1.1. Let $(R,\mathfrak{m},k)$ be a local ring and $M$ a finite (meaning finitely generated) $R$-module. Fix a minimal generating set of $\mathfrak{m}$ and let $\operatorname{K}^{R}$ denote the Koszul complex on this set. Let $\operatorname{H}^{R}$ denote the homology algebra of $\operatorname{K}^{R}$. The complex $\operatorname{K}^{R}$ has a natural structure of a graded commutative algebra, and this structure is inherited by $\operatorname{H}^{R}$. We denote by $\operatorname{K}^{M}$ the Koszul complex $\operatorname{K}^{R}\otimes _{R}M$.

The Poincaré series $\operatorname{P}_{M}^{R}(z)$ of $M$ is defined as

$$\begin{eqnarray}\operatorname{P}_{M}^{R}(z)=\mathop{\sum }_{i\geqslant 0}\operatorname{rank}_{k}(\operatorname{Tor}_{i}^{R}(M,k))z^{i}.\end{eqnarray}$$

If $\unicode[STIX]{x1D719}:(R,\mathfrak{m},k)\rightarrow (S,\mathfrak{n},k)$ is a surjective homomorphism of local rings then the following coefficientwise inequality holds

$$\begin{eqnarray}\operatorname{P}_{M}^{S}(z)\preccurlyeq \frac{\operatorname{P}_{M}^{R}(z)}{1-z(\operatorname{P}_{S}^{R}(z)-1)}.\end{eqnarray}$$

If equality holds for $M=k$ then we say that $\unicode[STIX]{x1D719}$ is a Golod homomorphism.

The homomorphism $\unicode[STIX]{x1D719}$ induces maps

$$\begin{eqnarray}\operatorname{Ext}_{\unicode[STIX]{x1D719}}^{i}(k,k):\operatorname{Ext}_{S}^{i}(k,k)\rightarrow \operatorname{Ext}_{R}^{i}(k,k).\end{eqnarray}$$

If $\operatorname{Ext}_{\unicode[STIX]{x1D719}}^{\ast }(k,k)$ is surjective then we say $\unicode[STIX]{x1D719}$ is small. Recall that if $\unicode[STIX]{x1D719}$ is Golod then $\unicode[STIX]{x1D719}$ is small (cf. Avramov [Reference Avramov1, 3.5]).

Proof. As $\mathfrak{m}^{s}\cong k^{a}$, we have $\operatorname{P}_{R/\mathfrak{m}^{s}}^{R}(z)=az\operatorname{P}_{k}^{R}(z)+1$.

Since $R/\mathfrak{m}^{s}$ is Golod and $R\rightarrow R/\mathfrak{m}^{s}$ is small, $R\rightarrow R/\mathfrak{m}^{s}$ is a Golod homomorphism by [Reference Şega26, 6.7]. Therefore, we have

$$\begin{eqnarray}\operatorname{P}_{k}^{R}(z)=\operatorname{P}_{k}^{R/\mathfrak{m}^{s}}(z)(1-z(\operatorname{P}_{R/\mathfrak{m}^{s}}^{R}(z)-1))=\operatorname{P}_{k}^{R/\mathfrak{m}^{s}}(z)(1-az^{2}\operatorname{P}_{k}^{R}(z)).\end{eqnarray}$$

By rearranging, we have that

(1.2.2)$$\begin{eqnarray}\operatorname{P}_{k}^{R}(z)=\frac{\operatorname{P}_{k}^{R/\mathfrak{m}^{s}}(z)}{1+az^{2}\operatorname{P}_{k}^{R/\mathfrak{m}^{s}}(z)}.\end{eqnarray}$$

Finally, since $R/\mathfrak{m}^{s}$ is Golod, we have that

(1.2.3)$$\begin{eqnarray}\operatorname{P}_{k}^{R/\mathfrak{m}^{s}}(z)=\frac{(1+z)^{n}}{1-z(\operatorname{H}^{R/\mathfrak{m}^{s}}(z)-1)}.\end{eqnarray}$$

The conclusion follows from (1.2.2) and (1.2.3).

1.3. Let $\unicode[STIX]{x1D711}:R\rightarrow S$ be a homomorphism of local rings. The usual products of $\operatorname{H}^{R}$ are understood to be the set of products $\operatorname{H}_{{\geqslant}1}^{R}\cdot \operatorname{H}_{{\geqslant}1}^{R}$. We denote by $MH(\operatorname{K}^{R})$ the set of matric Massey products of $\operatorname{H}_{{\geqslant}1}^{R}$, as defined in [Reference May20]. This set of higher order homology operations is a submodule of $\operatorname{H}_{{\geqslant}1}^{R}$ and contains the usual products; see also [Reference Avramov2, (1.4.1)] for a more concise definition.

2.1. Adjunction of variables. Let $B$ be a DG algebra over $R$ and suppose $z$ is a cycle in $B$. We embed $B$ into a DG algebra $B^{\prime }=B\langle y\rangle$ by freely adjoining a variable $y$ such that $\unicode[STIX]{x2202}(y)=z$ as follows:

If $|z|$ is even, the variable $y$ such that $\unicode[STIX]{x2202}(y)=z$ is called an exterior variable and satisfies $y^{2}=0$. Denote by $k\langle y\rangle$ the exterior algebra over $k$ of a free $k$-module on a generator of degree $|z|+1$. The differential on $B\langle y\rangle =B\otimes _{k}k\langle y\rangle$ is given by

$$\begin{eqnarray}\unicode[STIX]{x2202}(b_{0}+b_{1}y)=\unicode[STIX]{x2202}(b_{0})+\unicode[STIX]{x2202}(b_{1})y+(-1)^{|b_{1}|}b_{1}z.\end{eqnarray}$$

If $|z|>0$ is odd, $y$ is a divided powers variable. The $k$-algebra $k\langle y\rangle$ on a divided powers variable $y$ is the free $k$-module with basis $\{y^{(i)}:|y^{(i)}|=i|y|\}_{i\geqslant 0}$ and multiplication table

$$\begin{eqnarray}y^{(i)}y^{(j)}=\binom{i+j}{i}y^{(i+j)},\quad \text{for }i,j\geqslant 0.\end{eqnarray}$$

We set $y^{(1)}=y$, $y^{(0)}=1$, and $y^{(i)}=0$ for $i<0$. Forgetting the differentials, $B\langle y\rangle =B\otimes _{k}k\langle y\rangle$. If $z\in B$ is a cycle of positive odd degree, then

$$\begin{eqnarray}\unicode[STIX]{x2202}\left(\mathop{\sum }_{i}b_{i}y^{(i)}\right)=\mathop{\sum }_{i}\unicode[STIX]{x2202}(b_{i})y^{(i)}+\mathop{\sum }_{i}(-1)^{|b_{i}|}b_{i}zy^{(i-1)}\end{eqnarray}$$

is a differential on $B\langle y\rangle$ that extends that of $B$ and satisfies the Leibniz rule.

2.2. The Tate resolution. Let $x_{1},\ldots ,x_{n}$ be a minimal generating set for $\mathfrak{m}$ and $\operatorname{K}^{R}$ the Koszul complex on $x_{1},\ldots ,x_{n}$. Note that we can interpret $\operatorname{K}^{R}$ as the DG algebra

$$\begin{eqnarray}\operatorname{K}^{R}=R\langle T_{1},\ldots ,T_{n}\rangle ,\end{eqnarray}$$

where $T_{i}$ are degree 1 exterior variables (these exterior variables in degree 1 will be referred to as Koszul variables) with $\unicode[STIX]{x2202}(T_{i})=x_{i}$. One can continue to “kill” homology by adjoining variables to $\operatorname{K}^{R}$, following the construction in [Reference Avramov3, 6.3.1]. The resulting DG algebra ${\mathcal{A}}$ is a minimal free resolution of $k$ over $R$, often referred to as the Tate resolution of $k$ over $R$.

2.3. Notation. We write $\operatorname{K}$ for $\operatorname{K}^{R}$ when the ring $R$ is understood. Since ${\mathcal{A}}$ is a free algebra over $\operatorname{K}$, we consider a homogeneous $\operatorname{K}$-basis of ${\mathcal{A}}$. For each $j$, let $\unicode[STIX]{x1D712}_{i,j}$ with $1\leqslant i\leqslant q_{j}$ denote the elements of homological degree $j$ in this basis. If $z\in {\mathcal{A}}_{p}$, then we write $z$ in terms of this basis as

(2.3.1)$$\begin{eqnarray}z=\mathop{\sum }_{j=0}^{p}\mathop{\sum }_{i=1}^{q_{j}}z_{i,j}\unicode[STIX]{x1D712}_{i,p-j}\end{eqnarray}$$

with $z_{i,j}\in \operatorname{K}_{j}$ for each $j$.

As usual, $Z({\mathcal{A}})$ denotes the set of cycles of ${\mathcal{A}}$ and $B({\mathcal{A}})$ denotes the set of boundaries.

Proof. Since $z\in Z({\mathcal{A}})$, we have $\unicode[STIX]{x2202}(z)=0$. On the other hand we can compute $\unicode[STIX]{x2202}(z)$ from (2.3.1), using the Leibniz rule; this yields:

(2.4.1)$$\begin{eqnarray}0=\mathop{\sum }_{j=0}^{p}\mathop{\sum }_{i=1}^{q_{j}}(-1)^{j}z_{i,j}\unicode[STIX]{x2202}(\unicode[STIX]{x1D712}_{i,p-j})+\unicode[STIX]{x2202}(z_{i,j})\unicode[STIX]{x1D712}_{i,p-j}.\end{eqnarray}$$

Let $i$ be such that $1\leqslant i\leqslant q_{a}$. We express all terms in the right-hand side of (2.4.1) in terms of the $\operatorname{K}$-basis of ${\mathcal{A}}$, and collect the terms to compute the coefficient of each basis element in the sum. We see that the coefficient of $\unicode[STIX]{x1D712}_{i,p-a}$ in this sum is

$$\begin{eqnarray}\unicode[STIX]{x2202}(z_{i,a})+\mathop{\sum }_{j=0}^{a-1}\mathop{\sum }_{i^{\prime }=1}^{q_{j}}z_{i^{\prime },j}w_{i^{\prime },j}\quad \text{with }w_{i^{\prime },j}\in \mathfrak{m}\operatorname{K}_{a-j}.\end{eqnarray}$$

The coefficients $w_{i^{\prime },j}$ come from expressing $\unicode[STIX]{x2202}(\unicode[STIX]{x1D712}_{i,p-j})$ in terms of the $\operatorname{K}$-basis, for $j\leqslant a-1$. In particular, $w_{i^{\prime },j}\in \mathfrak{m}\operatorname{K}$ since ${\mathcal{A}}$ is minimal. (Note that if $j\geqslant a$ then $z_{i,j}\unicode[STIX]{x2202}(\unicode[STIX]{x1D712}_{i,p-j})$ does not have any contribution to the coefficient of $\unicode[STIX]{x1D712}_{i,p-a}$, for degree reasons.) These coefficients of $\unicode[STIX]{x1D712}_{i,p-a}$ must equal $0$, hence

$$\begin{eqnarray}\unicode[STIX]{x2202}(z_{i,a})=-\mathop{\sum }_{j=0}^{a-1}\mathop{\sum }_{i^{\prime }=1}^{q_{j}}z_{i^{\prime },j}w_{i^{\prime },j}\in (\mathfrak{m}^{t}\operatorname{K})(\mathfrak{m}\operatorname{K})\subseteq \mathfrak{m}^{t+1}\operatorname{K}\end{eqnarray}$$

for all $i$ with $1\leqslant i\leqslant q_{a}$.

2.5. Let $\widehat{R}$ denote the completion of $R$ with respect to $\mathfrak{m}$. We may write $\widehat{R}=Q/I$, with $(Q,\mathfrak{n},k)$ a regular local ring and $I\subseteq \mathfrak{n}^{2}$; this presentation is called a minimal Cohen presentation. We set

$$\begin{eqnarray}v(R)=\max \{j\mid I\subseteq \mathfrak{n}^{j}\}.\end{eqnarray}$$

As noted in [Reference Hoffmeier and Şega13], this integer is independent of the choice of the minimal Cohen presentation.

Proof. Let $x\in {\mathcal{A}}_{p+1}$. If $p=0$, we may take $y=x$ and the result follows trivially. Now assume $p\geqslant 1$. Since ${\mathcal{A}}$ is minimal, we have $\unicode[STIX]{x2202}(x)=\sum x_{i}g_{i}$ with $x_{i}\in \mathfrak{m}$ and $g_{i}\in {\mathcal{A}}_{p}$. Choosing $A_{i}\in \operatorname{K}_{1}$ such that $\unicode[STIX]{x2202}(A_{i})=x_{i}$, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x2202}\left(x-\sum A_{i}g_{i}\right) & = & \displaystyle \unicode[STIX]{x2202}(x)-\sum \unicode[STIX]{x2202}(A_{i}g_{i})\nonumber\\ \displaystyle & = & \displaystyle \sum x_{i}g_{i}-\sum \unicode[STIX]{x2202}(A_{i})g_{i}+\sum A_{i}\unicode[STIX]{x2202}(g_{i})\nonumber\\ \displaystyle & = & \displaystyle \sum A_{i}\unicode[STIX]{x2202}(g_{i}).\nonumber\end{eqnarray}$$

Set $y^{\prime }=\sum A_{i}g_{i}$. Then $y^{\prime }\in \operatorname{K}_{1}{\mathcal{A}}$ and $\unicode[STIX]{x2202}(x-y^{\prime })\in \operatorname{K}_{1}{\mathcal{A}}$.

Apply Lemma 2.4 with $z=\unicode[STIX]{x2202}(x-y^{\prime })$ and $a=1$ (noting that $z_{i,0}=0$ for all $i$ with $1\leqslant i\leqslant q_{0}$, so the hypothesis is satisfied). For all $i$ with $0\leqslant i\leqslant q_{1}$ this yields $\unicode[STIX]{x2202}(z_{i,1})\in \mathfrak{m}^{t+1}\operatorname{K}$ (indeed, this holds for all $t$ in Lemma 2.4, hence $\unicode[STIX]{x2202}(z_{i,1})=0$), and then Remark 2.6 shows

$$\begin{eqnarray}z_{i,1}=\unicode[STIX]{x2202}(e_{i,1})+f_{i,1}\end{eqnarray}$$

for some $e_{i,1}\in \operatorname{K}_{2}$ and $f_{i,1}\in \mathfrak{m}^{t}\operatorname{K}$. Consequently, we have:

$$\begin{eqnarray}\unicode[STIX]{x2202}(x-y^{\prime })=\mathop{\sum }_{i=1}^{q_{1}}(\unicode[STIX]{x2202}(e_{i,1})+f_{i,1})\unicode[STIX]{x1D712}_{i,p-1}+V,\quad \text{with }V\in \operatorname{K}_{2}{\mathcal{A}}.\end{eqnarray}$$

Now take $y_{1}=\sum _{i=1}^{q_{1}}e_{i,1}\unicode[STIX]{x1D712}_{i,p-1}$ and we have:

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x2202}(x-y^{\prime }-y_{1}) & = & \displaystyle \mathop{\sum }_{i=1}^{q_{1}}(\unicode[STIX]{x2202}(e_{i,1})+f_{i,1})\unicode[STIX]{x1D712}_{i,p-1}+V-\mathop{\sum }_{i=1}^{q_{1}}\unicode[STIX]{x2202}(e_{i,1}\unicode[STIX]{x1D712}_{i,p-1})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{i=1}^{q_{1}}f_{i,1}\unicode[STIX]{x1D712}_{i,p-1}+\left(V+\mathop{\sum }_{i=1}^{q_{1}}e_{i,1}\unicode[STIX]{x2202}(\unicode[STIX]{x1D712}_{i,p-1})\right)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{i=1}^{q_{1}}f_{i,1}\unicode[STIX]{x1D712}_{i,p-1}+V_{1},\quad \text{with }V_{1}\in \operatorname{K}_{2}{\mathcal{A}}.\nonumber\end{eqnarray}$$

Set $m=\min \{p,n\}$, and let us assume inductively, for $a-1<m$, that we constructed $y_{1},y_{2},\ldots ,y_{a-1}\in \operatorname{K}_{1}{\mathcal{A}}$ such that

$$\begin{eqnarray}\unicode[STIX]{x2202}(x-y^{\prime }-y_{1}-\cdots -y_{a-1})=\mathop{\sum }_{j=1}^{a-1}\mathop{\sum }_{i=1}^{q_{j}}f_{i,j}\unicode[STIX]{x1D712}_{i,p-j}+V_{a-1},\end{eqnarray}$$

with $f_{i,j}\in \mathfrak{m}^{t}\operatorname{K}$ and $V_{a-1}\in \operatorname{K}_{a}{\mathcal{A}}$. Applying again Lemma 2.4 and Remark 2.6, with $z=\unicode[STIX]{x2202}(x-y^{\prime }-y_{1}-\cdots -y_{a-1})$ we have that

$$\begin{eqnarray}z_{i,a}=\unicode[STIX]{x2202}(e_{i,a})+f_{i,a}\end{eqnarray}$$

with $e_{i,a}\in \operatorname{K}_{a+1}$ and $f_{i,a}\in \mathfrak{m}^{t}\operatorname{K}_{a}$. Consequently, we can write

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x2202}(x-y^{\prime }-y_{1}-\cdots -y_{a-1}) & = & \displaystyle \mathop{\sum }_{j=1}^{a-1}\mathop{\sum }_{i=1}^{q_{j}}f_{i,j}\unicode[STIX]{x1D712}_{i,p-j}\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{i=1}^{q_{a}}(\unicode[STIX]{x2202}(e_{i,a})+f_{i,a})\unicode[STIX]{x1D712}_{i,p-a}+V,\nonumber\end{eqnarray}$$

with $V\in \operatorname{K}_{a+1}{\mathcal{A}}$. Now take $y_{a}=\sum _{i=1}^{q_{a}}e_{i,a}\unicode[STIX]{x1D712}_{i,p-a}$ and, as above, we get:

$$\begin{eqnarray}\unicode[STIX]{x2202}(x-y^{\prime }-y_{1}-\cdots -y_{a})=\mathop{\sum }_{j=1}^{a}\mathop{\sum }_{i=1}^{q_{j}}f_{i,j}\unicode[STIX]{x1D712}_{i,p-j}+V_{a},\end{eqnarray}$$

with $V_{a}\in \operatorname{K}_{a+1}{\mathcal{A}}$. Note that $V_{a}=0$ when $a\geqslant m$, by degree reasons (if $p<n$) and since $K_{{>}n}=0$.

Set $y=y^{\prime }+y_{1}+\cdots +y_{m}$. Then the cycle $z=\unicode[STIX]{x2202}(x-y)$ satisfies the conclusions of our statement.

Proof. For $x\in {\mathcal{A}}$, Lemma 2.7 provides an element $y\in \operatorname{K}_{1}{\mathcal{A}}$ such that $x-y\in {\mathcal{A}}^{\prime }$. A reformulation of Lemma 2.7 therefore gives that ${\mathcal{A}}={\mathcal{A}}^{\prime }+\operatorname{K}_{1}{\mathcal{A}}$. Applying this fact repeatedly, and noting that $\operatorname{K}_{n+1}=0$, we get:

$$\begin{eqnarray}\displaystyle {\mathcal{A}} & = & \displaystyle {\mathcal{A}}^{\prime }+\operatorname{K}_{1}{\mathcal{A}}={\mathcal{A}}^{\prime }+\operatorname{K}_{1}({\mathcal{A}}^{\prime }+\operatorname{K}_{1}{\mathcal{A}})={\mathcal{A}}^{\prime }+\operatorname{K}_{1}{\mathcal{A}}^{\prime }+\operatorname{K}_{2}{\mathcal{A}}=\cdots \nonumber\\ \displaystyle & = & \displaystyle {\mathcal{A}}^{\prime }+\operatorname{K}_{1}{\mathcal{A}}^{\prime }+\operatorname{K}_{2}{\mathcal{A}}^{\prime }+\cdots +\operatorname{K}_{n}({\mathcal{A}}^{\prime }+\operatorname{K}_{1}{\mathcal{A}})\nonumber\\ \displaystyle & = & \displaystyle {\mathcal{A}}^{\prime }+\operatorname{K}_{1}{\mathcal{A}}^{\prime }+\operatorname{K}_{2}{\mathcal{A}}^{\prime }+\cdots +\operatorname{K}_{n}{\mathcal{A}}^{\prime }\nonumber\\ \displaystyle & = & \displaystyle \operatorname{K}{\mathcal{A}}^{\prime }.\nonumber\end{eqnarray}$$

3.2. [Reference Rossi and Şega24, Lemma 1.2]

Let $\unicode[STIX]{x1D705}:(R,\mathfrak{m},k)\rightarrow (\overline{R},\overline{\mathfrak{m}},k)$ be a surjective homomorphism of local rings. If there exists a positive integer $a$ such that:

1. (a) the map $\operatorname{Tor}_{i}^{R}(\overline{R},k)\rightarrow \operatorname{Tor}_{i}^{R}(\overline{R}/\overline{\mathfrak{m}}^{a},k)$ induced by the canonical quotient map $\overline{R}\rightarrow \overline{R}/\overline{\mathfrak{m}}^{a}$ is zero for all positive $i$, and

2. (b) the map $\operatorname{Tor}_{i}^{\overline{R}}(\overline{\mathfrak{m}}^{2a},k)\rightarrow \operatorname{Tor}_{i}^{\overline{R}}(\overline{\mathfrak{m}}^{a},k)$ induced by the inclusion $\overline{\mathfrak{m}}^{2a}\subseteq \overline{\mathfrak{m}}^{a}$ is zero for all non-negative integers $i$,

then $\unicode[STIX]{x1D705}$ is a Golod homomorphism.

Proof. We apply 3.2 to the natural projection $\unicode[STIX]{x1D705}:R\rightarrow R/\mathfrak{m}^{s}$, with $\overline{R}=R/\mathfrak{m}^{s}$, $\overline{\mathfrak{m}}=\mathfrak{m}/\mathfrak{m}^{s}$ and $a=b$. We need to check that conditions (a) and (b) hold. Since $2a=2b\geqslant s$, we have $\overline{\mathfrak{m}}^{2a}=0$, so condition 3.2(b) holds trivially. Theorem 3.1 gives that the induced maps $\operatorname{Tor}_{i}^{R}(\mathfrak{m}^{s},k)\rightarrow \operatorname{Tor}_{i}^{R}(\mathfrak{m}^{b},k)$ are zero for all $i>0$, and this implies condition 3.2(a), since $R/\mathfrak{m}^{s}=\overline{R}$ and $R/\mathfrak{m}^{b}=\overline{R}/\overline{\mathfrak{m}}^{a}$.

Proof. The homomorphism $R\rightarrow R/\mathfrak{m}^{s}$ is Golod by Theorem 3.1, and thus small (see 1.1). The ring $R/\mathfrak{m}^{s}$ is Golod by Remark 3.4.

Proof of Theorem 3.1.

Let $|Z|$ denote the cardinality of $Z$. Let $\{z_{1},\ldots ,z_{|Z|}\}$ be the cycles in $Z$ and let ${\mathcal{I}}$ denote the set of all finite ordered lists of elements in $\{1,\ldots ,|Z|\}$, including the empty set.

Let $I\in {\mathcal{I}}$. If $I=(i)$ has length $1$, we set $I^{-}=\emptyset$. If $I=(i_{1},\ldots ,i_{r})$ has length $r\geqslant 2$, we set $I^{-}=(i_{1},\ldots ,i_{r-1})$. We now define for each $I\in {\mathcal{I}}$ an element $y_{I}\in {\mathcal{A}}$, where ${\mathcal{A}}$ is the Tate resolution of $k$ as in 2.2, such that

1. (1) $y_{I}=1\in {\mathcal{A}}_{0}$, if $I=\emptyset$;

2. (2) $\unicode[STIX]{x2202}(y_{I})=z_{i_{r}}y_{I^{-}}$, if $I=(i_{1},\ldots ,i_{r})$ with $r\geqslant 1$.

The details of constructing these elements are as follows. If $I=\emptyset$, we choose $y_{I}$ as in (1). If $r=1$ and $I=(i)$, we can choose $y_{I}\in {\mathcal{A}}$ such that $\unicode[STIX]{x2202}(y_{I})=z_{i}$, since $z_{i}\in Z$ is a cycle. Assuming that $r\geqslant 2$ and the elements $y_{I}$ have been defined for all $I\in {\mathcal{I}}$ of length $r-1$, we can construct elements $y_{I}$ satisfying (2) for $I=(i_{1},\ldots ,i_{r})$ by noting that $z_{i_{r}}y_{I^{-}}$ is a cycle, so such a $y_{I}$ exists since ${\mathcal{A}}$ is acyclic. Indeed, we have

$$\begin{eqnarray}\unicode[STIX]{x2202}(z_{i_{r}}y_{I^{-}})=(-1)^{|z_{i_{r}}|}z_{i_{r}}\unicode[STIX]{x2202}(y_{I^{-}})=(-1)^{|z_{i_{r}}|}z_{i_{r}}z_{i_{r-1}}y_{(I^{-})^{-}}=0,\end{eqnarray}$$

where the last equality is due to the hypothesis on the set $Z$.

We identify the map $\operatorname{Tor}_{i}^{R}(\mathfrak{m}^{s},k)\rightarrow \operatorname{Tor}_{i}^{R}(\mathfrak{m}^{b},k)$ with the map $H_{i}(\mathfrak{m}^{s}{\mathcal{A}})\rightarrow H_{i}(\mathfrak{m}^{b}{\mathcal{A}})$. In order to show this map is trivial for $i\geqslant 0$, we let $x\in \mathfrak{m}^{s}{\mathcal{A}}_{c}$ for some $c$ and must show that $x\in \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})$.

Set $Y=\{y_{I}\in {\mathcal{A}}|I\in {\mathcal{I}}\}$. For $i,j\geqslant 0,$ define

$$\begin{eqnarray}{\mathcal{A}}(i,j):=\left(\mathfrak{m}^{s}\operatorname{K}_{i}Y_{j}{\mathcal{A}}^{\prime }\right)\cap {\mathcal{A}}_{c},\end{eqnarray}$$

where $Y_{j}$ is the set of elements of $Y$ of degree $j$.

Claim.

For $i,j\geqslant 0$,

(3.5.1)$$\begin{eqnarray}{\mathcal{A}}(i,j)\subseteq \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})+\mathop{\sum }_{\stackrel{p+q=i+j}{p>i}}{\mathcal{A}}(p,q)+\mathop{\sum }_{p+q>i+j}{\mathcal{A}}(p,q),\end{eqnarray}$$

where only finitely many terms in this sum are not zero for degree reasons.

Proof of Claim.

To prove the inclusion, it suffices to consider elements of ${\mathcal{A}}(i,j)$ of the form $vy_{I}a^{\prime }$ for some $v\in \mathfrak{m}^{s}\operatorname{K}_{i}$, $a^{\prime }\in {\mathcal{A}}^{\prime }$ and $y_{I}\in Y_{j}$, with $I=(i_{1},\ldots ,i_{r})\in {\mathcal{I}}$ or $I=\emptyset$; in the last case, we set $r=0$. (Note that every element of ${\mathcal{A}}(i,j)$ can be written as a sum of elements of this form.)

By assumption there exist $z_{\unicode[STIX]{x1D704},v}\in Z$ and $u_{\unicode[STIX]{x1D704},v}\in Z(\mathfrak{m}^{b}\operatorname{K})$ with $1\leqslant \unicode[STIX]{x1D704}\leqslant m$ such that

$$\begin{eqnarray}v-\mathop{\sum }_{\unicode[STIX]{x1D704}=1}^{m}z_{\unicode[STIX]{x1D704},v}u_{\unicode[STIX]{x1D704},v}\in \unicode[STIX]{x2202}(\mathfrak{m}^{s-1}\operatorname{K}_{i+1}).\end{eqnarray}$$

We need to show thus that $(\sum _{\unicode[STIX]{x1D704}=1}^{m}z_{\unicode[STIX]{x1D704},v}u_{\unicode[STIX]{x1D704},v}+w)y_{I}{\mathcal{A}}^{\prime }$ is contained in the right-hand side of (3.5.1), for all $w\in \unicode[STIX]{x2202}(\mathfrak{m}^{s-1}\operatorname{K}_{i+1})$, $z_{\unicode[STIX]{x1D704},v}\in Z$ and $u_{\unicode[STIX]{x1D704},v}\in Z(\mathfrak{m}^{b}\operatorname{K})$. Note that it suffices to prove this statement when $m=1$. We assume thus that $v-z_{i_{r+1}}u\in \unicode[STIX]{x2202}(\mathfrak{m}^{s-1}\operatorname{K}_{i+1})$ for some $z_{i_{r+1}}\in Z$ and $u\in Z(\mathfrak{m}^{b}\operatorname{K})$ and we show that $vy_{I}{\mathcal{A}}^{\prime }$ is contained in the right-hand side of (3.5.1).

In what follows, we simplify the notation when $I$ has length $1$ and we write $y_{i}=y_{(i)}$. Define $I^{+}=(i_{1},\ldots ,i_{r},i_{r+1})$. Recall that we defined a set $I^{-}$ for any nonempty $I\in {\mathcal{I}}$. Although we do not define a set $\emptyset ^{-}$, we agree to set $y_{I^{-}}=z_{i_{r}}=y_{i_{r}}=0$ when $I=\emptyset$. With this convention, note that the formula $\unicode[STIX]{x2202}(y_{I})=z_{i_{r}}y_{I^{-}}$ also holds when $I=\emptyset$, so we will not treat this case separately.

Note that

$$\begin{eqnarray}\displaystyle |u| & = & \displaystyle i-|z_{i_{r+1}}|=i+1-|y_{i_{r+1}}|,\nonumber\\ \displaystyle |y_{I^{-}}| & = & \displaystyle j-|z_{i_{r}}|-1=j-|y_{i_{r}}|\quad \text{when }I\neq \emptyset ,\;\text{and}\nonumber\\ \displaystyle |y_{I^{+}}| & = & \displaystyle j+|z_{i_{r+1}}|+1=j+|y_{i_{r+1}}|.\nonumber\end{eqnarray}$$

We now have

$$\begin{eqnarray}\displaystyle vy_{I}{\mathcal{A}}^{\prime } & \subseteq & \displaystyle \left(\unicode[STIX]{x2202}(\mathfrak{m}^{s-1}\operatorname{K}_{i+1})+z_{i_{r+1}}u\right)y_{I}{\mathcal{A}}^{\prime }\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x2202}(\mathfrak{m}^{s-1}\operatorname{K}_{i+1})y_{I}{\mathcal{A}}^{\prime }+u\unicode[STIX]{x2202}(y_{I^{+}}){\mathcal{A}}^{\prime }\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x2202}(\mathfrak{m}^{s-1}\operatorname{K}_{i+1}y_{I}{\mathcal{A}}^{\prime })+u\unicode[STIX]{x2202}(y_{I^{+}}){\mathcal{A}}^{\prime }\nonumber\\ \displaystyle & & \displaystyle +\,\mathfrak{m}^{s-1}\operatorname{K}_{i+1}z_{i_{r}}y_{I^{-}}{\mathcal{A}}^{\prime }+\mathfrak{m}^{s-1}\operatorname{K}_{i+1}y_{I}\unicode[STIX]{x2202}({\mathcal{A}}^{\prime })\nonumber\\ \displaystyle & \subseteq & \displaystyle \unicode[STIX]{x2202}(\mathfrak{m}^{s-1}{\mathcal{A}})+\unicode[STIX]{x2202}(uy_{I^{+}}{\mathcal{A}}^{\prime })+uy_{I^{+}}\unicode[STIX]{x2202}({\mathcal{A}}^{\prime })\nonumber\\ \displaystyle & & \displaystyle +\,\mathfrak{m}^{s-1}\operatorname{K}_{i+1}z_{i_{r}}y_{I^{-}}{\mathcal{A}}^{\prime }+\mathfrak{m}^{s-1}\operatorname{K}_{i+1}y_{I}\unicode[STIX]{x2202}({\mathcal{A}}^{\prime }),\nonumber\end{eqnarray}$$

where in line 2 we used the formula $z_{i_{r+1}}y_{I}=\unicode[STIX]{x2202}(y_{I^{+}})$, in line 3 we used the Leibniz rule and the formula $\unicode[STIX]{x2202}(y_{I})=z_{i_{r}}y_{I^{-}}$, and in line 4 we used again the Leibniz rule and the fact that $u$ is a cycle.

Consider the terms from the last line in the previous display. Since $u\in Z(\mathfrak{m}^{b}\operatorname{K})$ and $b\leqslant s-1$, we have $\unicode[STIX]{x2202}(\mathfrak{m}^{s-1}{\mathcal{A}})+\unicode[STIX]{x2202}(uy_{I^{+}}{\mathcal{A}}^{\prime })\subseteq \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})$. Additionally, since $\unicode[STIX]{x2202}({\mathcal{A}}^{\prime })\subseteq \mathfrak{m}^{t}\operatorname{K}_{1}{\mathcal{A}}$ by the definition of ${\mathcal{A}}^{\prime }$, we have $uy_{I^{+}}\unicode[STIX]{x2202}({\mathcal{A}}^{\prime })\subseteq \mathfrak{m}^{b+t}\operatorname{K}_{|u|+1}y_{I^{+}}{\mathcal{A}}$ and $\mathfrak{m}^{s-1}\operatorname{K}_{i+1}y_{I}\unicode[STIX]{x2202}({\mathcal{A}}^{\prime })\subseteq \mathfrak{m}^{s-1+t}\operatorname{K}_{i+2}y_{I}{\mathcal{A}}$. Furthermore, since $z_{i_{r}}=\unicode[STIX]{x2202}(y_{i_{r}})$, we have $\mathfrak{m}^{s-1}\operatorname{K}_{i+1}z_{i_{r}}y_{I^{-}}{\mathcal{A}}^{\prime }\subseteq \mathfrak{m}^{s}\operatorname{K}_{i+|y_{i_{r}}|}y_{I^{-}}{\mathcal{A}}^{\prime }$. Thus

$$\begin{eqnarray}vy_{I}{\mathcal{A}}^{\prime }\subseteq \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})+\mathfrak{m}^{b+t}\!\operatorname{K}_{|u|+1}y_{I^{+}}{\mathcal{A}}+\mathfrak{m}^{s}\operatorname{K}_{i+|y_{i_{r}}|}y_{I^{-}}{\mathcal{A}}^{\prime }+\mathfrak{m}^{s-1+t}\!\operatorname{K}_{i+2}y_{I}{\mathcal{A}}.\end{eqnarray}$$

Using Proposition 2.8 and the facts $b+t\geqslant s$ and $t\geqslant 1$, we have

$$\begin{eqnarray}vy_{I}{\mathcal{A}}^{\prime }\subseteq \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})+\mathfrak{m}^{s}\operatorname{K}_{|u|+1}y_{I^{+}}{\mathcal{A}}^{\prime }+\mathfrak{m}^{s}\operatorname{K}_{i+|y_{i_{r}}|}y_{I^{-}}{\mathcal{A}}^{\prime }+\mathfrak{m}^{s}\operatorname{K}_{i+2}y_{I}{\mathcal{A}}^{\prime },\end{eqnarray}$$

with the provision that if $I=\emptyset$, the third term on the right of this inclusion is $0$ by convention. Since $|u|=i+1-|y_{i_{r+1}}|$, we conclude $vy_{I}{\mathcal{A}}^{\prime }$ is contained in

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})+\mathop{\sum }_{i^{\prime }\geqslant i+2-|y_{i_{r+1}}|}{\mathcal{A}}(i^{\prime },j+|y_{i_{r+1}}|)+{\mathcal{A}}(i+|y_{i_{r}}|,j-|y_{i_{r}}|)\nonumber\\ \displaystyle & & \displaystyle \quad +\,\mathop{\sum }_{i^{\prime }\geqslant i+1}{\mathcal{A}}(i^{\prime }+1,j),\nonumber\end{eqnarray}$$

with the caveat that we remove the term ${\mathcal{A}}(i+|y_{i_{r}}|,j-|y_{i_{r}}|)$ from the right-hand sum when $I=\emptyset$. In the display above, the second and last terms of the right-hand side are sums of the form ${\mathcal{A}}(p,q)$ with $p+q>i+j$ and the third term is of the form ${\mathcal{A}}(p,q)$ with $p+q=i+j$ and $p>i$. The Claim is thus proved.

To finish the proof of the theorem, order the set $M=\{(i,j)|i,j\geqslant 0\}$ as follows: Order the elements by $i+j$ first, then by $i$ as a tiebreak. In other words:

$$\begin{eqnarray}(i,j)>(i^{\prime },j^{\prime })\;\Longleftrightarrow \;i+j>i^{\prime }+j^{\prime }\qquad \text{or}\qquad (i+j=i^{\prime }+j^{\prime }~\text{and}~i>i^{\prime }).\end{eqnarray}$$

Recall that $x\in \mathfrak{m}^{s}{\mathcal{A}}_{c}$ and we need to show $x\in \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})$. Using Proposition 2.8, we know that $x\in \sum _{i=0}^{n}\mathfrak{m}^{s}\operatorname{K}_{i}{\mathcal{A}}^{\prime }$. Thus if $M_{0}=\{(i,0)\mid 0\leqslant i\leqslant n\}$, then

$$\begin{eqnarray}x\in \mathop{\sum }_{(i,j)\in M_{0}}{\mathcal{A}}(i,j).\end{eqnarray}$$

Using the Claim, we see that there exists a finite set $M_{1}$ and an element

$$\begin{eqnarray}x_{1}\in \mathop{\sum }_{(i,j)\in M_{1}}{\mathcal{A}}(i,j)\end{eqnarray}$$

such that $x-x_{1}\in \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})$, and such that the smallest element of $M_{1}$ is strictly larger (in the order described above) than the smallest element of $M_{0}$.

Applying the Claim again, this time using $x_{1}$, we see that there exists a finite set $M_{2}$ and an element $x_{2}$ such that

$$\begin{eqnarray}x_{2}\in \mathop{\sum }_{(i,j)\in M_{2}}{\mathcal{A}}(i,j)\end{eqnarray}$$

such that $x_{1}-x_{2}\in \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})$, hence $x-x_{2}\in \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})$, and such that the smallest element of $M_{2}$ is strictly larger (in the order described above) than the smallest element of $M_{1}$. A repeated use of the argument ensures the construction of elements $x_{a}$ and sets $M_{a}$ such that for each integer $a$ the smallest element of $M_{a}$ is greater than the smallest element of $M_{a-1}$ and such that $x-x_{a}\in \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})$. When $a$ is sufficiently large (so that $p+q>c$ for all $(p,q)\in M_{a}$) one sees that, for degree reasons, we have ${\mathcal{A}}(p,q)=0$ for all $(p,q)\in M_{a}$, since the elements of ${\mathcal{A}}(p,q)$ are in ${\mathcal{A}}_{c}$, with $c$ fixed. We conclude that $x_{a}=0$ for $a$ sufficiently large, hence $x\in \unicode[STIX]{x2202}(\mathfrak{m}^{b}{\mathcal{A}})$.

Proof. We first prove two claims.

Claim 1.

For a cycle $z\in K$, if $z=z^{\prime }+w$ with $z^{\prime }\in \mathfrak{m}^{t}\operatorname{K}_{p}y^{(q)}$ and $w\in \mathfrak{m}^{t}\operatorname{K}_{p+1}{\mathcal{B}}$, then $z^{\prime }\in Z_{p}(\mathfrak{m}^{t}\operatorname{K})y^{(q)}$.

Proof of Claim 1.

Set $|z|=c$, hence $p+(r+1)q=c$. Write $z^{\prime }=vy^{(q)}$ with $v\in \mathfrak{m}^{t}\operatorname{K}_{p}$. We have

(4.2.1)$$\begin{eqnarray}0=\unicode[STIX]{x2202}(z)=\unicode[STIX]{x2202}(vy^{(q)})+\unicode[STIX]{x2202}(w)=\unicode[STIX]{x2202}(v)y^{(q)}+(-1)^{p}vly^{(q-1)}+\unicode[STIX]{x2202}(w).\end{eqnarray}$$

Since $w\in \mathfrak{m}^{t}\operatorname{K}_{p+1}{\mathcal{B}}$ and $|w|=c$ we can write $\unicode[STIX]{x2202}(w)=\sum _{i,j}k_{i}y^{(j)}$ with $k_{i}\in \mathfrak{m}^{t}\operatorname{K}_{i}$, where the sum is taken over non-negative $i,j$ with $i+(r+1)j=c-1$, and $i\geqslant p$. Since $p+(r+1)q=c$, we must have $j<q$ for all such $j$. Since $1,y,y^{(2)},y^{(3)},\ldots \,$ is a basis of ${\mathcal{B}}$ over $\operatorname{K}$, we conclude from (4.2.1) that $\unicode[STIX]{x2202}(v)=0$, and hence $v\in Z_{p}(\mathfrak{m}^{t}\operatorname{K})$.

Claim 2.

If $z^{\prime }\in Z_{p}(\mathfrak{m}^{t}\operatorname{K})y^{(q)}$, then

$$\begin{eqnarray}z^{\prime }\in \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}{\mathcal{B}})+\mathfrak{m}^{t}\operatorname{K}_{p+r+1}y^{(q-1)}.\end{eqnarray}$$

Proof of Claim 2.

Let $z^{\prime }=vy^{(q)}$ with $v\in Z_{p}(\mathfrak{m}^{t}\operatorname{K})$. The hypothesis of the theorem gives $v-v^{\prime }l\in \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}\operatorname{K}_{p+1})$ for some $v^{\prime }\in Z_{p-r}(\mathfrak{m}^{t-1}\operatorname{K})$, and so $z^{\prime }-v^{\prime }ly^{(q)}\in \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}\operatorname{K}_{p+1})y^{(q)}$. An application of the Leibniz rule yields

$$\begin{eqnarray}z^{\prime }-v^{\prime }ly^{(q)}=z^{\prime }-v^{\prime }\unicode[STIX]{x2202}(y^{(q+1)})=z^{\prime }-(-1)^{|v^{\prime }|}\unicode[STIX]{x2202}(v^{\prime }y^{(q+1)}),\end{eqnarray}$$

and therefore

$$\begin{eqnarray}\displaystyle z^{\prime } & \in & \displaystyle \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}{\mathcal{B}})+\unicode[STIX]{x2202}(\mathfrak{m}^{t-1}\operatorname{K}_{p+1})y^{(q)}\nonumber\\ \displaystyle & \subseteq & \displaystyle \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}{\mathcal{B}})+\unicode[STIX]{x2202}(\mathfrak{m}^{t-1}\!\operatorname{K}_{p+1}y^{(q)})+\mathfrak{m}^{t-1}\operatorname{K}_{p+1}\unicode[STIX]{x2202}(y^{(q)}),\text{by the Leibniz rule,}\nonumber\\ \displaystyle & \subseteq & \displaystyle \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}{\mathcal{B}})+\mathfrak{m}^{t}\operatorname{K}_{p+1+r}y^{(q-1)},\nonumber\end{eqnarray}$$

which verifies Claim 2.

Now suppose $z\in {\mathcal{B}}$ represents a nontrivial class in $\operatorname{H}_{c}(\mathfrak{m}^{t}{\mathcal{B}})$ and let $p$ be the largest integer such that $z\in \mathfrak{m}^{t}\operatorname{K}_{p}{\mathcal{B}}$. Write $z=\sum k_{i}y^{(j)}$, where the sum is taken over integers $i,j$ with $i+(r+1)j=c$ and $i\geqslant p$, and $k_{i}\in \mathfrak{m}^{t}\operatorname{K}_{i}$. Set $z^{\prime }=k_{p}y^{(c-p/r+1)}$ and $w=z-z^{\prime }$. Then $z=z^{\prime }+w$ and the hypotheses of Claim 1 are satisfied. Putting together Claim 1 and Claim 2, we have:

(4.2.2)$$\begin{eqnarray}z\in \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}{\mathcal{B}})+z_{1},\quad \text{for some }z_{1}\in \mathfrak{m}^{t}\operatorname{K}_{p+1}{\mathcal{B}}.\end{eqnarray}$$

As $z_{1}$ is also a cycle, we may repeat the argument for $z_{1}$, and so on. Inductively, we obtain $z\in \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}{\mathcal{B}})+\mathfrak{m}^{t}\operatorname{K}_{i}{\mathcal{B}}$ for all $i>p$. Since $\operatorname{K}_{i}=0$ for $i\gg 0$, we get $z\in \unicode[STIX]{x2202}(\mathfrak{m}^{t-1}{\mathcal{B}})$.

4.3. Let ${\mathcal{F}}$ be a differential graded $R$-algebra and ${\mathcal{E}}$ a differential graded ${\mathcal{F}}$-module which is free as a graded ${\mathcal{F}}$-module (i. e. forgetting differentials) and such that $\unicode[STIX]{x2202}({\mathcal{E}})\subseteq \mathfrak{m}{\mathcal{E}}$. Let $M$, $N$ be $R$-modules such that $\mathfrak{m}M\subseteq N\subseteq M$, and such that the canonical map $N\otimes _{R}{\mathcal{E}}\rightarrow M\otimes _{R}{\mathcal{E}}$ is injective. If the induced homomorphism $\operatorname{H}(N\otimes _{R}{\mathcal{F}})\rightarrow \operatorname{H}(M\otimes _{R}{\mathcal{F}})$ is zero, then so is the induced homomorphism $\operatorname{H}(N\otimes _{R}{\mathcal{E}})\rightarrow \operatorname{H}(M\otimes _{R}{\mathcal{E}})$.

Proof. If $l\notin \mathfrak{m}Z_{1}(\operatorname{K})$, we can then construct a minimal Tate resolution ${\mathcal{A}}$ of $k$ over $R$ by starting with the Koszul complex K, then adjoining a variable $y$ with $\unicode[STIX]{x2202}(y)=l$, and then the rest of the needed variables as described in 2.2. The description of the basis of ${\mathcal{A}}$ in [Reference Avramov3, Remark 6.2.1] shows that, forgetting differentials, ${\mathcal{A}}$ is free over the algebra ${\mathcal{B}}=\operatorname{K}\langle y\mid \unicode[STIX]{x2202}(y)=l\rangle$. Using 4.3 and Theorem 4.2, we conclude that the induced map

$$\begin{eqnarray}\operatorname{H}_{i}(\mathfrak{m}^{t}{\mathcal{A}})\rightarrow \operatorname{H}_{i}(\mathfrak{m}^{t-1}{\mathcal{A}})\end{eqnarray}$$

is zero for all $i\geqslant 0$, and this yields the desired conclusion.

If $l\in \mathfrak{m}Z_{1}(\operatorname{K})$, then by Remark 4.1 the induced map $\operatorname{H}_{i}(\mathfrak{m}^{t}\operatorname{K})\rightarrow \operatorname{H}_{i}(\mathfrak{m}^{t-1}\operatorname{K})$ is zero for all $i\geqslant 0$, and the conclusion follows again by applying Levin’s result in 4.3.

In view of (1), part (2) follows from [Reference Hoffmeier and Şega13], and part (3) from [Reference Roos.22, Corollary 1].

Proof. The hypothesis implies that $h$ is part of a minimal generating set of $I$.

Set ${\mathcal{B}}=\operatorname{K}\langle y\mid \unicode[STIX]{x2202}(y)=l\rangle$ and ${\mathcal{B}}^{\prime }=\operatorname{K}^{Q}\langle y\mid \unicode[STIX]{x2202}(y)=L\rangle$. Note that ${\mathcal{B}}=$$({\mathcal{B}}^{\prime }\otimes _{Q}P)\otimes _{P}R$ and ${\mathcal{B}}^{\prime }\otimes _{Q}P$ is a minimal free resolution of $k$ over $P$. We can identify thus the induced map $\operatorname{Tor}_{i}^{P}(\mathfrak{m}^{t},k)\rightarrow \operatorname{Tor}_{i}^{P}(\mathfrak{m}^{t-1},k)$ with the induced map $\operatorname{H}_{i}(\mathfrak{m}^{t}{\mathcal{B}})\rightarrow \operatorname{H}_{i}(\mathfrak{m}^{t-1}{\mathcal{B}})$, and the latter is zero, by the theorem. Under the additional hypotheses in the last paragraph of the statement, it also follows that the induced map $\operatorname{Tor}_{i}^{P}(\mathfrak{m}^{2a},k)\rightarrow \operatorname{Tor}_{i}^{P}(\mathfrak{m}^{a},k)$ is zero for all $i\geqslant 0$, since $t\leqslant 2a$ and $t-1\geqslant a$. Since $I\subseteq \mathfrak{n}^{a+1}$, [Reference Rossi and Şega24, Lemma 1.4] gives that the induced map $\operatorname{Tor}_{i}^{P}(R,k)\rightarrow \operatorname{Tor}_{i}^{P}(R/\mathfrak{m}^{a},k)$ is zero for all positive $i$, and then applying 3.2 to the canonical projection $\unicode[STIX]{x1D705}:P\rightarrow R$ gives the final conclusion.

5.1. Structure of stretched artinian local rings. Assume that $(R,\mathfrak{m},k)$ is a stretched artinian local ring, not a field. We further set

$$\begin{eqnarray}p=v-r\qquad \text{and}\qquad q=r-1.\end{eqnarray}$$

Assume that $h\geqslant 3$. (The case $h\leqslant 2$ has been treated in [Reference Avramov, Iyengar and Şega5] and will be recalled later.) As described in [Reference Sally25], we choose elements $t,z_{1},\ldots ,z_{p},w_{1},\ldots ,w_{q}$ forming a minimal generating system of $\mathfrak{m}$ such that the following hold:

1. (1) $\mathfrak{m}^{i}=(t^{i})$ for all $i\geqslant 2$;

2. (2) The elements $w_{1},\ldots ,w_{r-1},t^{h}$ form a basis of $(0:\mathfrak{m})$; in particular, $tw_{j}=0$ and $z_{i}w_{j}=0$ for all $i$, $j$ with $1\leqslant i\leqslant p$ and $1\leqslant j\leqslant q$;

3. (3) $tz_{i}=0$ for all $i$ with $1\leqslant i\leqslant p$;

4. (4) $z_{i}z_{j}=a_{ij}t^{h}$, with $a_{ij}=0$ or $a_{ij}$ a unit of $R$, for all $i$, $j$ with $1\leqslant i,j\leqslant p$.

(Note that there are no elements $w_{i}$ if $r=1$ and there are no elements $z_{i}$ if $r=v$.)

If $r\neq v$, let $\overline{a_{ij}}$ denote the image of $a_{ij}$ in $k=R/\mathfrak{m}$ and note that the matrix $(\overline{a_{ij}})$ is invertible. Indeed, if this matrix is not invertible, then there exists an element $z=\sum _{i=1}^{p}b_{i}z_{i}$, with $b_{i}\in R$ such that $b_{i}$ is a unit for at least one index $i$, and such that $zz_{j}=0$ for all $j$ for all $1\leqslant j\leqslant p$. This implies that $z\in (0:\mathfrak{m})$, hence $z\in (w_{1},\ldots ,w_{r-1},t^{h})$, contradicting the fact that $t,z_{1},\ldots ,z_{p},w_{1},\ldots ,w_{q}$ is a minimal generating set of $\mathfrak{m}$.

5.2. Structure of Koszul homology. Let $R$ be as in 5.1. We consider the Koszul complex $K^{R}$ on the set $\{w_{1},\ldots ,w_{q},z_{1},\ldots ,z_{p},t\}$. As a DG algebra it can be described as the complex

$$\begin{eqnarray}K^{R}=R\langle W_{1},\ldots ,W_{q},Z_{1},\ldots ,Z_{p},T\rangle\end{eqnarray}$$

with $\unicode[STIX]{x2202}(W_{i})=w_{i}$, $\unicode[STIX]{x2202}(Z_{i})=z_{i}$ and $\unicode[STIX]{x2202}(T)=t$.

In what follows, we consider products of some of the variables $W_{i}$ or $Z_{i}$, with the index $i$ ranging over certain sets $I$. We adopt the convention that the product is equal to $1$ if $I=\emptyset$; for example $Z_{1}\cdots Z_{p}=1$ if $p=0$. We set

$$\begin{eqnarray}{\mathcal{W}}=\{W_{j_{1}}W_{j_{2}}\cdots W_{j_{i}}\mid 1\leqslant j_{1}<j_{2}<\cdots <j_{i}\leqslant q,\,0\leqslant i\leqslant q\},\end{eqnarray}$$

where $W_{j_{1}}W_{j_{2}}\cdots W_{j_{i}}=1$ when $i=0$.

Proof. If $L=Z_{i_{1}}Z_{i_{2}}\cdots Z_{i_{m}}W_{j_{1}}W_{j_{2}}\cdots W_{j_{n}}$ with indices satisfying the inequalities $1\leqslant i_{1}<i_{2}<\cdots <i_{m}\leqslant p$ and $1\leqslant j_{1}<j_{2}<\cdots <j_{n}\leqslant q$, then the Leibniz rule and the fact that $tw_{j}=0=tz_{i}$ for all $1\leqslant i\leqslant p$ and $1\leqslant j\leqslant q$ imply that $t\unicode[STIX]{x2202}(L)=0$. Consequently, another application of the Leibniz rule gives

$$\begin{eqnarray}t^{2}L=\unicode[STIX]{x2202}(tTL).\end{eqnarray}$$

Since $\mathfrak{m}^{2}=(t^{2})$, it suffices to consider cycles of the form $z=\sum _{L}a_{L}t^{2}TL$ with $L$ ranging over all possible products $L$ as above (if $p=0=q$, then we only have one term where $L=1$). Taking differentials, we have

$$\begin{eqnarray}0=\unicode[STIX]{x2202}(z)=\sum a_{L}t^{3}L,\end{eqnarray}$$

where the last equality follows again from the Leibniz rule and the fact that $t\unicode[STIX]{x2202}(L)=0$, as noted above. Since the products $L$ are linearly independent over $R$, we have $a_{L}t^{3}=0$ for all $L$, and hence $a_{L}=t^{h-2}b_{L}+c_{L}$, with $b_{L}\in R$ and $c_{L}\in (w_{1},\ldots ,w_{q},z_{1},\ldots ,z_{p})$. We have thus $z=\sum _{L}b_{L}t^{h}TL$.

Consider a cycle $t^{h}TL$, with $L$ as above. Set $Z=Z_{i_{1}}Z_{i_{2}}\cdots Z_{i_{m}}$ and $W=W_{j_{1}}W_{j_{2}}\cdots W_{j_{n}}$, so that $t^{h}TL=t^{h}TZW$. We show that $t^{h}TZW\in \unicode[STIX]{x2202}(\mathfrak{m}K^{R})$ whenever $Z\neq Z_{1}Z_{2}\cdots Z_{p}$.

Indeed, assume $Z\neq Z_{1}Z_{2}\cdots Z_{p}$. In particular, we have $p\neq 0$. Without loss of generality, we may assume $Z_{1}$ is not a factor in $Z$. Then for any such $Z$ and any $W$ as above we have:

$$\begin{eqnarray}t^{h}TZW=z_{1}y_{1}TZW=\unicode[STIX]{x2202}(y_{1}Z_{1}TZW),\end{eqnarray}$$

where the element $y_{1}$ is as defined above, and the last equality follows from the Leibniz rule, in view of the relations in (5.1.1) and (5.1.2).

5.3. Generation of Koszul homology. We continue with the notation and hypotheses of 5.1 and 5.2. In addition, we assume $r\neq v$, hence $p\neq 0$. We set:

$$\begin{eqnarray}{\mathcal{J}}_{1}=\{(i,j)\mid 1\leqslant i\leqslant j\leqslant p,a_{ij}\neq 0\}\text{; }{\mathcal{J}}_{2}=\{(i,j)\mid 1\leqslant i\leqslant j\leqslant p,a_{ij}=0\}.\end{eqnarray}$$

Consider the following element in $K_{1}^{R}$:

(5.3.1)$$\begin{eqnarray}\displaystyle F & = & \displaystyle \mathop{\sum }_{1\leqslant i<j\leqslant q}\unicode[STIX]{x1D6FC}_{ij}w_{i}W_{j}+\mathop{\sum }_{1\leqslant j\leqslant q,1\leqslant i\leqslant p}\unicode[STIX]{x1D6FD}_{ij}w_{j}Z_{i}+\mathop{\sum }_{1\leqslant i\leqslant q}\unicode[STIX]{x1D6FE}_{i}w_{i}T\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{1\leqslant i\leqslant p}\unicode[STIX]{x1D6FF}_{i}tZ_{i}+\mathop{\sum }_{(i,j)\in {\mathcal{J}}_{1}}\unicode[STIX]{x1D702}_{ij}(t^{h-1}T-a_{ij}^{-1}z_{i}Z_{j})+\mathop{\sum }_{(i,j)\in {\mathcal{J}}_{2}}\unicode[STIX]{x1D703}_{ij}z_{i}Z_{j},\end{eqnarray}$$

where $\unicode[STIX]{x1D6FC}_{ij}$, $\unicode[STIX]{x1D6FD}_{ij}$, $\unicode[STIX]{x1D6FE}_{i}$, $\unicode[STIX]{x1D6FF}_{i}$, $\unicode[STIX]{x1D702}_{ij}$, $\unicode[STIX]{x1D703}_{ij}\in R$. Recall that the elements $a_{ij}$ were introduced in 5.1. The relations in 5.1 yield that $F$ is a cycle.

Proof. Using Lemma 5.2, we see that it suffices to consider cycles of the form $t^{h}TZ_{1}\cdots Z_{p}W$, with $W\in {\mathcal{W}}$. Now consider the following element, which can be seen to be a cycle in $\mathfrak{m}K^{R}$:

$$\begin{eqnarray}\displaystyle C & = & \displaystyle \mathop{\sum }_{1\leqslant i<j\leqslant p}d_{ij}\left(y_{i}TZ_{1}\cdots \widehat{Z}_{j}\cdots Z_{p}W+(-1)^{j-i}y_{j}TZ_{1}\cdots \widehat{Z}_{i}\cdots Z_{p}W\right)\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{1\leqslant i\leqslant p}d_{ii}(y_{i}TZ_{1}\cdots \widehat{Z}_{i}\cdots Z_{p}W)\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{1\leqslant i\leqslant p}b_{i}\left(t^{h-1}TZ_{1}\cdots \widehat{Z}_{i}\cdots Z_{p}W+(-1)^{i}y_{i}Z_{1}\cdots Z_{p}W\right)\nonumber\end{eqnarray}$$

with $d_{ij},b_{i}\in R$, where the notation $\widehat{Z}_{j}$ indicates that the element $Z_{j}$ is missing from the product. Set

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\mathop{\sum }_{1\leqslant i\leqslant p}(-1)^{i}b_{i}\unicode[STIX]{x1D6FF}_{i}-\mathop{\sum }_{(i,j)\in {\mathcal{J}}_{1}}(-1)^{j}d_{ij}\unicode[STIX]{x1D702}_{ij}a_{ij}^{-1}+\mathop{\sum }_{(i,j)\in {\mathcal{J}}_{2}}(-1)^{j}d_{ij}\unicode[STIX]{x1D703}_{ij}.\end{eqnarray}$$

Then we have

$$\begin{eqnarray}FC=\unicode[STIX]{x1D6FC}(t^{h}TZ_{1}\cdots Z_{p}W).\end{eqnarray}$$

Since at least one of the coefficients $\unicode[STIX]{x1D6FF}_{i}$, $\unicode[STIX]{x1D702}_{ij}$, or $\unicode[STIX]{x1D703}_{ij}$ is a unit, we can choose the coefficients $d_{ij}$ and $b_{i}$ such that $\unicode[STIX]{x1D6FC}$ is a unit. Then, if we take $A=\unicode[STIX]{x1D6FC}^{-1}C$, we have $t^{h}TZ_{1}\cdots Z_{p}W=AF$.

Proof. Choose $\widetilde{w}_{i}$, $\widetilde{t}$, $\widetilde{z}_{i}$ preimages of the elements $w_{i}$, $t$, $z_{i}$ in $Q$. Let $K^{R}$ be the Koszul complex described earlier, and let $K^{Q}$ be the Koszul complex over $Q$ on the set $\{\widetilde{w}_{1},\ldots ,\widetilde{w}_{q},\widetilde{z}_{1},\ldots ,\widetilde{z}_{p},\widetilde{t}\}$, and note that $K^{R}=K^{Q}\otimes _{Q}R$, hence we can identify $K^{R}$ with $K^{Q}/IK^{Q}$.

(1) Assume $r\neq v$ and $\mathfrak{m}^{3}\neq 0$. In this case, the ideal $I$ is generated by the elements

$$\begin{eqnarray}\displaystyle & \displaystyle \widetilde{w}_{j}\widetilde{w}_{l},\qquad \widetilde{w}_{j}\widetilde{z}_{i},\qquad \widetilde{w}_{j}\widetilde{t},\qquad \widetilde{z}_{i}\widetilde{t}\quad \text{with }1\leqslant j\leqslant l\leqslant q\text{ and }1\leqslant i\leqslant p; & \displaystyle \nonumber\\ \displaystyle & \displaystyle \widetilde{t}^{h}-\widetilde{a}_{ij}^{-1}\widetilde{z}_{i}\widetilde{z}_{j}\quad \text{with }(i,j)\in {\mathcal{J}}_{1};\qquad \,\widetilde{z}_{i}\widetilde{z}_{j}\quad \text{with }(i,j)\in {\mathcal{J}}_{2}. & \displaystyle \nonumber\end{eqnarray}$$

We also have

$$\begin{eqnarray}(I:\mathfrak{n})=(\widetilde{w}_{1},\ldots ,\widetilde{w}_{q},\widetilde{t}^{h})+I\qquad \text{and}\qquad \mathfrak{n}(I:\mathfrak{n})=\mathfrak{n}(\widetilde{w}_{1},\ldots ,\widetilde{w}_{q})+\mathfrak{n}I.\end{eqnarray}$$

(Note that, if $R$ is Gorenstein, then $q=0$ and $\mathfrak{n}(I:\mathfrak{n})=\mathfrak{n}I$.) In particular, note that $I\neq \mathfrak{n}(I:\mathfrak{n})$.

Let $f\in I\smallsetminus \mathfrak{n}(I:\mathfrak{n})$. Then we can choose elements $\unicode[STIX]{x1D6FC}_{ij}$, $\unicode[STIX]{x1D6FD}_{ij}$, $\unicode[STIX]{x1D6FE}_{i}$, $\unicode[STIX]{x1D6FF}_{i}$, $\unicode[STIX]{x1D702}_{ij}$, $\unicode[STIX]{x1D703}_{ij}\in R$ in (5.3.1) such that at least one of the elements $\unicode[STIX]{x1D6FF}_{i}$, $\unicode[STIX]{x1D702}_{ij}$, or $\unicode[STIX]{x1D703}_{ij}$ is a unit, and such that we can lift the element $F$ in (5.3.1) to an element $L$ in $K^{Q}$ with $\unicode[STIX]{x2202}(L)=f$. Then Lemma 5.3 shows that the ring $R$ satisfies ${\mathcal{P}}_{2,1}$, with $l=F$. Part (1a) follows then from Corollary 4.5 and (1b) follows from Corollary 4.4. Furthermore, the existence of a surjective Golod homomorphism onto $R$ from a hypersurface ring, together with formula (5.0.1) and a result of Levin, see [Reference Avramov, Kustin and Miller6, Proposition 5.18], prove (1c).

Remark 5.5 explains the statements (1b) and (1c) when $\mathfrak{m}^{3}=0$.

(2) Assume now that $r=v$, hence $p=0$. We see that the ideal $I$ is then generated by the elements

$$\begin{eqnarray}\widetilde{w}_{j}\widetilde{w}_{l},\qquad \,\widetilde{w}_{j}\widetilde{t},\qquad \widetilde{t}^{h+1}\quad \text{with }1\leqslant j\leqslant l\leqslant q.\end{eqnarray}$$

We set $A=R/(w_{1},\ldots ,w_{q})$ and $B=R/(t)$. Note that $(A,\mathfrak{a},k)$ is a local ring with $A\cong Q^{\prime }/(\unicode[STIX]{x1D70F}^{h+1})$, where $Q^{\prime }$ is the regular local ring $Q^{\prime }=Q/(\widetilde{w}_{1},\ldots ,\widetilde{w}_{q})$ and $\unicode[STIX]{x1D70F}$ is the image of $\widetilde{t}$ in $Q^{\prime }$; in particular, $A$ is a hypersurface. Also, note that $(B,\mathfrak{b},k)$ is a local ring whose maximal ideal satisfies $\mathfrak{b}^{2}=0$. Then $R$ is isomorphic to the fiber product $A\times _{k}B$. Since $A$ and $B$ are both Golod rings, [Reference Lescot16, Théorème 4.1] implies that $R$ is a Golod ring, establishing (2a). Also, by [Reference Moore21], the algebra $\operatorname{Ext}_{R}(k,k)$ is then the coproduct of the algebras $\operatorname{Ext}_{A}(k,k)$ and $\operatorname{Ext}_{B}(k,k)$; see loc. cit. for the definition of the coproduct. Since $\mathfrak{b}^{2}=0$, the Yoneda algebra $\operatorname{Ext}_{B}(k,k)$ is generated by its elements of degree $1$. The Yoneda algebra of the hypersurface $A$ is generated in degrees $1$ and $2$. It follows that the coproduct is generated in degrees $1$ and $2$, establishing (2b).

By a result of Ghione and Gulliksen [Reference Ghione and Gulliksen9] and formula 5.0.1, the fact that $R$ is Golod implies that $P_{M}^{R}(t)$ is rational, with denominator $(1+t)^{v}(1-vt)$. The more precise denominator $1-vt$ in (2c) requires additional discussion. To compute $P_{M}^{R}(t)$ we use a method employed in the proof of [Reference Avramov, Iyengar and Şega5, Corollary 4.4]. By [Reference Dress and Krämer8, Rem. 3] one has $\unicode[STIX]{x1D6FA}_{2}^{R}(M)=K\oplus L$, where $K$ is an $A$-module and $L$ is a $B$-module. Using further a formula in [Reference Dress and Krämer8, Thm. 2], we have

(5.5.1)$$\begin{eqnarray}\displaystyle P_{M}^{R}(t)-\unicode[STIX]{x1D6FD}_{0}^{R}(M)-\unicode[STIX]{x1D6FD}_{1}^{R}(M)\cdot t & = & \displaystyle P_{K}^{A}(t)\cdot t^{2}+P_{L}^{B}(t)\cdot t^{2}\nonumber\\ \displaystyle & = & \displaystyle \frac{P_{K}^{A}(t)\cdot P_{k}^{B}(t)+P_{k}^{A}(t)\cdot P_{L}^{B}(t)}{P_{k}^{A}(t)+P_{k}^{B}(t)-P_{k}^{A}(t)\cdot P_{k}^{B}(t)}\cdot t^{2}.\end{eqnarray}$$

Since $A$ is a hypersurface, we have that $P_{k}^{A}(t)=1/(1-t)$, and the Poincaré series of every finitely generated $A$-module can be written as a rational function with denominator $1-t$, since every $A$-module $M$ has an eventually periodic resolution. Since $\mathfrak{b}^{2}=0$ and $\operatorname{rank}_{k}(\mathfrak{b}/\mathfrak{b}^{2})=v-1$, we have that $P_{k}^{B}(t)=1/(1-(v-1)t)$ and the Poincaré series of every finitely generated $B$-module can be written as a rational function with denominator $1-(v-1)t$. Plugging these formulas into (5.5.1) we obtain that $P_{M}^{R}(t)$ can be written as a fraction with denominator $1-vt$, establishing (2c).

Proof. If (1) holds, then ${\mathcal{P}}_{2,1}$ holds. By Corollary 4.4, the induced maps $\operatorname{Tor}_{\ast }^{R}(\mathfrak{m}^{2},k)\rightarrow \operatorname{Tor}_{\ast }^{R}(\mathfrak{m},k)$ are zero. This implies that $R$ is Koszul.

One of the consequences of the hypothesis in (1) is that there are no elements in bidegree $(1,i)$ with $i>2$. Consequently, the ideal $I$ is quadratic and $v(R)=2$. Let $L$ denote a preimage of $l$ in $\operatorname{K}^{Q}$ and set $h=\unicode[STIX]{x2202}(L)$. Note that $h\in I\smallsetminus \mathfrak{n}^{3}$. We can apply then Corollary 4.5 with $a=1$ and $t=2$ to conclude that $R$ is a homomorphic image of a quadratic hypersurface via a Golod homomorphism, hence $R$ is absolutely Koszul by [Reference Herzog and Iyengar12, Theorem 5.9].

If (2) holds, then ${\mathcal{Z}}_{1,1,2}$ holds. We apply then the graded version of Theorem 3.1 to conclude that the induced map $\operatorname{Tor}_{i}^{R}(\mathfrak{m}^{2},k)\rightarrow \operatorname{Tor}_{i}^{R}(\mathfrak{m},k)$ is zero, hence $R$ is Koszul.

Proof. If $R$ is Koszul, then the induced map $\operatorname{Tor}_{\ast }^{R}(\mathfrak{m}^{2},k)\rightarrow \operatorname{Tor}_{\ast }^{R}(\mathfrak{m},k)$ is zero, hence the canonical projection $R\rightarrow R/\mathfrak{m}^{2}$ is Golod, and thus small. Apply then Lemma 1.3 to see that $\operatorname{H}_{{\geqslant}1}(\mathfrak{m}^{2}\operatorname{K})\subseteq MH(\operatorname{K})$.

7.1. Consider the ring $R$ which is case 66 in [Reference Roos23, Table C]:

$$\begin{eqnarray}\displaystyle R=k[X,Y,Z,U]/(XZ,Y^{2},YU,Z^{2},ZU,U^{2}). & & \displaystyle \nonumber\end{eqnarray}$$

We claim that the nonlinear strand of $\operatorname{H}^{R}$ is generated by a single element in bidegree $(1,2)$, that is, it satisfies Theorem 6.1(1). As a result, we can conclude that this ring is absolutely Koszul, hence Koszul. To do this, we use the Macaulay2 package DGAlgebras.

In the following code, the set $\mathtt{C}$ is a list of cycles whose images (given by $\mathtt{G}$) generate $\mathtt{H}$ ($=\operatorname{H}^{R}$) as a $k$-algebra and $\mathtt{P}$ is the list of generators in $\mathtt{G}$ which are not in the linear strand. We define an ideal $\mathtt{I}$ in $\mathtt{H}$ to be generated by the elements in $\mathtt{P}$ and all products $hk$ with $h,k\in \mathtt{G}$; this ideal is equal to the union of all nonlinear strands of $\mathtt{H}$, together with the zero element. Showing generation of the nonlinear strands by an element in bidegree $(1,2)$ now reduces to checking ideal containment.

In this example, we see that the homology class of the cycle $N=zT_{1}+(y+u)T_{2}+(z+u)T_{3}+uT_{4}$ generates the nonlinear strands of $\operatorname{H}^{R}$, hence by Theorem 6.1(1), $R$ is absolutely Koszul.

Moreover, a similar argument shows that 40 of the rings from [Reference Roos23, Tables A–D] have the nonlinear strand of their Koszul homology algebra generated by a single element in bidegree $(1,2)$, and hence are absolutely Koszul rings; these are cases 1–4, 8–10, 23, 25–28, 49, 50, 52, 53, 66–68, 70, 72, 75–83, and isotopes 46va, 66v5, 68v, 71v4, 72v2e, 75v2, 78v1, 78v2e, 78v3v, and 81va from [Reference Roos23, Tables A–D]. Indeed, in all these cases we show that the homology class of the sum of generating cycles in bidegree $(1,2)$ found by DGAlgebras can be taken to be such a generator.

7.2. Let $R$ be the ring which is case 54 in [Reference Roos23]:

$$\begin{eqnarray}\displaystyle R=k[X,Y,Z,U]/(X^{2},XZ,Y^{2},Z^{2},YU+ZU,U^{2}). & & \displaystyle \nonumber\end{eqnarray}$$

The nonlinear strand of $\operatorname{H}^{R}$ cannot be generated by a single element in bidegree $(1,2)$ as in the previous example, which can be seen by computing the Betti table for $R$ over $Q$:

and observing there is no way for an element in bidegree $(4,6)$ to be a multiple of an element in bidegree $(1,2)$.

Next, we use the package DGAlgebras again to show that the nonlinear strand of $\operatorname{H}^{R}$ is generated by a finite set of classes of cycles with trivial self-multiplication, that is, the ring $R$ satisfies condition (2) of Theorem 6.1.

In this example, the set Cyc is the desired generating set contained in the linear strand and its elements correspond to the following cycles:

$$\begin{eqnarray}\{xT_{1},zT_{3},zT_{1},zT_{1}T_{3},xT_{1}T_{3}\}.\end{eqnarray}$$

The Macaulay2 code first checks $\mathfrak{m}^{3}=0$, and then verifies that all the products of elements in Cyc are 0 and that the nonlinear strand I is contained in the ideal generated by Cls (the images in H of cycles in Cyc). A similar argument shows that the ring in case 71 of [Reference Roos23, Tables A–D] also satisfies Theorem 6.1(2); for that ring, one can show that the set Cyc=$\{C_{0},C_{1},C_{2},C_{7},C_{8},C_{9},C_{11},C_{15}\}$ is the desired generating set with trivial self-multiplication.

7.4. Consider the ring which is case 55 in [Reference Roos23]:

$$\begin{eqnarray}\displaystyle R=k[X,Y,Z,U]/(X^{2}+XY,XZ+YU,XU,Y^{2},Z^{2},ZU+U^{2}). & & \displaystyle \nonumber\end{eqnarray}$$

The graded Betti table of $R$ over $Q$ is the same as for the ring in 7.2. Moreover, we see that $\operatorname{H}^{R}$ is generated by the linear strand. This code can also be used to show the ring in 7.2 is generated by its linear strand.

We see that $\operatorname{H}^{R}$ has 6 generators in bidegree $(1,2)$ and 4 generators in bidegree $(2,3)$, all in the linear strand; further, the resolution of $k$ over $R$ is not linear, hence $R$ is not Koszul.