjective local homomorphism $Q\to R$ such that the composition map $Q\to R/I$ is a large homomorphism then $R\to R/I$ is large. Indeed, since the composition of the maps $Q\to R\to R/I$ is large, the induced composition of homomorphisms $\operatorname{Tor}^{Q}_{*}(k,k)\to\operatorname{Tor}^{R}_{*}(k,k)\to\operatorname{Tor}^{R/I}_{*}(k,k)$ is surjective. Hence $\operatorname{Tor}^{R}_{*}(k,k)\to\operatorname{Tor}^{R/I}_{*}(k,k)$ is surjective.

By using Example 2.4(6) and Remark 2.5 we get the following.

Example 2.6**.**

Let $R=S\#_{k}T$ be the connected sum of local rings $S$ and $T$; see [2]. Then $R\to S$ is a large homomorphism. To see that, let $Q=S\times_{k}T$ be the fiber product of $S$ and $T$. Then the composition of homomorphisms $Q\to R\to S$ is large by Example 2.4 (6). Hence $R\to S$ is large by Remark 2.5.

** 2.7****.**

Change of ring spectral sequence. Let $f:R\to S$ be a local homomorphism. Consider the change of ring spectral sequence

[TABLE]

The spectral sequence is derived from the double complex $C_{p,q}=G_{p}\otimes_{R}F_{q}$ where $F$ and $G$ are free resolutions of $k$ over $R$ and $S$, respectively. The edge homomorphisms $\operatorname{Tor}^{R}_{p}(k,k)\rightarrow\operatorname{Tor}^{S}_{p}(k,k)\cong\operatorname{E}^{2}_{p,0}$ and $\operatorname{E}^{2}_{0,q}\cong\operatorname{Tor}^{R}_{q}(S,k)\rightarrow\operatorname{Tor}^{R}_{q}(k,k)$ are the homomorphisms induced by $f:R\to S$ and $\varphi:S\to k$, respectively; see [10, page 348].

** 2.8****.**

Koszul homology Let $(R,\mathfrak{m},k)$ be a local ring and let $I$ be an ideal of $R$. We denote $\operatorname{K}(I)$ the Koszul complex with respect to a minimal generating set of $I$, and $\operatorname{H}_{i}(I):=\operatorname{H}_{i}(\operatorname{K}(I))$. If $I=\mathfrak{m}$ we write $\operatorname{K}(R):=\operatorname{K}(\mathfrak{m})$ and $\operatorname{H}_{i}(R):=\operatorname{H}_{i}(\mathfrak{m})$. If $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$ then a minimal generating set of $I$ can be completed to one for $\mathfrak{m}$. In this case, $\operatorname{K}(I)$ is a subcomplex of $\operatorname{K}(R)$. Therefore the inclusion map induces a natural homomorphism $\operatorname{H}_{*}(I)\to\operatorname{H}_{*}(R)$ of homology algebras.

Large homomorphisms are well understood over complete intersection local rings due to Gulliksen and Levin [12]. It is worth to bring here those conditions which characterize large homomorphisms over complete intersection local rings. First we need the following lemmas.

Lemma 2.9**.**

Let $(R,\mathfrak{m},k)$ be a local ring and let $I$ be an ideal of $R$ such that $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. Then the induced map $\operatorname{H}_{1}(R)\rightarrow\operatorname{H}_{1}(R/I)$ is surjective.

Proof.

Set $S=R/I$. The assumption $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$ implies that $\operatorname{Tor}^{R}_{1}(S,k)\to\operatorname{Tor}^{R}_{1}(k,k)$ is injective. Hence by 2.7, $d^{2}_{2,0}=0$ and therefore the induced map $\operatorname{Tor}^{R}_{i}(k,k)\rightarrow\operatorname{Tor}^{S}_{i}(k,k)$ is surjective for $i\leq 2$. Let $\operatorname{K}(R)$ and $\operatorname{K}(S)$ respectively be the Koszul complexes of $R$ and $S$. Let $\sigma_{1},\dots,\sigma_{r}\in\operatorname{K}(R)$ and $\delta_{1},\dots,\delta_{s}\in\operatorname{K}_{1}(S)$ be cycles whose classes are basis for the vector spaces $\operatorname{H}_{1}(R)$ and $\operatorname{H}_{1}(S)$, respectively. Then by the Tate construction, there exists a commutative diagram

[TABLE]

where the rows are beginning of the free resolutions of $k$ over $R$ and $S$, $T_{1},\dots,T_{r}$ and $\tau_{1},\dots\tau_{s}$ are divided variables of homological degree $2$ with $\partial T_{i}=\sigma_{i}$ and $\partial\tau_{i}=\delta_{i}$; see 2.1. Since $\operatorname{Tor}^{R}_{2}(k,k)\rightarrow\operatorname{Tor}^{S}_{2}(k,k)$ is surjective, $f_{2}$ is surjective too, and hence, $\tau_{j}=f_{2}(\eta_{j})$ for some $\eta_{j}\in F_{2}$. Thus $\delta_{j}=f_{1}(\partial\eta_{j})$. Since the map $\operatorname{H}_{1}(R)\rightarrow\operatorname{H}_{1}(S)$ is induced by $f_{1}$, we are done. ∎

Lemma 2.10**.**

Let $I$ be an ideal of $R$ such that $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. If the natural map $\operatorname{H}_{1}(I)\otimes_{R}k\to\operatorname{H}_{1}(R)$ is injective, then the induced map $\operatorname{Tor}^{R}_{2}(R/I,k)\to\operatorname{Tor}^{R}_{2}(k,k)$ is injective.

Proof.

Let $\mathfrak{m}=(x_{1},\dots,x_{n})$ and $I=(x_{1},\dots,x_{l})$ with $l\leq n$. Let $z_{1},\dots,z_{q}\in\operatorname{K}_{1}(I)$ and $z^{\prime}_{1},\dots,z^{\prime}_{r}\in\operatorname{K}_{1}(R)$ be cycles whose classes minimally generate $\operatorname{H}_{1}({I})$ and $\operatorname{H}_{1}(R)$, respectively. Then there exists a commutative diagram

[TABLE]

where the rows are beginning of Tate resolutions of $R/I$ and $k$, $\phi_{1}$ is natural inclusion, $S_{1},\dots S_{l}$ and $T_{1},\dots,T_{r}$ are divided variables of homological degree $2$ with $\partial S_{i}=z_{i}$ and $\partial T_{i}=z^{\prime}_{i}$. Note that $\operatorname{H}_{1}(I)\to\operatorname{H}_{1}(R)$ is induced by $\phi_{1}$ and one has $\phi_{2}|_{\oplus_{i<j\leq n}Re_{i}\wedge e_{j}}$ is the natural injection $\operatorname{K}(I)\to\operatorname{K}(R)$. Assume $\operatorname{H}_{1}(I)\otimes_{R}k\to\operatorname{H}_{1}(R)$ is injective and let $\bar{\phi_{1}}$ be the induced map $\operatorname{H}_{1}(I)\to\operatorname{H}_{1}(R)$. Then for any choice of $\lambda_{1},\dots,\lambda_{q}$ such that $\lambda_{j}\in R\setminus\mathfrak{m}$ for some $j$, one has $\bar{\phi_{1}}(\lambda_{1}\bar{z_{1}}+\cdots+\lambda_{q}\bar{z_{q}})$ is non zero in $\operatorname{H}_{1}(R)$ where $\bar{z_{i}}$ is the class of $z_{i}$ in $\operatorname{H}_{1}(I)$. This means $\phi_{1}(\lambda_{1}z_{1}+\cdots+\lambda_{q}z_{q})\notin\operatorname{Im}\partial^{K(R)}_{2}$ and therefore one has $\phi_{2}(\lambda_{1}S_{1}+\cdots+\lambda_{q}S_{q})\notin(\oplus_{i<j\leq n}Re_{i}\wedge e_{j})\oplus(\oplus_{\ell=1}^{r}\mathfrak{m}T_{\ell})$. This implies that $\phi_{2}$ is split injective and hence the induced map $\operatorname{Tor}^{R}_{2}(R/I,k)\to\operatorname{Tor}^{R}_{2}(k,k)$ is injective. ∎

Theorem 2.11**.**

Let $(R,\mathfrak{m},k)$ be a complete intersection local ring, and let $I$ be an ideal of $R$ with $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. Let $S=R/I$. Then the following are equivalent.

1. (1)

The induced map $R\rightarrow S$ is large. 2. (2)

The ring $S$ is complete intersection. 3. (3)

The induced map $\operatorname{Tor}^{R}_{2}(S,k)\rightarrow\operatorname{Tor}^{R}_{2}(k,k)$ is injective. 4. (4)

The induced map $\operatorname{Tor}^{R}_{3}(k,k)\rightarrow\operatorname{Tor}^{S}_{3}(k,k)$ is surjective. 5. (5)

The induced map $\operatorname{H}_{1}(I)\otimes_{R}k\to\operatorname{H}_{1}(R)$ is injective. 6. (6)

The induced map $\operatorname{H}_{2}(R)\rightarrow\operatorname{H}_{2}(S)$ is surjective.

Proof.

(1)$\Rightarrow$(2) Since $R$ is complete intersection, we have $\varepsilon_{3}(R)=0$ by [3, Theorem 7.3.3]. Since $R\rightarrow S$ is large, we get $\varepsilon_{3}(S)=0$. Therefore $S$ is complete intersection by [3, Theorem 7.3.3].

(2)$\Rightarrow$(1) This follows from Example 2.4(3).

(2)$\Rightarrow$(5) Assume $S$ is complete intersection. Then the map $R\rightarrow S$ is a quasi-complete intersection homomorphism by [5, Proposition 7.7]. By [5, Theorem 5.3] there exists an exact sequence $0\to\operatorname{H}_{1}(I)\otimes_{R}k\to\pi_{2}(R)$ of $k$-vector spaces. Since $\pi_{2}(R)\cong\operatorname{H}_{1}(R)$, we have the induced map $\operatorname{H}_{1}(I)\otimes_{R}k\to\operatorname{H}_{1}(R)$ is injective; see 2.1.

(5)$\Rightarrow$(3) This follows from Lemma 2.10.

(3)$\Leftrightarrow$(4) This follows from the change of rings spectral sequence; see 2.7.

(4)$\Rightarrow$(2) Same argument as (1)$\Rightarrow$(2) applies here as well.

(2)$\Leftrightarrow$(6) Since $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$, we have $\operatorname{H}_{1}(R)\to\operatorname{H}_{1}(S)$ is surjective by Lemma 2.9. Since $R$ is complete intersection, by the Tate-Assmus Theorem [9, Theorem 2.3.11] $\operatorname{H}_{2}(R)=\operatorname{H}_{1}(R)^{2}$. Therefore $S$ is complete intersection if and only if $\operatorname{H}_{2}(S)=\operatorname{H}_{1}(S)^{2}$ if and only if $\operatorname{H}_{2}(R)\rightarrow\operatorname{H}_{2}(S)$ is surjective. ∎

Remark 2.12**.**

Let $f:R\rightarrow S$ be a surjective homomorphism of local rings. Then $f$ is large if $f_{i}:\operatorname{Tor}^{R}_{i}(k,k)\rightarrow\operatorname{Tor}^{S}_{i}(k,k)$ is surjective for all $i\gg 0$. Indeed, this is easy to see when $S$ is a regular ring. In this case, for example, the sujectivity of $\operatorname{K}(R)\to\operatorname{K}(S)$ implies that $f$ is large. Suppose $S$ is singular. One has $f_{i}$ is surjective if and only if $f^{i}:\operatorname{Ext}^{i}_{S}(k,k)\rightarrow\operatorname{Ext}^{i}_{R}(k,k)$ is injective. It is well-known that $\operatorname{Ext}^{*}_{S}(k,k)$ is the universal enveloping algebra of the homotopy Lie algebra $\pi^{*}(S)$; see [3, Theorem 10.2.1]. Then by [8, Lemma 5.1.7], any element $\chi\in\pi^{2}(S)$ (and hence any powers of $\chi$) is a non zero divisor on $\operatorname{Ext}^{*}_{S}(k,k)$. Let $\alpha\in\ker{f^{j}}$ for some $j$. One has $\chi^{i}\alpha\in\ker f^{2i+j}$. Since $f^{i}$ is injective for all $i\gg 0$, we have $\chi^{i}\alpha=0$ and hence $\alpha=0$.

Regarding Remark 2.12, one may ask whether $R\to S$ is large if the induced map $\operatorname{Tor}^{R}_{i}(S,k)\to\operatorname{Tor}^{R}_{i}(k,k)$ is injective for all $i\gg 0$. This is not true in general. For example, if $I\subseteq\mathfrak{m}^{2}$ is any non zero ideal of finite projective dimension then $R\to R/I$ is not large, while $\operatorname{Tor}^{R}_{i}(R/I,k)\to\operatorname{Tor}^{R}_{i}(k,k)$ is injective for all $i\gg 0$. We don’t know whether $R\to R/I$ is large if $\mathrm{pd}_{R}(I)=\infty$ and $\operatorname{Tor}^{R}_{i}(R/I,k)\to\operatorname{Tor}^{R}_{i}(k,k)$ is injective for all $i\gg 0$. However, we are able to prove the following.

Proposition 2.13**.**

Let $(R,\mathfrak{m},k)$ be a local ring and let $I$ be an ideal of $R$ with $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. Assume the induced map $\operatorname{H}_{1}(I)\to\operatorname{H}_{1}(R)$ is non zero. If $\varphi_{i}:\operatorname{Tor}^{R}_{i}(R/I,k)\to\operatorname{Tor}^{R}_{i}(k,k)$ is injective for all $i\gg 0$ then $R\to R/I$ is large.

Proof.

Let $\zeta$ be an element in $\operatorname{H}_{1}(I)$ whose image under the natural map $\operatorname{H}_{1}(I)\to\operatorname{H}_{1}(R)$ is non zero, and let $z\in\operatorname{K}_{1}(I)$ be a cycle whose class in $\operatorname{H}_{1}(I)$ is $\zeta$. By using the natural injection $\operatorname{K}_{1}(I)\hookrightarrow\operatorname{K}_{1}(R)$ we consider $z$ as an element of $\operatorname{K}_{1}(R)$. Then one checks that $z=z^{\prime}+\partial(a)$ for some $z^{\prime}\in I\operatorname{K}_{1}(R)$ and $a\in\operatorname{K}_{2}(R)$. Therefore $[z]=[z^{\prime}]$ in $\operatorname{H}_{1}(R)$. Let $U=R\langle X_{i}\rangle_{i\geq 1}$ be an acyclic closure of $k$ over $R$, and let ${X}\in U$ be a divided variable of degree 2 such that $\partial({X})=z$; see 2.1. By [3, Lemma 6.3.3], there exists a chain $\Gamma$-derivation $d:U\to U$ of degree $-2$ which is trivial on $\operatorname{K}(R)$ and $d({X}^{(i)})={X}^{(i-1)}$.

Set $S=R/I$ and $A=U\otimes_{R}S$. Then ${X}\otimes 1_{S}$ is a cycle in $A$ whose class $[{X}\otimes 1_{S}]$ is non zero in $\operatorname{Tor}^{R}_{2}(S,k)$ (because $\varphi_{2}([{X}\otimes 1_{S}])={X}\otimes 1_{k}\neq 0$). Let $\alpha\otimes 1_{S}\in A$ be such that $[\alpha\otimes 1_{S}]\in\operatorname{Tor}^{R}_{j}(R/I,k)$ and $\varphi_{j}([\alpha\otimes 1_{S}])=0$. Since $\varphi_{*}$ is homomorphism of graded algebras, we have $\varphi_{j+2i}([{X}^{(i)}\alpha\otimes 1_{S}])=0$. Thus by assumption, $[{X}^{(i)}\alpha\otimes 1_{S}]=0$ for $i\gg 0$. If $[d(\alpha)\otimes 1_{S}]=0$ then we have $[\alpha\otimes 1_{S}]=[d^{i}({X}^{(i)}\alpha)\otimes 1_{S}]=0$ and we are done.

Suppose $[d(\alpha)\otimes 1_{S}]\neq 0$, and let $r$ be the biggest integer such that $[d^{r}(\alpha)\otimes 1_{S}]\neq 0$. Since $\varphi_{j}([\alpha\otimes 1_{S}])=0$ and $d$ commutes with differentials, one has $\varphi_{j-2r}([d^{r}(\alpha)\otimes 1_{S}])=0$. By replacing $[\alpha\otimes 1_{S}]$ with $[d^{r}(\alpha)\otimes 1_{S}]$, the same argument as above shows that $[d^{r}(\alpha)\otimes 1_{S}]=0$ which is a contradiction. ∎

3. Large homomorphisms over Koszul rings and Golod rings

Let $(R,\mathfrak{m},k)$ be a local ring and let $M$ be a finitely generated $R$-module. Let $F$ be a minimal free resolution of $M$ over $R$ and let $\operatorname{lin}^{R}(F)$ be the associated graded complex of $F$; see [14, §1] and [20, §2] for more details. The linearity defect $\operatorname{ld}_{R}(M)$ of $M$ over $R$ is defined by

[TABLE]

** 3.1****.**

Koszul Modules. An $R$-module $M$ is called a Koszul module (or is said to have a linear resolution) if $\operatorname{lin}^{R}(F)$ is acyclic, equivalently $\operatorname{ld}_{R}(M)=0$. $R$ is called a Koszul ring if $k$ is a Koszul $R$-module.

Let $\operatorname{gr}_{\mathfrak{m}}(R)=\oplus_{i\geq 0}\mathfrak{m}^{i}/\mathfrak{m}^{i+1}$ be the associated graded ring of $R$ and $\operatorname{gr}_{\mathfrak{m}}(M)=\oplus_{i\geq 0}\mathfrak{m}^{i}M/\mathfrak{m}^{i+1}M$ be the associated graded module of $M$. If $M$ is a Koszul module then $\operatorname{lin}^{R}(F)$ is a minimal free resolution of $\operatorname{gr}_{\mathfrak{m}}(M)$ over $\operatorname{gr}_{\mathfrak{m}}(R)$; see [14, Proposition 1.5].

The regularity $\operatorname{reg}_{R}(M)$ is defined to be the regularity of the graded module $\operatorname{gr}_{\mathfrak{m}}(M)$ over the graded ring $\operatorname{gr}_{\mathfrak{m}}(R)$; see [20]. It follows that $M$ is Koszul if and only if $\operatorname{reg}_{R}(M)=0$.

Definition 3.2**.**

Let $k$ be a field, and let $A$ be a DG algebra with $\operatorname{H}_{0}(A)\cong k$. The algebra $A$ is said to admit a trivial Massey operation, if for some $k$-basis ${\bf h}=\{h_{\lambda}\}_{\lambda\in\Lambda}$ of $\operatorname{H}_{\geq 1}(A)$ there exists $\mu:\coprod_{i\geq 1}{\bf h}^{i}\to A$ such that

1. (1)

$\mu(h_{\lambda})=z_{\lambda}$ where $z_{\lambda}$ is a cycle in $A$ with class $[z_{\lambda}]=h_{\lambda}\in\operatorname{H}_{\geq 1}(A)$, and 2. (2)

$\partial\mu(h_{\lambda_{1}},\dots,h_{\lambda_{n}})=\displaystyle\sum^{n-1}_{i=1}\overline{\mu(h_{\lambda_{1}},\dots,h_{\lambda_{i}})}\mu(h_{\lambda_{i+1}},\dots,h_{\lambda_{n}})$, where $\overline{a}=(-1)^{|a|+1}a$.

Definition 3.3**.**

A surjective local homomorphism $f:R\to S$ is called Golod if

[TABLE]

or equivalently, the DG algebra $A=U\otimes_{R}S$ admits a trivial Massey operation, where $U$ is a minimal DG algebra resolution of $k$ over $R$; see [16, Theorem 1.5].

Theorem 3.4**.**

Let $(R,\mathfrak{m},k)$ be a local ring, and let $I$ be an ideal of $R$ such that $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. Then the following conditions are equivalent.

1. (1)

The map $\operatorname{Tor}^{R}_{i}(\mathfrak{m}I,k)\rightarrow\operatorname{Tor}^{R}_{i}(I,k)$ induced by the inclusion $\mathfrak{m}I\hookrightarrow I$ is zero for all $i\geq 0$. 2. (2)

The homomorphism $\phi:R\rightarrow R/I$ is large, and the homomorphism $\psi:R\rightarrow R/\mathfrak{m}I$ is small.

Moreover, under these equivalent conditions $\psi$ is a Golod homomorphism.

Proof.

The assumption $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$ implies an exact sequence $0\rightarrow I/\mathfrak{m}I\overset{\iota}{\rightarrow}\mathfrak{m}/\mathfrak{m}^{2}\rightarrow\mathfrak{m}/(\mathfrak{m}^{2}+I)\rightarrow 0$ of $k$-vector spaces. By applying $-\otimes_{R}k$ to the commutative diagram

[TABLE]

we get a commutative diagram

[TABLE]

with exact rows, where $\iota_{i}$ is injective.

(1)$\Rightarrow$ (2): Since $f_{i}$ is zero map and $\iota_{i}$ is split injection, the composition $\iota_{i}\circ\gamma_{i}$ is injective. Therefore, the map $g_{i}$ is injective for all $i\geq 0$. Therefore $\operatorname{Tor}^{R}_{i}(R/I,k)\rightarrow\operatorname{Tor}^{R}_{i}(k,k)$ is injective for all $i\geq 0$.

Next, we show that $\psi$ is Golod and hence it is small; see [17, Definition 1.1]. The argument is similar to the proofs of [19, Lemma 1.2] and [1, Lemma 2.1] but we bring it for the reader’s convenience. Let $U\rightarrow k$ be a minimal DG algebra resolution of $k$ over $R$, and set $A=U\otimes_{R}R/\mathfrak{m}I$. Then $\operatorname{H}_{i}(A)\cong\operatorname{Tor}^{R}_{i}(R/\mathfrak{m}I,k)$. The map $\operatorname{Tor}^{R}_{i}(R/\mathfrak{m}I,k)\rightarrow\operatorname{Tor}^{R}_{i}(R/I,k)$ is identified with the natural map $\operatorname{H}_{i}(A)\rightarrow\operatorname{H}_{i}(A/IA)$ for all $i$. Since $f_{i}=0$, every element in $\operatorname{H}_{i>0}(A)$ can be represented by $[x]$ for some $x\in IA$. Thus for any choice of $[z],[w]\in\operatorname{H}_{>0}(A)$ one has $z\cdot w\in I^{2}A=0$. Hence $A$ admits a trivial Massey operation; see [16, Lemma 1.2].

(2) $\Rightarrow$ (1): Suppose the map $\phi:R\rightarrow R/I$ is large, and the map $\psi:R\rightarrow R/\mathfrak{m}I$ is small homomorphisms. There exist a commutative diagram

[TABLE]

for all $i\geq 1$. Since $\psi_{i}$ is injective for all $i\geq 0$, the commutative diagram implies that $h_{i}=0$ for all $i\geq 1$. Therefore, in the commutative diagram above, we have $g_{i}\circ f_{i}=0$ for all $i\geq 0$. Since $\phi$ is large, $g_{i}$ is injective and therefore $f_{i}=0$ as desired. ∎

The following is an immediate consequence of Theorem 3.4.

Corollary 3.5**.**

Let $(R,\mathfrak{m},k)$ be a local ring, and let $I$ be an ideal of $R$ with $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. If $R/I$ is a Koszul $R$-module then $R\rightarrow R/I$ is a large homomorphism and $R\to R/\mathfrak{m}I$ is a Golod homomorphism.

Proof.

Since $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$ and $R/I$ is a Koszul $R$-module, $\operatorname{reg}_{R}(R/I)=0$. Therefore the maps $\operatorname{Tor}^{R}_{i}(\mathfrak{m}I,k)\rightarrow\operatorname{Tor}^{R}_{i}(I,k)$ are zero for all $i\geq 0$ by [20, Theorem 3.2]. Now the result follows from Theorem 3.4. ∎

Corollary 3.6**.**

Let $(R,\mathfrak{m},k)$ be a graded Koszul local ring and let $I$ be a homogeneous ideal of $R$. If the map ${R\to}R/I$ is large, then $R\to R/\mathfrak{m}I$ is a Golod homomorohism and $R/\mathfrak{m}I$ is a Koszul ring.

Proof.

Set $S=R/\mathfrak{m}I$. Since $R$ is Koszul, we have $\operatorname{reg}_{R}(k)=0$, and since $R\to R/I$ is large, we have the induced map $\operatorname{Tor}^{R}_{*}(R/I,k)\to\operatorname{Tor}^{R}_{*}(k,k)$ is injective. This implies that $\operatorname{reg}_{R}(R/I)=0$ and hence it is a Koszul $R$-module. Therefore the map $R\to S$ is a Golod homomorphism by Corollary 3.5.

Next, we show that $S$ is Koszul. Since $R\to R/I$ is large, we have $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$ by 2.3. Then we may assume $\mathfrak{m}=(x_{1},\dots,x_{r},x_{r+1},\dots,x_{n})$ such that $I=(x_{1},\dots,x_{r})$. Let $J=(x_{r+1},\dots,x_{n})$, and let $\bar{I}$ and $\bar{J}$ be ideals of $S$ generated by images of $I$ and $J$, respectively. Then one easily checks that $\bar{I}\cap\bar{J}=0$ and therefore $S$ is the fiber product of $S/\bar{I}$ and $S/\bar{J}$. We have $S/\bar{I}\cong R/I$. Since $R$ is Koszul and $R\to R/I$ is large, the surjectivity of $\operatorname{Tor}^{R}_{*}(k,k)\to\operatorname{Tor}^{R/I}_{*}(k,k)$ shows that $R/I$ is Koszul. On the other hand, $S/\bar{J}$ is isomorphic to a graded local ring $(T,\mathfrak{n})$ with $\mathfrak{n}^{2}=0$ which is Koszul. Now, by [18, Proposition 3.11], $S$ is Koszul. ∎

A Koszul local ring $R$ is called absolutely Koszul if $\operatorname{ld}_{R}(M)<\infty$ for every finitely generated $R$-module $M$.

Corollary 3.7**.**

Let $(R,\mathfrak{m},k)$ be a graded Koszul complete intersection local ring and let $I$ be an ideal of $R$. If ${R\to}R/I$ is large then $R/\mathfrak{m}I$ is absolutely Koszul.

Proof.

This follows by Corollary 3.6 and [14, Theorem 5.9]. ∎

Remark 3.8**.**

Let $(R,\mathfrak{m},k)$ be a local Gorenstein ring with $\mathfrak{m}^{3}=0$. Let $I$ be a non zero ideal of $R$ such that $I\nsubseteq\mathfrak{m}^{2}$. Then by [6, Corollary 4.7] $R/I$ is a Koszul $R$-module. Therefore the map $R\rightarrow R/I$ is large and $R\to R/(0:_{R}\mathfrak{m})$ is Golod by Corollary 3.5. The later has been proven for all Artinian Gorenstein rings; see [7, Theorem 2]. Note that $R\to R/I$ may not be large if $R$ is not Gorenstein; see [11, Example 3.12].

Example 3.9**.**

Let $R=k[x,y,z]/(x^{2},y^{2},z^{2})$, where $k$ is a field. Then $R$ is an Artinian Koszul complete intersection with $\mathfrak{m}^{4}=0$. Consider the ideal $I=(x+y+z)$ of $R$. Then $R/I\cong k[y,z]/(y^{2},z^{2},yz)$ which is not a complete intersection. Therefore $R\rightarrow R/I$ is not large by Proposition 2.11. If $k$ has characteristic different than $2$ then $\mathfrak{m}I=(xy,xz,yz)$ and therefore $R/\mathfrak{m}I=R/\mathfrak{m}^{2}$. This shows that the converse of Corollary 3.7 is not true. Also, one checks that $R\to R/\mathfrak{m}I$ is a Golod homomorphism. Therefore in Theorem 3.4, Golodness of $R\to R/\mathfrak{m}I$ does not guarantee that $R\to R/I$ is large.

Proposition 3.10**.**

Let $(R,\mathfrak{m},k)$ be a local ring and let $I$ be an ideal of $R$. Assume $\operatorname{ld}_{R}(R/I)=l<\infty$. If the induced maps $\operatorname{H}_{1}(I)\to\operatorname{H}_{1}(R)$ and $\operatorname{Tor}^{R}_{l}(R/I,k)\to\operatorname{Tor}^{R}_{l}(k,k)$ are respectively non zero and injective, then the map $R\to R/I$ is large.

Proof.

Let $F$ and $G$ respectively be minimal free resolutions of $R/I$ and $k$ over $R$. Let $f:F\to G$ be the comparison homomorphism induced by $R/I\to k$. We show by induction that $f_{i}$ is split injective for all $i\geq l$. The injectivity of $\operatorname{Tor}^{R}_{l}(R/I,k)\to\operatorname{Tor}^{R}_{l}(k,k)$ settles the case $i=l$. Let $n>l$ and suppose $f_{i}$ is split injective for $l\leq i<n$. Consider the commutative diagram

[TABLE]

and let $e\in F_{n}\setminus\mathfrak{m}F_{n}$. If $f_{n}(e)\in\mathfrak{m}G_{n}$ then $f_{n-1}\partial^{F}_{n}(e)\in\mathfrak{m}^{2}G_{n-1}$. Since $f_{n-1}$ is split injective, we have $\partial^{F}_{n}(e)\in\mathfrak{m}^{2}F_{n-1}$. As $\operatorname{ld}_{R}(R/I)<n$, this is a contradiction and hence $f_{n}(e)\in G_{n}\backslash\mathfrak{m}G_{n}$. Therefore the induced map $\operatorname{Tor}^{R}_{i}(R/I,k)\to\operatorname{Tor}^{R}_{i}(k,k)$ is injective for all $i\geq l$. Now by using Proposition 2.13, the induced map $R\to R/I$ is large. ∎

Definition 3.11**.**

A local ring $(R,\mathfrak{m},k)$ is called Golod if the Poincaré series of $k$ over $R$ has presentation

[TABLE]

where $\nu_{R}(\mathfrak{m})$ is the number of minimal generating set of $\mathfrak{m}$.

The following gives a sufficient condition for large homomorphisms over Golod rings by using the Koszul homologies.

Proposition 3.12**.**

Let $(R,\mathfrak{m},k)$ be a Golod local ring, and let $f:R\to S$ be a surjective local homomorphism. If the natural map $\operatorname{H}_{i}(R)\rightarrow\operatorname{H}_{i}(S)$ is surjective for all $i\geq 1$, then $f$ is a large homomorphism.

Proof.

Note that $R$ is Golod if and only if $\operatorname{K}(R)$ admits a trivial Massey operation; see [3, Theorem 5.2.2] and [12, Corollary 4.2.4]. Since $R$ is Golod, $\operatorname{K}(R)$ admits a trivial Massey operation. The surjectivity of $\operatorname{H}_{i}(R)\rightarrow\operatorname{H}_{i}(S)$ implies that $\operatorname{K}(S)$ admits a trivial Massey operation too and therefore it is a Golod ring. Let $F$ and $G$ respectively be minimal free resolutions of $k$ over $R$ and $S$ described in [3, Theorem 5.2.2]. Then there exists a natural chain map $f:F\rightarrow G$ which only depends on the choice of $k$-bases of $\operatorname{H}_{i\geq 1}(R)$ and $\operatorname{H}_{i\geq 1}(S)$. Since $\operatorname{H}_{i}(R)\rightarrow\operatorname{H}_{i}(S)$ is surjective, $f$ is surjective as well. Therefore the induced map $\operatorname{Tor}_{*}^{R}(k,k)\rightarrow\operatorname{Tor}^{S}_{*}(k,k)$ is surjective as well. This finishes the proof. ∎

Recently, Gupta proved that if $R$ is a Golod local ring and a homomorphism $f:R\to S$ is large, then $S$ is Golod; see [13, Theorem 1.5]. This result provides a useful tool to detect Golod rings by using large homomorphism.

Example 3.13**.**

Let $R=\displaystyle\frac{k[x,y,z]}{(x^{2},xy,xz,y^{2},z^{2})}$. Then $R$ is an Artinian local ring with the maximal ideal $\mathfrak{m}=(x,y,z)$. We have $R/(x)\cong k[y,z]/(y^{2},z^{2})$ is a complete intersection of codimension two. Therefore $R\rightarrow R/(x)$ is large by Example 2.4(3), and $R/(x)$ is not Golod. Hence $R$ is not Golod.

Lemma 3.14**.**

Let $(Q,\mathfrak{n},k)$ be a regular local ring, $R=Q/\mathfrak{n}^{p}$ with $p\geq 2$, and let $\mathfrak{m}$ be the maximal ideal of $R$. Let $I$ be an ideal of $R$ such that $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$. Then the natural map $\mathfrak{m}^{p-1}\operatorname{K}_{i}(I)\rightarrow\operatorname{H}_{i}(I)$ is surjective and therefore splits for all $i\geq 1$.

Proof.

Let $x_{1},\dots,x_{n}$ be a minimal generating set of $\mathfrak{n}$, and let $\overline{x}_{i}$ be the image of $x_{i}$ in $R$. We may assume that $I=(\overline{x}_{1},\dots,\overline{x}_{r})$ for some $1\leq r\leq n$. If $r=1$ then $I=(\overline{x}_{1})$ and we have $\operatorname{H}_{1}(I)=(0:_{R}\overline{x}_{1})\operatorname{K}_{1}(I)=\mathfrak{m}^{p-1}\operatorname{K}_{1}(I)$.

Assume $r>1$ and set $J=(\overline{x}_{1},\dots,\overline{x}_{r-1})$. There exists an exact sequence

[TABLE]

By induction hypothesis $f_{i}:\mathfrak{m}^{p-1}\operatorname{K}_{i}(J)\rightarrow\operatorname{H}_{i}(J)$ is surjective and so $\operatorname{H}_{i}(J)$ is a $k$-vector space for all $i\geq 1$. Thus the multiplication with $\overline{x}_{r}$ is zero map when $i\geq 1$. Therefore for $i\geq 2$ there is a commutative diagram

[TABLE]

By induction $f_{i-1}$ and $f_{i}$ are surjective and therefore $g_{i}$ is surjective for $i\geq 2$. When $i=1$ the last diagram turns to

[TABLE]

Note that $R/J\cong Q^{\prime}/\mathfrak{n}^{\prime p}$ where $Q^{\prime}=Q/(x_{1},\dots,x_{r-1})$ is a regular local ring whose maximal ideal is $\mathfrak{n}^{\prime}=(\overline{x}_{r},\dots,\overline{x}_{n})$. One observes that $f_{0}$ is a natural map of $k$-vector spaces such that $f_{0}(\overline{x}_{r}^{\alpha_{r}}\cdots\overline{x}_{n}^{\alpha_{n}})=\overline{x}_{r}^{\alpha_{r}}\cdots\overline{x}_{n}^{\alpha_{n}}$ where $\alpha_{r}+\dots+\alpha_{n}=p-1$. Therefore $f_{0}$ is surjective and since $f_{1}$ is surjective by induction, $g_{1}$ is surjective as well. ∎

Corollary 3.15**.**

Let $(Q,\mathfrak{n},k)$ be a regular local ring. Let $R=Q/\mathfrak{n}^{p}$ where $p\geq 2$, and let $\mathfrak{m}$ be the maximal ideal of $R$. Then for any ideal $I$ of $R$ such that $I\cap\mathfrak{m}^{2}=\mathfrak{m}I$ one has $R\rightarrow R/I$ is a large homomorphism.

Proof.

It is well-known that $R$ is a Golod ring; see [12, Theorem 4.2.6]. Set $S=R/I$ and let $\overline{\mathfrak{m}}$ be the maximal ideal of $S$. By Lemma 3.14, $\mathfrak{m}^{p-1}\operatorname{K}_{i}(R)\rightarrow\operatorname{H}_{i}(R)$ and $\overline{\mathfrak{m}}^{p-1}\operatorname{K}_{i}(S)\rightarrow\operatorname{H}_{i}(S)$ are surjective for all $i\geq 1$. Since $\mathfrak{m}^{p-1}\operatorname{K}_{i}(R)\rightarrow\overline{\mathfrak{m}}^{p-1}\operatorname{K}_{i}(S)$ is also surjective, the commutative diagram

[TABLE]

shows that the induced map $\operatorname{H}_{i}(R)\rightarrow\operatorname{H}_{i}(S)$ is surjective for all $i\geq 1$. Now the result follows from Proposition 3.12 . ∎

Acknowledgments**.**

The authors thank Rasoul Ahangari Maleki, David Jorgensen, and Liana Şega for their valuable suggestions and comments on the manuscript. The authors also thank the referee for providing very useful comments and suggestions that improved this article.