3.1 Componentwise linear ideals

It might be an interesting problem to find necessary and/or sufficient conditions for a $\mathbb{Z}$ -graded ideal to have a trivial $\unicode[STIX]{x1D708}$ -table. In this direction, we have the following relation to the notion of componentwise linear ideals.

Römer [Reference Römer20] and the second author [Reference Yanagawa25, Theorem 4.1] independently showed that $I$ is componentwise linear if and only if $H_{i}(\mathbb{L}_{\bullet }^{{<}r>}(I))=0$ for all $r$ and all $i\geqslant 1$ .

The converse of the above proposition is not true. For example, in Corollary 3.13 below, we show that if $I_{1}\neq 0$ , then it has a trivial $\unicode[STIX]{x1D708}$ -table. However, there is no relation between being componentwise linear and $I_{1}\neq 0$ .

3.2 Consecutiveness of nontrivial columns of the $\unicode[STIX]{x1D708}$ -tables

For a $\mathbb{Z}$ -graded ideal $I\subseteq R$ and $i\in \mathbb{N}$ , set

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{i}(I)=\mathop{\sum }_{j\in \mathbb{N}}\unicode[STIX]{x1D708}_{i,j}(I).\end{eqnarray}$$

If we denote $\mathbb{L}_{\bullet }(I):=\bigoplus _{r\in \mathbb{N}}\mathbb{L}_{\bullet }^{{<}r>}(I)$ , then

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{i}(I)=\dim _{\mathbb{K}}H_{i}(\mathbb{L}_{\bullet }(I)\otimes _{R}\mathbb{K}).\end{eqnarray}$$

By Lemma 3.2, we have $\sum _{i=0}^{n}(-1)^{i}\unicode[STIX]{x1D708}_{i}(I)=1$ . We say that $\unicode[STIX]{x1D708}_{i}(I)$ is nontrivial if

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{i}(I)\geqslant \left\{\begin{array}{@{}ll@{}}2,\quad & \text{if }i=0,\\ 1,\quad & \text{if }i\geqslant 1.\end{array}\right.\end{eqnarray}$$

Clearly, the $\unicode[STIX]{x1D708}$ -table of $I$ is nontrivial if and only if $\unicode[STIX]{x1D708}_{i}(I)$ is nontrivial for some  $i$ . If $n\geqslant 1$ , we have $\text{proj.dim}_{R}I\leqslant n-1$ , and hence $\unicode[STIX]{x1D708}_{n}(I)=0$ . In particular, $\unicode[STIX]{x1D708}_{n}(I)$ is always trivial.

The main result of this subsection is the following.

In order to prove the theorem, we reconstruct $\mathbb{L}_{\bullet }(I)$ using a spectral sequence. Let $L_{\bullet }(I)$ be the minimal free resolution of $I$ as before. Consider the $\mathfrak{m}$ -adic filtration $L_{\bullet }(I)=F_{0}L_{\bullet }\supset F_{1}L_{\bullet }\supset \cdots \,$ of $L_{\bullet }(I)$ , where $F_{i}L_{\bullet }$ is a subcomplex whose component of homological degree $j$ is $\mathfrak{m}^{i}L_{j}$ . For any given $R$ -module $M$ , we regard $\text{gr}\,(M):=\bigoplus _{i\in \mathbb{N}}\mathfrak{m}^{i}M/\mathfrak{m}^{i+1}M$ as an $R$ -module via the isomorphism $\text{gr}\,R=\bigoplus _{i\in \mathbb{N}}\mathfrak{m}^{i}/\mathfrak{m}^{i+1}\cong R=\Bbbk [x_{1},\ldots ,x_{n}]$ . Since each $L_{j}$ is a free $R$ -module,

$$\begin{eqnarray}\bigoplus _{p+q=-j}E_{0}^{p,q}=\left(\bigoplus _{p\geqslant 0}\mathfrak{m}^{p}L_{j}/\mathfrak{m}^{p+1}L_{j}\right)=\text{gr}\,L_{j}\end{eqnarray}$$

is isomorphic to $L_{j}$ (if we identify $\text{gr}\,R$ with $R$ ), while we have to forget the original $\mathbb{Z}$ -grading of $L_{j}$ . Since $L_{\bullet }(I)$ is a minimal free resolution, $d_{0}^{p,q}:E_{0}^{p,q}\rightarrow E_{0}^{p,q+1}$ is the zero map for all $p,q$ , and hence $E_{0}^{p,q}=E_{1}^{p,q}$ . It follows that

$$\begin{eqnarray}\mathbb{E}_{j}^{(1)}:=\bigoplus _{p+q=-j}E_{1}^{p,q}=\bigoplus _{p+q=-j}E_{0}^{p,q}\end{eqnarray}$$

is isomorphic to $L_{j}$ under the identification $R\cong \text{gr}\,R$ . Collecting the maps

$$\begin{eqnarray}d_{1}^{p,q}:E_{1}^{p,q}(=\mathfrak{m}^{p}L_{j}/\mathfrak{m}^{p+1}L_{j})\longrightarrow E_{1}^{p+1,q}(=\mathfrak{m}^{p+1}L_{j-1}/\mathfrak{m}^{p+2}L_{j-1})\end{eqnarray}$$

for $p,q$ with $p+q=-j$ , we have the $R$ -morphism $d_{j}^{(1)}:\mathbb{E}_{j}^{(1)}\rightarrow \mathbb{E}_{j-1}^{(1)}$ , and these morphisms make $\mathbb{E}_{\bullet }^{(1)}$ a chain complex of $R$ -modules. Under the isomorphism $\mathbb{E}_{j}^{(1)}\cong L_{j}$ , $\mathbb{E}_{\bullet }^{(1)}$ is isomorphic to $\mathbb{L}_{\bullet }(I)=\bigoplus _{r\in \mathbb{N}}\mathbb{L}_{\bullet }^{{<}r>}(I)$ . Hence, we have

$$\begin{eqnarray}\mathbb{E}_{j}^{(2)}:=\bigoplus _{p+q=-j}E_{2}^{p,q}\cong H_{j}(\mathbb{L}_{\bullet }(I))\end{eqnarray}$$

and $\unicode[STIX]{x1D708}_{j}(I)=\dim _{\mathbb{K}}(\mathbb{E}_{j}^{(2)}\otimes _{R}\mathbb{K})$ . Collecting the maps $d_{2}^{p,q}:E_{2}^{p,q}\rightarrow E_{2}^{p+2,q-1}$ , we have the $R$ -morphism

$$\begin{eqnarray}d_{j}^{(2)}:\mathbb{E}_{j}^{(2)}\,(\cong H_{j}(\mathbb{L}_{\bullet }(I)))\longrightarrow \mathbb{E}_{j-1}^{(2)}\,(\cong H_{j-1}(\mathbb{L}_{\bullet }(I))).\end{eqnarray}$$

Moreover, we have the chain complex

$$\begin{eqnarray}\cdots \longrightarrow \mathbb{E}_{j+1}^{(2)}\stackrel{d_{j+1}^{(2)}}{\longrightarrow }\mathbb{E}_{j}^{(2)}\stackrel{d_{j}^{(2)}}{\longrightarrow }\mathbb{E}_{j-1}^{(2)}\longrightarrow \cdots\end{eqnarray}$$

of $R$ -modules whose $j$ th homology is isomorphic to $\mathbb{E}_{j}^{(3)}:=\bigoplus _{p+q=-j}E_{3}^{p,q}.$ For all $r\geqslant 4$ , $\mathbb{E}_{j}^{(r)}:=\bigoplus _{p+q=-j}E_{r}^{p,q}$ satisfies the same property.

By the construction of spectral sequences, if

$$\begin{eqnarray}r>\max \{k\mid \unicode[STIX]{x1D6FD}_{j,k}(I)\neq 0\}-\min \{k\mid \unicode[STIX]{x1D6FD}_{j-1,k}(I)\neq 0\},\end{eqnarray}$$

then the map $d_{r}^{p,q}:E_{r}^{p,q}\rightarrow E_{r}^{p+r,q-r+1}$ is zero for all $p,q$ with $p+q=-j$ , and hence $d_{j}^{(r)}:\mathbb{E}_{j}^{(r)}\rightarrow \mathbb{E}_{j-1}^{(r)}$ is zero. This implies that $\mathbb{E}_{j}^{(r)}$ is isomorphic to

$$\begin{eqnarray}\mathbb{E}_{j}^{(\infty )}:=\bigoplus _{p+q=-j}E_{\infty }^{p,q}\end{eqnarray}$$

for $r\gg 0$ .

Proof of Theorem 3.7.

We prove the assertion by contradiction using the spectral sequence introduced above. First, assume that $\unicode[STIX]{x1D708}_{j}(I)$ is nontrivial for some $2\leqslant j\leqslant n-1$ , but both $\unicode[STIX]{x1D708}_{j-1}(I)$ and $\unicode[STIX]{x1D708}_{j+1}(I)$ are trivial. (The cases $j=0,1$ can be proved using similar arguments, and we make a few remarks later.) Then, we have $\mathbb{E}_{j}^{(2)}\otimes _{R}\mathbb{K}\neq 0$ and $\mathbb{E}_{j+1}^{(2)}\otimes _{R}\mathbb{K}=\mathbb{E}_{j-1}^{(2)}\otimes _{R}\mathbb{K}=0.$ It follows that $\mathbb{E}_{j}^{(3)}\otimes _{R}\mathbb{K}\neq 0,$ since it is the homology of

$$\begin{eqnarray}\mathbb{E}_{j+1}^{(2)}\otimes _{R}\mathbb{K}\longrightarrow \mathbb{E}_{j}^{(2)}\otimes _{R}\mathbb{K}\longrightarrow \mathbb{E}_{j-1}^{(2)}\otimes _{R}\mathbb{K}.\end{eqnarray}$$

Similarly, we have $\mathbb{E}_{j-1}^{(3)}\otimes _{R}\mathbb{K}=\mathbb{E}_{j+1}^{(3)}\otimes _{R}\mathbb{K}=0.$ Repeating this argument, we have $\mathbb{E}_{j}^{(r)}\otimes _{R}\mathbb{K}\neq 0$ for all $r\geqslant 4$ . Hence, $E_{\infty }^{p,q}\neq 0$ for some $p,q$ with $p+q=-j$ . However, this contradicts the facts that

$$\begin{eqnarray}E_{r}^{p,q}\Longrightarrow H_{-p-q}(L_{\bullet }(I))\end{eqnarray}$$

and $H_{j}(L_{\bullet }(I))=0$ . (Recall that $j>0$ now.)

Next, we assume that $\unicode[STIX]{x1D708}_{1}(I)$ is nontrivial, but $\unicode[STIX]{x1D708}_{0}(I)$ and $\unicode[STIX]{x1D708}_{2}(I)$ are trivial. That is,

$$\begin{eqnarray}\mathbb{E}_{1}^{(2)}\otimes _{R}\mathbb{K}\neq 0,\qquad \mathbb{E}_{0}^{(2)}\otimes _{R}\mathbb{K}\cong \mathbb{K}\qquad \text{and}\qquad \mathbb{E}_{2}^{(2)}\otimes _{R}\mathbb{K}=0.\end{eqnarray}$$

As we have seen above, we must have $\mathbb{E}_{1}^{(r)}=0$ for $r\gg 0$ . Since $\mathbb{E}_{2}^{(r)}\otimes _{R}\mathbb{K}=0$ for all $r$ now, if $d_{1}^{(r)}\otimes _{R}\mathbb{K}:\mathbb{E}_{1}^{(r)}\otimes _{R}\mathbb{K}\longrightarrow \mathbb{E}_{0}^{(r)}\otimes _{R}\mathbb{K}$ are the zero maps for all $r$ , then $\mathbb{E}_{1}^{(r)}\otimes _{R}\mathbb{K}\cong \mathbb{E}_{1}^{(2)}\otimes _{R}\mathbb{K}\neq 0$ for all $r$ , and this is a contradiction. Therefore, there is some $r\geqslant 2$ such that $d_{1}^{(r)}\otimes _{R}\mathbb{K}$ is not zero. If $s$ is the minimum among these $r$ , $d_{1}^{(s)}\otimes _{R}\mathbb{K}:\mathbb{E}_{1}^{(s)}\otimes _{R}\mathbb{K}\longrightarrow (\mathbb{E}_{0}^{(s)}\otimes _{R}\mathbb{K})\cong \mathbb{K}$ is surjective. Hence, $\mathbb{E}_{0}^{(r)}\otimes _{R}\mathbb{K}=0$ for all $r>s$ , and $\mathbb{E}_{0}^{(\infty )}\otimes _{R}\mathbb{K}=0$ . However, since $\mathbb{E}_{0}^{(\infty )}\cong \text{gr}\,H_{0}(L_{\bullet }(I))\cong \text{gr}\,(I)$ and $\dim _{R}I=n$ , we have $\dim _{R}(\text{gr}\,(I))=n$ and hence $\mathbb{E}_{0}^{(\infty )}\otimes _{R}\mathbb{K}\neq 0$ . This is a contradiction. The case when $\unicode[STIX]{x1D708}_{0}(I)$ is nontrivial can be proved in a similar way.◻

3.3 Thom–Sebastiani type formulas

Let $I,J$ be $\mathbb{Z}$ -graded ideals in two disjoint sets of variables, say $I\subseteq R=\Bbbk [x_{1},\ldots ,x_{m}]$ and $J\subseteq S=\Bbbk [y_{1},\ldots ,y_{n}]$ . The aim of this subsection is to describe the $\unicode[STIX]{x1D708}$ -numbers of $IT+JT$ , where $T=R\otimes _{\Bbbk }S=\Bbbk [x_{1},\ldots ,x_{m},y_{1},\ldots ,y_{n}]$ , in terms of those of $I$ and $J$ , respectively. When we just consider Betti numbers, we have the following results due to Jacques and Katzman [Reference Jacques and Katzman11].

Proposition 3.8. (Cf. [Reference Jacques and Katzman11, Lemma 2.1])

Let $L_{\bullet }(R/I)$ and $L_{\bullet }(S/J)$ be minimal graded free resolutions of $R/I$ and $S/J$ , respectively. Then,

$$\begin{eqnarray}(L_{\bullet }(R/I)\otimes _{R}T)\otimes _{T}(L_{\bullet }(S/J)\otimes _{S}T)\end{eqnarray}$$

is a minimal graded free resolution of $T/IT+JT$ .

Hence, Betti numbers satisfy the following relation.

Corollary 3.9. (Cf. [Reference Jacques and Katzman11, Corollary 2.2])

The Betti numbers of $T/IT+JT$ have the following form:

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{i,j}(T/IT+JT)=\mathop{\sum }_{\substack{ k+k^{\prime }=i \\ l+l^{\prime }=j}}\unicode[STIX]{x1D6FD}_{k,l}(T/IT)\unicode[STIX]{x1D6FD}_{k^{\prime },l^{\prime }}(T/JT).\end{eqnarray}$$

Hence, we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{i,j}(IT+JT)=\unicode[STIX]{x1D6FD}_{i,j}(IT)+\unicode[STIX]{x1D6FD}_{i,j}(JT)+\mathop{\sum }_{\substack{ k+k^{\prime }=i-1 \\ l+l^{\prime }=j}}\unicode[STIX]{x1D6FD}_{k,l}(IT)\unicode[STIX]{x1D6FD}_{k^{\prime },l^{\prime }}(JT).\end{eqnarray}$$

Our aim is to extend the result in [Reference Jacques and Katzman11] to the case of $\unicode[STIX]{x1D708}$ -numbers. To such purpose, it is more convenient to consider separately the case of ideals with degree one elements. Thus, let $I\subseteq R$ be any $\mathbb{Z}$ -graded ideal and assume for simplicity that $J$ is principally generated by an element of degree 1; for example, $J=(y)\subseteq S$ .

Lemma 3.10. Let $I\subseteq R=\Bbbk [x_{1},\ldots ,x_{m}]$ and $J=(y)\subseteq S=\Bbbk [y]$ be $\mathbb{Z}$ -graded ideals, and set $T=R\otimes _{\Bbbk }S=\Bbbk [x_{1},\ldots ,x_{m},y]$ . For $r\geqslant 2$ , the $r$ -linear strand $\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)$ is the mapping cone of the chain map

$$\begin{eqnarray}\times y:(\mathbb{L}_{\bullet }^{{<}r>}(IT))(-1)\rightarrow \mathbb{L}_{\bullet }^{{<}r>}(IT).\end{eqnarray}$$

The general case is more involved. Assume now that $I\subseteq R=\Bbbk [x_{1},\ldots ,x_{m}]$ and $J\subseteq S=\Bbbk [y_{1},\ldots ,y_{n}]$ are $\mathbb{Z}$ -graded ideals such that $I_{1}=0$ and $J_{1}=0$ . Let $L_{\bullet }(I)$ be a minimal graded $R$ -free resolution of $I$ , and let $L_{\bullet }(J)$ be a minimal graded $S$ -free resolution of $J$ , and consider their extensions $L_{\bullet }(IT)$ and $L_{\bullet }(JT)$ to $T=R\otimes _{\Bbbk }S=\Bbbk [x_{1},\ldots ,x_{m},y_{1},\ldots ,y_{n}]$ .

Lemma 3.11. Under the previous assumptions, the $r$ -linear strand $\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)$ is

$$\begin{eqnarray}\displaystyle \mathbb{L}_{\bullet }^{{<}r>}(IT+JT) & = & \displaystyle \mathbb{L}_{\bullet }^{{<}r>}(IT)\oplus \mathbb{L}_{\bullet }^{{<}r>}(JT)\nonumber\\ \displaystyle & \oplus & \displaystyle \left(\bigoplus _{a+b=r+1}(\mathbb{L}_{\bullet }^{{<}a>}(IT)\otimes _{T}\mathbb{L}_{\bullet }^{{<}b>}(JT))[-1]\right).\nonumber\end{eqnarray}$$

Here, for a chain complex $C_{\bullet }$ , $C_{\bullet }[-1]$ denotes the translated complex whose component of homological degree $j$ is $C_{j-1}$ .

Proof. Consider the minimal $\mathbb{Z}$ -graded free resolutions of $R/I$ and $S/J$ , respectively,

where $L_{0}=R$ and ${L^{\prime }}_{0}=S$ . According to Proposition 3.8, the minimal $\mathbb{Z}$ -graded free resolution $L_{\bullet }(T/IT+JT)$ has the formFootnote ²

where, for any given $x_{i}\otimes y_{p-i}\in L_{i}\otimes L_{p-i}^{\prime }$ , we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x2202}_{p}(x_{i}\otimes y_{p-i}) & = & \displaystyle d_{i}(x_{i})\otimes y_{p-i}+(-1)^{i}x_{i}\otimes d_{p-i}^{\prime }(y_{p-i})\in (L_{i-1}\otimes L_{p-i}^{\prime })\nonumber\\ \displaystyle & \oplus & \displaystyle (L_{i}\otimes L_{p-i-1}^{\prime }).\nonumber\end{eqnarray}$$

To describe the $r$ -linear strand $\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)$ of the ideal $IT+JT$ , we must consider the truncation at the first term of the above resolution and take a close look at the free modules and the components of the corresponding differentials. Recall that $L_{\bullet }^{{<}r-1>}(R/I)$ corresponds to $L_{\bullet }^{{<}r>}(I)$ for all $r\geqslant 2$ . It is easy to see that both

$$\begin{eqnarray}\displaystyle \mathbb{L}_{\bullet }^{{<}r>}(IT):0 & \longrightarrow & \displaystyle L_{m}^{{<}r-1>}\otimes L_{0}^{\prime }\longrightarrow \cdots \longrightarrow L_{2}^{{<}r-1>}\otimes L_{0}^{\prime }\nonumber\\ \displaystyle & \longrightarrow & \displaystyle L_{1}^{{<}r-1>}\otimes L_{0}^{\prime }\longrightarrow 0\nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle \mathbb{L}_{\bullet }^{{<}r>}(JT):0 & \longrightarrow & \displaystyle L_{0}\otimes {L^{\prime }}_{n}^{{<}r-1>}\longrightarrow \cdots \longrightarrow L_{0}\otimes {L^{\prime }}_{2}^{{<}r-1>}\nonumber\\ \displaystyle & \longrightarrow & \displaystyle L_{0}\otimes {L^{\prime }}_{1}^{{<}r-1>}\longrightarrow 0\nonumber\end{eqnarray}$$

are subcomplexes of $\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)$ . Moreover, $\mathbb{L}_{\bullet }^{{<}r>}(IT)$ and $\mathbb{L}_{\bullet }^{{<}r>}(JT)$ are direct summands of $\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)$ . In fact, since $I_{1}=J_{1}=0$ , the linear parts of the maps $L_{i}\otimes L_{1}^{\prime }\rightarrow L_{i}\otimes L_{0}^{\prime }$ and $L_{1}\otimes L_{j}^{\prime }\rightarrow L_{0}\otimes L_{j}^{\prime }$ vanish.

In order to obtain the remaining components of $\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)$ , we must consider the $r$ -linear strand of

This complex starts at the second term (i.e., the term of homological degree 1), and the first term of the $r$ -linear strand is $\bigoplus _{a+b=r+1}\mathbb{L}_{0}^{{<}a>}(IT)\otimes _{T}\mathbb{L}_{0}^{{<}b>}(JT).$ If we take a close look at the free summands of these components and their differentials, we obtain the following description:

$$\begin{eqnarray}\bigoplus _{a+b=r+1}(\mathbb{L}_{\bullet }^{{<}a>}(IT)\otimes _{T}\mathbb{L}_{\bullet }^{{<}b>}(JT))[-1].\end{eqnarray}$$

Therefore, we are done. ◻

The main result of this subsection is the following.

Proof. (i) If $J_{1}\neq 0$ , we may assume that $y_{n}\in J$ without loss of generality. Now, we have $J=(f_{1},\ldots ,f_{r},y_{n})$ , where $f_{1}\ldots ,f_{r}$ are homogeneous polynomials in $\Bbbk [y_{1},\ldots .y_{n-1}]$ . Set $R^{\prime }:=\Bbbk [x_{1},\ldots ,x_{m},y_{1},\ldots ,y_{n-1}]$ , $S^{\prime }=\Bbbk [y_{n}]$ , and let $I^{\prime }=IR^{\prime }+(f_{1},\ldots ,f_{r})$ be an ideal in $R^{\prime }$ (note that $f_{1}\ldots ,f_{r}$ are elements in $R^{\prime }$ ), and let $J^{\prime }=(y_{n})$ be an ideal in $S^{\prime }$ . Then, we have $T=R\otimes _{\Bbbk }S=R^{\prime }\otimes _{\Bbbk }S^{\prime }$ , and $IT+JT=I^{\prime }T+J^{\prime }T$ . This means that we may assume that $J=(y)\subseteq S=\Bbbk [y]$ from the beginning. For $r\geqslant 2$ , the $r$ -linear strand $\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)$ is given by the mapping cone of the chain map $\times y:(\mathbb{L}_{\bullet }^{{<}r>}(IT))(-1)\rightarrow \mathbb{L}_{\bullet }^{{<}r>}(IT)$ by Lemma 3.10. Hence, $\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)\otimes _{T}\mathbb{K}$ is given by the mapping cone of the chain map

$$\begin{eqnarray}\times y:\mathbb{L}_{\bullet }^{{<}r>}(IT)\otimes _{T}\mathbb{K}\longrightarrow \mathbb{L}_{\bullet }^{{<}r>}(IT)\otimes _{T}\mathbb{K},\end{eqnarray}$$

where $\mathbb{K}$ is the field of fractions of $T$ . Clearly, this is the identity map, and its mapping cone is exact. This means that $H_{i}(\mathbb{L}_{\bullet }^{{<}r>}(IT+JT)\otimes _{T}\mathbb{K})=0$ for all $r\geqslant 2$ and all $i$ .

On the other hand, $(IT+JT)_{{<}1>}$ is a complete intersection ideal generated by degree-1 elements, and hence we have $\dim _{\mathbb{K}}H_{i}(\mathbb{L}_{\bullet }^{{<}1>}(I)\otimes _{T}\mathbb{K})=\unicode[STIX]{x1D6FF}_{0,i}$ . Summing up, we see that $IT+JT$ has a trivial $\unicode[STIX]{x1D708}$ -table.

(ii) Follows immediately from Lemma 3.11. ◻

The following is just a rephrasing of part (i) of the previous result.

The following is another corollary of Proposition 3.12.

4.1 Sequentially Cohen–Macaulay rings

Let $M$ be a finitely generated graded module over the polynomial ring $R=\Bbbk [x_{1},\ldots ,x_{n}]$ . We say that $M$ is sequentially Cohen–Macaulay if $\text{Ext}_{R}^{n-i}(M,R)$ is either a Cohen–Macaulay module of dimension $i$ or the 0 module for all $i$ . The original definition is given by the existence of a certain filtration (see [Reference Stanley22, III, Definition 2.9]). However, it is equivalent to the above one by [Reference Stanley22, III, Theorem 2.11]. The sequentially Cohen–Macaulay property of a finitely generated module over a regular local ring is defined/characterized in the same way.

In [Reference Àlvarez Montaner2], the first author showed that the sequentially Cohen–Macaulay property implies the triviality of Lyubeznik tables in positive characteristic as well as in the case of squarefree monomial ideals. Using Proposition 3.6, we can give a new proof/interpretation of this result for the case of monomial ideals.

The converse of Proposition 4.4 is not true. That is, even if $R/I$ has a trivial Lyubeznik table, it need not be sequentially Cohen–Macaulay. For example, if $I$ is the monomial ideal

$$\begin{eqnarray}(x_{1},x_{2})\cap (x_{3},x_{4})\cap (x_{1},x_{5})\cap (x_{2},x_{5})\cap (x_{3},x_{5})\cap (x_{4},x_{5})\end{eqnarray}$$

in $R=\Bbbk [x_{1},\ldots ,x_{5}]$ , then $R/I$ has a trivial Lyubeznik table, but this ring is not sequentially Cohen–Macaulay. Since all associated primes of $I$ have the same height, it is the same to say that $R/I$ is not Cohen–Macaulay. However, $R/I$ does not even satisfy Serre’s condition $(S_{2})$ .

In Proposition 4.5 below, we see that if a monomial ideal $I$ has height 1 (i.e., admits a height one associated prime), then the Lyubeznik table of $R/I$ is trivial. Of course, $R/I$ need not be sequentially Cohen–Macaulay in this situation.

4.2 Thom–Sebastiani type formulas

Let $I\subseteq R=\Bbbk [x_{1},\ldots ,x_{m}]$ and $J\subseteq S=\Bbbk [y_{1},\ldots ,y_{n}]$ be squarefree monomial ideals in two disjoint sets of variables. Let $\unicode[STIX]{x1D6E5}_{1}$ and $\unicode[STIX]{x1D6E5}_{2}$ be the simplicial complexes associated to $I$ and $J$ by the Stanley–Reisner correspondence; that is, $I=I_{\unicode[STIX]{x1D6E5}_{1}}$ and $J=I_{\unicode[STIX]{x1D6E5}_{2}}$ . Then, the sum $IT+JT=I_{\unicode[STIX]{x1D6E5}_{1}\ast \unicode[STIX]{x1D6E5}_{2}}$ corresponds to the simplicial join of the two complexes. Let $\unicode[STIX]{x1D6E5}_{1}^{\vee }$ (resp. $\unicode[STIX]{x1D6E5}_{2}^{\vee }$ ) be the Alexander dual of $\unicode[STIX]{x1D6E5}_{1}$ (resp. $\unicode[STIX]{x1D6E5}_{2}$ ) as a simplicial complex on $\{1,2,\ldots ,m\}$ (resp. $\{1,2,\ldots ,n\}$ ). Set $I^{\vee }:=I_{\unicode[STIX]{x1D6E5}_{1}^{\vee }}\subseteq R$ and $J^{\vee }:=I_{\unicode[STIX]{x1D6E5}_{2}^{\vee }}\subseteq S$ . Then, it is easy to see that

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbf{A}(T/IT)\cong I^{\vee }T,\qquad \mathbf{A}(T/JT)\cong J^{\vee }T\qquad \text{and}\qquad \nonumber\\ \displaystyle & & \displaystyle \qquad \mathbf{A}(T/IT\cap JT)\cong I^{\vee }T+J^{\vee }T,\nonumber\end{eqnarray}$$

where $\mathbf{A}$ denotes the Alexander duality functor of $\operatorname{Sq}T$ .

Proof. The assertion easily follows from Proposition 3.12 and Theorem 4.3, but for completeness we will make a few remarks.

(i) Recall that, for a simplicial complex $\unicode[STIX]{x1D6E5}$ , the height of $I_{\unicode[STIX]{x1D6E5}}$ is 1 if and only if $[I_{\unicode[STIX]{x1D6E5}^{\vee }}]_{1}\neq 0$ .

(ii) The last equality follows from the fact that

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{p,i}(T/IT)=\unicode[STIX]{x1D706}_{p-n,i-n}(R/I)\qquad \text{and}\qquad \unicode[STIX]{x1D706}_{p,i}(T/JT)=\unicode[STIX]{x1D706}_{p-m,i-m}(S/J),\end{eqnarray}$$

which can be seen from Theorem 4.3 and the construction of linear strands. ◻

Example 4.6. It is well known that local cohomology modules as well as free resolutions depend on the characteristic of the base field, so Lyubeznik numbers depend on the characteristic as well. The most recurrent example is the Stanley–Reisner ideal associated to a minimal triangulation of $\mathbb{P}_{\mathbb{R}}^{2}$ ; that is, the ideal in $R=\Bbbk [x_{1},\ldots ,x_{6}]$ :

$$\begin{eqnarray}\displaystyle & & \displaystyle I=\big(x_{1}x_{2}x_{3},x_{1}x_{2}x_{4},x_{1}x_{3}x_{5},x_{2}x_{4}x_{5},x_{3}x_{4}x_{5},x_{2}x_{3}x_{6},\nonumber\\ \displaystyle & & \displaystyle \qquad x_{1}x_{4}x_{6},x_{3}x_{4}x_{6},x_{1}x_{5}x_{6},x_{2}x_{5}x_{6}\big).\nonumber\end{eqnarray}$$

Its Lyubeznik table has been computed in [Reference Àlvarez Montaner and Vahidi1, Example 4.8]. Namely, in characteristic zero and two respectively, we have

$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{\mathbb{Q}}(R/I)=\left(\begin{array}{@{}cccc@{}}0 & 0 & 0 & 0\\ & 0 & 0 & 0\\ & & 0 & 0\\ & & & 1\end{array}\right),\qquad \unicode[STIX]{x1D6EC}_{\mathbb{Z}/2\mathbb{Z}}(R/I)=\left(\begin{array}{@{}cccc@{}}0 & 0 & 1 & 0\\ & 0 & 0 & 0\\ & & 0 & 1\\ & & & 1\end{array}\right).\end{eqnarray}$$

One can slightly modify this example and use Proposition 4.5 to obtain some interesting behavior of Lyubeznik numbers.

$\bullet$ The ideal $J=I\cap (x_{7})$ in $R=\Bbbk [x_{1},\ldots ,x_{7}]$ has a trivial Lyubeznik table in any characteristic, so we obtain an example where the local cohomology modules depend on the characteristic but Lyubeznik numbers do not.

$\bullet$ The ideal $J=I\cap (x_{7},x_{8})\cap (x_{9},x_{10})$ in $R=\Bbbk [x_{1},\ldots ,x_{10}]$ satisfies

$$\begin{eqnarray}1=\unicode[STIX]{x1D706}_{6,7}^{\mathbb{Q}}(R/J)\neq \unicode[STIX]{x1D706}_{6,7}^{\mathbb{Z}/2\mathbb{Z}}(R/J)=2,\end{eqnarray}$$

and both Lyubeznik numbers are different from zero.

5.1 Lyubeznik numbers

Let $R=\Bbbk [C]$ be a normal simplicial semigroup ring that is Gorenstein, and let $I$ be a monomial ideal of $R$ . As in the polynomial ring case, we set the Lyubeznik numbers as

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{p,i}(R/I):=\unicode[STIX]{x1D707}^{p}(\mathfrak{m},H_{I}^{n-i}(R)).\end{eqnarray}$$

Work of the second author in [Reference Yanagawa27] states that this set of invariants are well defined in this framework. Namely, Theorem 4.2 holds verbatim in this situation.

Theorem 5.2. [Reference Yanagawa27, Corollary 5.12]

Let $R=\Bbbk [C]$ be a normal simplicial semigroup ring that is Gorenstein, and let $I_{\unicode[STIX]{x1D6E5}}$ be a radical monomial ideal. Then, we have

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{p,i}(R/I_{\unicode[STIX]{x1D6E5}})=\dim _{\Bbbk }[\text{Ext}_{R}^{n-p}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}<\infty .\end{eqnarray}$$

Notice that in this setting we have that whenever we have a multigraded isomorphism $\Bbbk [C]/I_{\unicode[STIX]{x1D6E5}}\cong \Bbbk [C^{\prime }]/I_{\unicode[STIX]{x1D6E5}^{\prime }}$ between quotients of Gorenstein normal simplicial semigroup rings by radical monomial ideals, then the corresponding Lyubeznik numbers coincide. This multigraded framework slightly differs from the original situation for regular local rings stated in [Reference Lyubeznik13]. However, as stated in [Reference Yanagawa27, Remark 5.14], if $\unicode[STIX]{x1D6E5}\cong \unicode[STIX]{x1D6E5}^{\prime }$ as simplicial complexes, then $R/I_{\unicode[STIX]{x1D6E5}}$ and $R^{\prime }/I_{\unicode[STIX]{x1D6E5}^{\prime }}$ have the same Lyubeznik numbers. In this sense, to study the Lyubeznik numbers of a quotient $R/I_{\unicode[STIX]{x1D6E5}}$ of a Gorenstein normal simplicial semigroup ring $R$ by a radical monomial ideal $I_{\unicode[STIX]{x1D6E5}}$ , we may assume that $R$ is a polynomial ring and $R/I_{\unicode[STIX]{x1D6E5}}$ is a Stanley–Reisner ring. In Theorem 5.3, we prove a stronger result.

It is also worth pointing out that several features of Lyubeznik numbers are still true in this setting. In what follows, we assume that $I$ is a monomial ideal of $R$ .

(1) As in the polynomial ring case, we have the Euler characteristic equation

$$\begin{eqnarray}\mathop{\sum }_{0\leqslant p,i\leqslant d}(-1)^{p-i}\unicode[STIX]{x1D706}_{p,i}(R/I)=1.\end{eqnarray}$$

Moreover, the statements corresponding to Theorem 2.1 (the consecutiveness of nontrivial lines) still hold. In fact, we may assume that $I$ is a radical ideal, and hence $I=I_{\unicode[STIX]{x1D6E5}}$ for some simplicial complex $\unicode[STIX]{x1D6E5}$ , and then reduce to the case when $R$ is a polynomial ring as in [Reference Yanagawa27, Remark 5.14 (b)].

If we assume that $I=\sqrt{I}$ , Proposition 4.4 also holds in the present situation. However, we cannot drop this assumption, since we have no idea whether the condition of being sequentially Cohen–Macaulay is preserved after taking radicals. What is known is that if $R/I$ is Cohen–Macaulay, then so is $R/\sqrt{I}$ (see [Reference Yanagawa30, Theorem 6.1]). Hence, if $R/I$ is Cohen–Macaulay, then the Lyubeznik table of $R/I$ is trivial.

(2) For a radical monomial ideal $I_{\unicode[STIX]{x1D6E5}}$ with $\dim R/I_{\unicode[STIX]{x1D6E5}}=d$ , the highest Lyubeznik number

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{d,d}(R/I_{\unicode[STIX]{x1D6E5}})=\dim _{\Bbbk }[\text{Ext}_{R}^{n-d}(\text{Ext}_{R}^{n-d}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}\end{eqnarray}$$

has a simple topological (or combinatorial) meaning. In fact, to study this number, we may assume that $R$ is a polynomial ring, and we can use a combinatorial description of

$$\begin{eqnarray}\text{Ext}_{R}^{n-d}(\text{Ext}_{R}^{n-d}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})\end{eqnarray}$$

given in [Reference Stanley22, p. 96]. Roughly speaking, $\unicode[STIX]{x1D706}_{d,d}(R/I_{\unicode[STIX]{x1D6E5}})$ is the number of “connected in codimension-one components” of $|\unicode[STIX]{x1D6E5}|$ . (This result holds in a much wider context; see [Reference Zhang31].) In particular, if $R/I_{\unicode[STIX]{x1D6E5}}$ satisfies Serre’s condition $(S_{2})$ , then $\unicode[STIX]{x1D706}_{d,d}(R/I_{\unicode[STIX]{x1D6E5}})=1$ , while the converse is not true.

5.2 Lyubeznik table is a topological invariant

Recall that if $R=\Bbbk [C]$ is simplicial, then an order ideal $\unicode[STIX]{x1D6E5}$ of $L$ is essentially a simplicial complex, and hence it has the geometric realization $|\unicode[STIX]{x1D6E5}|$ . It is natural to ask how Lyubeznik numbers of $R/I_{\unicode[STIX]{x1D6E5}}$ depend on $|\unicode[STIX]{x1D6E5}|$ . The next theorem shows that Lyubeznik numbers are not only algebraic invariants but also topological invariants.

Bearing in mind Theorem 5.2, it suffices to show that

$$\begin{eqnarray}\dim _{\Bbbk }[\text{Ext}_{R}^{n-p}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}\end{eqnarray}$$

depends only on the topology of $|\unicode[STIX]{x1D6E5}|$ and $\operatorname{char}(\Bbbk )$ . For this statement, the assumption that $R$ is simplicial and Gorenstein is irrelevant. (If $R$ is not simplicial, then $\unicode[STIX]{x1D6E5}$ is essentially a CW complex.) In [Reference Okazaki and Yanagawa19, Theorem 2.10], Okazaki and the second author showed that the invariant that is (essentially) equal to

$$\begin{eqnarray}\text{depth}_{R}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}))=\min \{j\mid \text{Ext}_{R}^{n-j}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})\neq 0\}\end{eqnarray}$$

depends only on $|\unicode[STIX]{x1D6E5}|$ and $\operatorname{char}(\Bbbk )$ for each $i$ . Our proof here uses similar arguments to the aforementioned result. To do so, we have to recall some previous work of the second author in [Reference Yanagawa28].

Recall that $P=\mathbb{R}_{{\geqslant}0}C$ is a polyhedral cone associated with the semigroup ring $R=\Bbbk [C]$ . We have a hyperplane $H\subset \mathbb{R}^{n}$ such that $B:=H\cap P$ is an $(n-1)$ -polytope (an $(n-1)$ -simplex, if $R$ is simplicial). For $F\in L$ , set $|F|$ to be the relative interior of the face $F\cap H$ of $B$ . We can regard an order ideal $\unicode[STIX]{x1D6E5}\subseteq L$ as a CW complex (a simplicial complex, if $R$ is simplicial) whose geometric realization is $|\unicode[STIX]{x1D6E5}|:=\bigcup _{F\in \unicode[STIX]{x1D6E5}}|F|\subseteq B$ .

For $F\in L$ ,

$$\begin{eqnarray}U_{F}:=\mathop{\bigcup }_{F^{\prime }\in L,\,F^{\prime }\supset F}|F^{\prime }|\end{eqnarray}$$

is an open set of $B$ . Note that $\{U_{F}\mid \{0\}\neq F\in L\}$ is an open covering of $B$ . In [Reference Yanagawa28], from $M\in \operatorname{Sq}R$ , we constructed a sheaf $M^{+}$ on $B$ . More precisely, the assignment

$$\begin{eqnarray}\unicode[STIX]{x1D6E4}(U_{F},M^{+})=M_{\boldsymbol{ c}\,(F)}\end{eqnarray}$$

for each $F\neq \{0\}$ and the map

$$\begin{eqnarray}\unicode[STIX]{x1D711}_{F,G}^{M}:\unicode[STIX]{x1D6E4}(U_{G},M^{+})=M_{\boldsymbol{ c}\,(G)}\longrightarrow M_{\mathbf{c}\,(F)}=\unicode[STIX]{x1D6E4}(U_{F},M^{+})\end{eqnarray}$$

for $F,G\neq \{0\}$ with $F\supset G$ (equivalently, $U_{G}\supset U_{F}$ ) define a sheaf. Note that $M_{0}$ is irrelevant to $M^{+}$ .

For example, $(R/I_{\unicode[STIX]{x1D6E5}})^{+}\cong j_{\ast }\text{}\underline{\Bbbk }_{|\unicode[STIX]{x1D6E5}|}$ , where $\text{}\underline{\Bbbk }_{|\unicode[STIX]{x1D6E5}|}$ is the constant sheaf on $|\unicode[STIX]{x1D6E5}|$ with coefficients in $\Bbbk$ , and $j$ is the embedding map $|\unicode[STIX]{x1D6E5}|{\hookrightarrow}B$ . Similarly, we have that $(\unicode[STIX]{x1D714}_{R})^{+}\cong h_{!}\text{}\underline{\Bbbk }_{B^{\circ }}$ , where $\text{}\underline{\Bbbk }_{B^{\circ }}$ is the constant sheaf on the relative interior $B^{\circ }$ of $B$ , and $h$ is the embedding map $B^{\circ }{\hookrightarrow}B$ . Note that $(\unicode[STIX]{x1D714}_{R})^{+}$ is the orientation sheaf of $B$ with coefficients in $\Bbbk$ .

Let $\unicode[STIX]{x1D6E5}\subseteq L$ be an order ideal, and set $X:=|\unicode[STIX]{x1D6E5}|\subseteq B$ . For $M\in \operatorname{Sq}R$ , $M$ is an $R/I_{\unicode[STIX]{x1D6E5}}$ -module (i.e., $\operatorname{ann}(M)\supset I_{\unicode[STIX]{x1D6E5}}$ ) if and only if $\text{Supp}(M^{+}):=\{x\in B\mid (M^{+})_{x}\neq 0\}\subseteq X$ . In this case, we have

$$\begin{eqnarray}H^{i}(B;M^{+})\cong H^{i}(X;M^{+}|_{X})\end{eqnarray}$$

for all $i$ . Here, $M^{+}|_{X}$ is the restriction of the sheaf $M^{+}$ to the closed set $X\subseteq B$ . Combining this fact with [Reference Yanagawa28, Theorem 3.3], we have the following.

Theorem 5.4. (Cf. [Reference Yanagawa28, Theorem 3.3])

With the above situation, we have

$$\begin{eqnarray}H^{i}(X;M^{+}|_{X})\cong [H_{\mathfrak{m}}^{i+1}(M)]_{0}\quad \text{for all }i\geqslant 1,\end{eqnarray}$$

and an exact sequence

(5.1) $$\begin{eqnarray}0\longrightarrow [H_{\mathfrak{m}}^{0}(M)]_{0}\longrightarrow M_{0}\longrightarrow H^{0}(X;M^{+}|_{X})\longrightarrow [H_{\mathfrak{m}}^{1}(M)]_{0}\longrightarrow 0.\end{eqnarray}$$

In particular, $[H_{\mathfrak{m}}^{i+1}(R/I_{\unicode[STIX]{x1D6E5}})]_{0}\cong \widetilde{H}^{i}(X;\Bbbk )$ for all $i\geqslant 0$ , where $\widetilde{H}^{i}(X;\Bbbk )$ denotes the $i$ th reduced cohomology of $X$ with coefficients in $\Bbbk$ .

Recall that $X$ admits Verdier’s dualizing complex ${\mathcal{D}}_{X}^{\bullet }$ with coefficients in $\Bbbk$ . For example, ${\mathcal{D}}_{B}^{\bullet }$ is quasi-isomorphic to $(\unicode[STIX]{x1D714}_{R})^{+}[n-1]$ . The first half of (1) of the next theorem is a restatement of [Reference Yanagawa28, Theorem 4.2], and the rest is that of [Reference Yanagawa30, Lemma 5.11].

The Proof of Theorem 5.3.

We show that the dimension of

$$\begin{eqnarray}[\text{Ext}_{R}^{n-p}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}\,(\cong [H_{\mathfrak{m}}^{p}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}))^{\ast }]_{0})\end{eqnarray}$$

depends only on $X$ and $\operatorname{char}(\Bbbk )$ . If $p\geqslant 2$ , then we have

$$\begin{eqnarray}[H_{\mathfrak{m}}^{p}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}))]_{0}\cong H^{p-1}(X;{\mathcal{E}}xt^{1-i}(\text{}\underline{\Bbbk }_{X},{\mathcal{D}}_{X}^{\bullet }))\end{eqnarray}$$

by Theorems 5.4 and 5.5 (1). The right-hand side of the equation clearly depends only on $X$ and $\operatorname{char}(\Bbbk )$ for each $p,i$ . Next, we consider the case $p=0,1$ . By Theorem 5.4, $H_{\mathfrak{m}}^{0}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}))$ and $H_{\mathfrak{m}}^{1}(\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}))$ are the kernel and the cokernel of the map

$$\begin{eqnarray}[\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})]_{0}\longrightarrow \unicode[STIX]{x1D6E4}(X;\text{Ext}_{R}^{n-i}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})^{+}|_{X}),\end{eqnarray}$$

respectively. If $i\geqslant 2$ , the above map is equivalent to the natural map

$$\begin{eqnarray}\text{Ext}^{1-i}(\text{}\underline{\Bbbk }_{X},{\mathcal{D}}_{X}^{\bullet })\longrightarrow \unicode[STIX]{x1D6E4}(X;{\mathcal{E}}xt^{1-i}(\text{}\underline{\Bbbk }_{X},{\mathcal{D}}_{X}^{\bullet }))\end{eqnarray}$$

by Theorem 5.5 (2), and the dimensions of its kernel and cokernel are invariants of $X$ .

It remains to show the case ( $p=0,1$ and) $i=0,1$ . Clearly, $\text{Ext}_{R}^{n}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})\neq 0$ if and only if $\text{Ext}_{R}^{n}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})=\Bbbk$ , if and only if $I_{\unicode[STIX]{x1D6E5}}=\mathfrak{m}$ , if and only if $X=\emptyset$ . Hence, $\unicode[STIX]{x1D706}_{0,0}(R/I_{\unicode[STIX]{x1D6E5}})\neq 0$ if and only if $\unicode[STIX]{x1D706}_{0,0}(R/I_{\unicode[STIX]{x1D6E5}})=1$ , if and only if $X=\emptyset$ . On the other hand, it is easy to check that $\unicode[STIX]{x1D706}_{1,1}(R/I_{\unicode[STIX]{x1D6E5}})$ is always trivial; that is,

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{1,1}(R/I_{\unicode[STIX]{x1D6E5}})=\left\{\begin{array}{@{}ll@{}}1,\quad & \text{if }\dim (R/I_{\unicode[STIX]{x1D6E5}})=1(\text{i.e.,}\dim |\unicode[STIX]{x1D6E5}|=0),\\ 0,\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

(The same is true for the local ring case using the spectral sequence argument as in the proof of Theorem 2.1 or adapting the techniques used in [Reference Walther24].) Hence, the remaining case is only $\unicode[STIX]{x1D706}_{0,1}(R/I_{\unicode[STIX]{x1D6E5}})$ , but the following fact holds.

Claim.

If $R=\Bbbk [C]$ is a simplicial normal semigroup ring that is Gorenstein, then we have

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{0,1}(R/I_{\unicode[STIX]{x1D6E5}})=\left\{\begin{array}{@{}ll@{}}c-1,\quad & \text{if }\dim (R/I_{\unicode[STIX]{x1D6E5}})\geqslant 2~(\text{i.e.},\dim |\unicode[STIX]{x1D6E5}|\geqslant 1),\\ 0,\quad & \text{otherwise,}\end{array}\right.\end{eqnarray}$$

where $c$ is the number of connected components of $|\unicode[STIX]{x1D6E5}^{\prime }|:=|\unicode[STIX]{x1D6E5}|\setminus \{\text{isolated points}\}$ .

Let us prove the claim. We may assume that $\dim (R/I_{\unicode[STIX]{x1D6E5}})>0$ . If $\dim (R/I_{\unicode[STIX]{x1D6E5}})=1$ , then $R/I_{\unicode[STIX]{x1D6E5}}$ is Cohen–Macaulay, and the assertion is clear. Therefore, we may assume that $\dim (R/I_{\unicode[STIX]{x1D6E5}})\geqslant 2$ . First, we consider the case when $I_{\unicode[STIX]{x1D6E5}}$ does not have one-dimensional associated primes; equivalently, $|\unicode[STIX]{x1D6E5}|$ does not admit isolated points (i.e., $|\unicode[STIX]{x1D6E5}|=|\unicode[STIX]{x1D6E5}^{\prime }|$ ). Then, we have

$$\begin{eqnarray}\dim _{R}(\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}))<1.\end{eqnarray}$$

Since $\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})$ is a squarefree module, we have

$$\begin{eqnarray}\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})=[\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})]_{0}.\end{eqnarray}$$

We also have

$$\begin{eqnarray}[\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})]_{0}\cong [H_{\mathfrak{m}}^{1}(R/I_{\unicode[STIX]{x1D6E5}})]_{0}\cong \widetilde{H}^{0}(X;\Bbbk )\cong \Bbbk ^{c-1},\end{eqnarray}$$

where the second isomorphism follows from the last statement of Theorem 5.4. Hence,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D706}_{0,1}(R/I_{\unicode[STIX]{x1D6E5}}) & = & \displaystyle \dim _{\Bbbk }[\text{Ext}_{R}^{n}(\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}\nonumber\\ \displaystyle & = & \displaystyle \dim _{\Bbbk }[\text{Ext}_{R}^{n}(\Bbbk ^{c-1},\unicode[STIX]{x1D714}_{R})]_{0}=c-1,\nonumber\end{eqnarray}$$

and we are done.

Therefore, we now consider the case where $I_{\unicode[STIX]{x1D6E5}}$ admits one-dimensional associated primes. Set $I:=I_{\unicode[STIX]{x1D6E5}^{\prime }}$ . Then, there is a monomial ideal $J$ of $R$ with $I_{\unicode[STIX]{x1D6E5}}=I\cap J$ and $\dim R/J=1$ . Note that $I+J=\mathfrak{m}$ . The short exact sequence $0\rightarrow R/I_{\unicode[STIX]{x1D6E5}}\rightarrow R/I\oplus R/J\rightarrow R/\mathfrak{m}\,(\cong \Bbbk )\rightarrow 0$ yields the exact sequence

(5.2) $$\begin{eqnarray}\displaystyle 0 & \longrightarrow & \displaystyle \text{Ext}_{R}^{n-1}(R/I,\unicode[STIX]{x1D714}_{R})\oplus \text{Ext}_{R}^{n-1}(R/J,\unicode[STIX]{x1D714}_{R})\longrightarrow \text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R})\nonumber\\ \displaystyle & \longrightarrow & \displaystyle \Bbbk \longrightarrow 0.\end{eqnarray}$$

Since Lyubeznik numbers of type $\unicode[STIX]{x1D706}_{1,1}(-)$ are always trivial, we have

$$\begin{eqnarray}[\text{Ext}_{R}^{n-1}(\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}=[\text{Ext}_{R}^{n-1}(\text{Ext}_{R}^{n-1}(R/I,\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}=0\end{eqnarray}$$

and $[\text{Ext}_{R}^{n-1}(\text{Ext}_{R}^{n-1}(R/J,\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}=\Bbbk$ . It is also clear that

$$\begin{eqnarray}\text{Ext}_{R}^{n}(\text{Ext}_{R}^{n-1}(R/J,\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})=0.\end{eqnarray}$$

Thus, applying $\text{Ext}_{R}^{\bullet }(-,\unicode[STIX]{x1D714}_{R})$ to (5.2), we obtain

$$\begin{eqnarray}\displaystyle 0 & \longrightarrow & \displaystyle [\text{Ext}_{R}^{n-1}(\text{Ext}_{R}^{n-1}(R/J,\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}~(\cong \Bbbk )\longrightarrow [\text{Ext}_{R}^{n}(\Bbbk ,\unicode[STIX]{x1D714}_{R})]_{0}~(\cong \Bbbk )\nonumber\\ \displaystyle & \longrightarrow & \displaystyle [\text{Ext}_{R}^{n}(\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}\nonumber\\ \displaystyle & \longrightarrow & \displaystyle [\text{Ext}_{R}^{n}(\text{Ext}_{R}^{n-1}(R/I,\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}\longrightarrow 0.\nonumber\end{eqnarray}$$

Since $[\text{Ext}_{R}^{n}(\text{Ext}_{R}^{n-1}(R/I,\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}\cong \Bbbk ^{c-1}$ , as we have shown above, it follows that

$$\begin{eqnarray}[\text{Ext}_{R}^{n}(\text{Ext}_{R}^{n-1}(R/I_{\unicode[STIX]{x1D6E5}},\unicode[STIX]{x1D714}_{R}),\unicode[STIX]{x1D714}_{R})]_{0}\cong \Bbbk ^{c-1},\end{eqnarray}$$

and we are done. ◻

Example 5.6. This example concerns the final step of the proof of Theorem 5.3. Let $R=\Bbbk [x_{1},\ldots ,x_{7}]$ be a polynomial ring, and consider the monomial ideal

$$\begin{eqnarray}\displaystyle I_{\unicode[STIX]{x1D6E5}} & = & \displaystyle (x_{2},x_{3},x_{4},x_{5},x_{6},x_{7})\cap (x_{1},x_{4},x_{5},x_{6},x_{7})\nonumber\\ \displaystyle & \cap & \displaystyle (x_{1},x_{2},x_{3},x_{6},x_{7})\cap (x_{1},x_{2},x_{3},x_{4},x_{5}).\nonumber\end{eqnarray}$$

Then, $|\unicode[STIX]{x1D6E5}|$ consists of one isolated point and three segments; see Figure 1. Therefore, $|\unicode[STIX]{x1D6E5}^{\prime }|$ , which is $|\unicode[STIX]{x1D6E5}|\setminus \{v_{1}\}$ , consists of three segments. We have $\unicode[STIX]{x1D706}_{0,1}(R/I_{\unicode[STIX]{x1D6E5}})=3-1=2$ .

Figure 1.

$\unicode[STIX]{x1D6E5}$ in Example 5.6.