Proof (1) For each $n\geq 0$ , we have

$$ \begin{align*}{\mathrm{\mathfrak{ca}}}^n(R)=\bigcap_{M\in {\mathrm{mod}}(R)}{\mathrm{ca}}_R^n(M)\subseteq \bigcap_{M\in {\mathrm{mod}}(R)}{\mathrm{ca}}_R(M)={\mathrm{ca}}_R({\mathrm{mod}}(R)). \end{align*} $$

Thus, $ {\mathrm {\mathfrak {ca}}}(R)\subseteq {\mathrm {ca}}_R( {\mathrm {mod}}(R))$ as $ {\mathrm {\mathfrak {ca}}}(R)= {\mathrm {\mathfrak {ca}}}^n(R)$ for $n\gg 0$ .

(2) By definition, $ {\mathrm {ca}}_R( {\mathrm {mod}}(R))=R$ if and only if $ {\mathrm {ca}}_R(M)=R$ for each $M\in {\mathrm {mod}}(R)$ . This is equivalent to $ {\mathrm {pd}}_R(M)<\infty $ for each $M\in {\mathrm {mod}}(R)$ (see 3.1). This is equivalent to that R is regular (see 2.2).

(3) By [Reference Iyengar and Takahashi32, Example 2.5], $ {\mathrm {\mathfrak {ca}}}(R)=R$ if and only if $ {\mathrm {gl.dim}}(R)<\infty $ . The desired result now follows immediately from 2.2.

Proof (1) This follows from [Reference Dey and Takahashi22, Lemma 3.8] and [Reference Iyengar and Takahashi32, Lemma 2.14].

(2) Combining with 3.7, $ {\mathrm {pd}}_{R_{\mathfrak {p}}}(M_{\mathfrak {p}})\geq n$ if and only if $\mathfrak {p}\in {\mathrm {Supp}}_R {\mathrm {Ext}}^n_R(M,\Omega ^n_R(M))$ . Since $ {\mathrm {Ext}}_R^n(M,\Omega ^n_R(M))$ is finitely generated over R,

$$ \begin{align*}{\mathrm{Supp}}_R{\mathrm{Ext}}^n_R(M,\Omega^n_R(M))=V({\mathrm{ann}}_R{\mathrm{Ext}}^n_R(M,\Omega^n_R(M)). \end{align*} $$

Combining this with (1), the desired statement follows.

(3) It is clear that

$$ \begin{align*}{\mathrm{ca}}_R^n(M)\subseteq {\mathrm{ann}}_R{\mathrm{Ext}}^n_R(M,{\mathrm{mod}}(R))\subseteq {\mathrm{ann}}_R{\mathrm{Ext}}^n_R(M,\Omega^n_R(M)). \end{align*} $$

By (1), these inclusions are equal.

(4) For each $X\in {\mathrm {mod}}(R)$ , consider the isomorphism

$$ \begin{align*}{\mathrm{Ext}}^n_R(\Omega_R^1(M),X)\cong {\mathrm{Ext}}^{n+1}_R(M,X). \end{align*} $$

Combining with (3), we have $ {\mathrm {ca}}_R^{n+1}(M)= {\mathrm {ca}}_R^n(\Omega ^1_R(M))$ . Then the first statement of (4) can be obtained by induction on the number i. Thus, for each $i\geq 1$ ,

$$ \begin{align*}{\mathrm{ca}}_R(M)=\bigcup_{n\geq 0}{\mathrm{ca}}_R^{n+i}(M)=\bigcup_{n\geq 0}{\mathrm{ca}}_R^n(\Omega^i_R(M))={\mathrm{ca}}_R(\Omega^i_R(M)). \end{align*} $$

This completes the proof.

(5) The second statement follows from the first one. For the first statement, combining with (4), it suffices to prove the case of $n=1$ . By (1), we get an inclusion $S^{-1} {\mathrm {ca}}^1_R(M)\subseteq {\mathrm {ca}}^1_{S^{-1}R}(S^{-1}M)$ . For the converse, assume $r/s\in {\mathrm {ca}}^1_{S^{-1}R}(S^{-1}M)$ , where $s\in S$ and $r\in R$ . It remains to prove that $r/s\in S^{-1} {\mathrm {ca}}^1_R(M)$ . By (1), we get that the multiplication $r/s\colon S^{-1}M\rightarrow S^{-1}M$ factors through a finitely generated projective $S^{-1}R$ -module, and hence $r/s\colon S^{-1}M\rightarrow S^{-1}M$ factors through a finitely generated free $S^{-1}R$ -module. Combining with the fact that the isomorphism $S^{-1} {\mathrm {Hom}}_R(X,Y)\cong {\mathrm {Hom}}_{S^{-1}R}(S^{-1}X,S^{-1}Y)$ holds for $X,Y\in {\mathrm {mod}}(R)$ , we conclude that there is a factorization

where F is a finitely generated free R-module, $s_1,s_2\in S$ , $\alpha \colon M\rightarrow F$ , and $\beta \colon F\rightarrow M$ . From the factorization, it follows that there exists $t\in S$ such that $ts_2s_1r=ts\beta \alpha $ . Note that $r/s=(ts_2s_1r)/(ts_2s_1s)$ and $ts_2s_1r=ts\beta \alpha $ factors through F. By (1), $ts_2s_1r\in {\mathrm {ca}}^1_R(M)$ , and hence $r/s\in S^{-1} {\mathrm {ca}}_R^1(M)$ .

Proof The first equality follows from the following:

$$ \begin{align*} {\mathrm{IPD}}(M)& =\bigcap_{n\geq 1}{\mathrm{Supp}}_R {\mathrm{Ext}}^n_R(M,\Omega^n_R(M))\\ & =\bigcap_{n\geq 1}V({\mathrm{ca}}_R^n(M))\\ &= V(\bigcup_{n\geq 1}{\mathrm{ca}}_R^n(M))\\ &=V({\mathrm{ca}}_R(M)), \end{align*} $$

where the first two equalities follow from Lemma 3.8; the first equality also follows directly from the proof of [Reference Iyengar and Takahashi33, Lemma 2.3].

Next, we prove the second equality. The inclusion $V( {\mathrm {ca}}_R(M))\subseteq V( {\mathrm {ca}}_R^{d+1}(M))$ is clear. Combining with the first equality, it remains to prove $V( {\mathrm {ca}}_R^{d+1}(M))\subseteq {\mathrm {IPD}}(M)$ . If not, let $\mathfrak {p}$ be a prime ideal containing $ {\mathrm {ca}}_R^{d+1}(M)$ and $\mathfrak {p}\notin {\mathrm {IPD}}(M)$ . By assumption, $ {\mathrm {pd}}_{R_{\mathfrak {p}}}(M_{\mathfrak {p}})\leq d$ . However, it follows from Lemma 3.8 that $ {\mathrm {pd}}_{R_{\mathfrak {p}}}(M_{\mathfrak {p}})\geq d+1$ . This is a contradiction.

Proof (1) For each $j\geq 0$ , consider the short exact sequence

$$ \begin{align*}0\rightarrow \Omega_R^{j+1}(M)\rightarrow P\rightarrow \Omega_R^{j}(M)\rightarrow 0, \end{align*} $$

where P is a finitely generated projective R-module. Applying $ {\mathrm {Hom}}_R(M,-)$ to this short exact sequence, we get an exact sequence

$$ \begin{align*}{\mathrm{Ext}}^{j}_{R}(M, P) \to {\mathrm{Ext}}^{j}_{R}(M, \Omega^{j}_{R} (M)) \to {\mathrm{Ext}}^{j+1}_{R}(M, \Omega^{j+1}_{R} (M)).\end{align*} $$

Set $j=m$ . The assumption yields that $ {\mathrm {Ext}}^m_R(M,P)=0$ . Thus, the above exact sequence yields that $ {\mathrm {ca}}^{m+1}_R(M)\subseteq {\mathrm {ca}}^m_R(M)$ (see Lemma 3.8), and hence $ {\mathrm {ca}}^m_R(M)= {\mathrm {ca}}^{m+1}_R(M)$ . The desired result follows similarly by choosing $j=m+1,\ldots ,n$ .

(2) For each $M\in {\mathrm {mod}}(R)$ and $i\geq d$ , $\Omega ^{i}_R(M)$ is in $ {\mathrm {CM}}(R)$ . Choose a minimal free resolution of $\Omega ^i_R(M)$ : $ 0\rightarrow \Omega ^{i+1}_R(M)\rightarrow F\rightarrow \Omega ^i_R(M)\rightarrow 0, $ where F is a finitely generated free R-module. In this case, $\Omega ^{i+1}_R(M)$ does not have a nonzero projective summand (see [Reference Herzog30, Lemma 1.4]), and hence $\Omega ^{i+1}_R(M)\in {\Omega \text{ CM}}^\times (R)$ . By assumption, $ {\mathrm {Hom}}_R(\Omega ^{i+1}_R(M),\omega )\in {\Omega \text{ CM}}(R)$ . It follows from [Reference Kobayashi and Takahashi38, Proposition 5.2] that there exists a short exact sequence

$$ \begin{align*}0\rightarrow K\rightarrow \omega^{\oplus s}\rightarrow \Omega^{i+1}_R(M)\rightarrow 0 \end{align*} $$

in $ {\mathrm {mod}}(R)$ . Note that $ {\mathrm {Ext}}^{>0}_R(\Omega ^i_R(M),\omega )=0$ as $\Omega ^i_R(M)\in {\mathrm {CM}}(R)$ . Applying $ {\mathrm {Hom}}_R(\Omega ^i_R(M),-)$ to the above short exact sequence, $ {\mathrm {Ext}}^1_R(\Omega ^i_R(M),\Omega ^{i+1}_R(M))\cong {\mathrm {Ext}}^2_R(\Omega ^i_R(M),K). $ We conclude from this that $ {\mathrm {Ext}}^{i+1}_R(M,\Omega ^{i+1}_R(M))\cong {\mathrm {Ext}}^{i+2}_R(M,K)$ . This isomorphism implies $ {\mathrm {ca}}_R^{i+2}(M)\subseteq {\mathrm {ann}}_{R} {\mathrm {Ext}}^{i+1}_R(M,\Omega ^{i+1}_R(M))= {\mathrm {ca}}_R^{i+1}(M), $ where the second equality follows from Lemma 3.8. Hence, $ {\mathrm {ca}}_R^{i+1}(M)= {\mathrm {ca}}_R^{i+2}(M)$ for each $i\geq d$ . This implies $ {\mathrm {ca}}_R(M)= {\mathrm {ca}}_R^{d+1}(M)$ .

For the second statement, it follows from Lemma 3.5 that $ {\mathrm {\mathfrak {ca}}}^{d+1}(R)\subseteq {\mathrm {\mathfrak {ca}}}(R)\subseteq {\mathrm {ca}}_R( {\mathrm {mod}}(R))$ . The desired result now follows from the following:

$$ \begin{align*}{\mathrm{\mathfrak{ca}}}^{d+1}(R)=\bigcap_{M\in {\mathrm{mod}}(R)}{\mathrm{ca}}_R^{d+1}(M)=\bigcap_{M\in {\mathrm{mod}}(R)}{\mathrm{ca}}_R(M)={\mathrm{ca}}_R({\mathrm{mod}}(R)). \end{align*} $$

This finishes the proof.

Proof (1) For each $M\in {\mathrm {CM}}(R)$ , we have $ {\mathrm {Ext}}^{>0}_R(M,R)=0$ (see 2.5). The desired result now follows from Lemma 3.11.

(2) Note that 3.13 yields that $l= \mathrm{Rfd}_R(M)$ is finite. By [Reference Bruns and Herzog12, Exercise 1.3.7], $\Omega ^l_R(M)$ is maximal Cohen–Macaulay over R. Let $i\geq 1$ be an integer. By Lemma 3.8, we have

$$ \begin{align*}{\mathrm{ca}}_R^{l+i}(M)={\mathrm{ann}}_R\,\ \underline{{\mathrm{End}}}_R(\Omega^{l+i-1}_R(M)). \end{align*} $$

Note that $\ \underline { {\mathrm {End}}}_R(\Omega ^{l+i-1}_R(M))\cong \ \underline { {\mathrm {End}}}_R(\Omega ^l_R(M))$ as $\Omega ^{l}_R(M)\in {\mathrm {CM}}(R)$ (see [Reference Buchweitz13, Theorem B.1.8]). Thus, $ {\mathrm {ca}}_R^{l+i}(M)= {\mathrm {ca}}_R^{l+1}(M)$ , and hence $ {\mathrm {ca}}_R(M)= {\mathrm {ca}}_R^{l+1}(M)$ .

(3) In [Reference Esentepe24, Lemma 2.3], Esentepe observed that $ {\mathrm {\mathfrak {ca}}}(R)=\displaystyle \bigcap \limits _{M\in {\mathrm {CM}}(R)} {\mathrm {ann}}_R\,\ \underline { {\mathrm {End}}}_R(M).$ By the same argument in the proof of Lemma 3.11, we have $ {\mathrm {\mathfrak {ca}}}^{d+1}(R)= {\mathrm {\mathfrak {ca}}}(R)= {\mathrm {ca}}_R( {\mathrm {mod}}(R))$ . This completes the proof.

Proof (1) The second statement follows from the first one and the fact that the localization functor $ {\mathrm {mod}}(R)\rightarrow {\mathrm {mod}}(R_{\mathfrak {p}})$ is dense for each $\mathfrak {p}\in {\mathrm {Spec}}(R)$ . For the first statement, let $n= \mathrm{n}_R(M)+1$ . It follows that $ {\mathrm {ca}}^n_R(M)= {\mathrm {ca}}_R(M)$ . By Lemma 3.8(5), for each $\mathfrak {p}\in {\mathrm {Spec}}(R)$ , we have

$$ \begin{align*}{\mathrm{ca}}^n_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})={\mathrm{ca}}_R^n(M)_{\mathfrak{p}}={\mathrm{ca}}_R(M)_{\mathfrak{p}}={\mathrm{ca}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}}).\end{align*} $$

It follows from this that $ \mathrm{n}_R(M)\geq \sup \{ \mathrm{n}_{R_{\mathfrak {p}}}(M_{\mathfrak {p}})\mid \mathfrak {p} \in {\mathrm {Spec}}(R)\} \geq \sup \{ \mathrm{n}_{R_{\mathfrak {m}}}(M_{\mathfrak {m}})\mid \mathfrak {m} \in {\mathrm {Max}}(R)\}$ . Assume $m=\sup \{ \mathrm{n}_{R_{\mathfrak {m}}}(M_{\mathfrak {m}})\mid \mathfrak {m} \in {\mathrm {Max}}(R)\}+1$ . Again by Lemma 3.8(5), we conclude that the localization of the inclusion $ {\mathrm {ca}}^m_R(M)\subseteq {\mathrm {ca}}_R(M)$ at each maximal ideal of R is equal, and hence $ {\mathrm {ca}}^m_R(M)= {\mathrm {ca}}_R(M)$ . It follows from this that $ \mathrm{n}_R(M)\leq m=\sup \{ \mathrm{n}_{R_{\mathfrak {m}}}(M_{\mathfrak {m}})\mid \mathfrak {m} \in {\mathrm {Max}}(R)\}$ .

(2) Let $\mathfrak {p} \in {\mathrm {Spec}}(R)$ , and let J be an ideal of $R_{\mathfrak {p}}$ generated by a maximal regular sequence of $R_{\mathfrak {p}}$ . Since J has finite projective dimension over $R_{\mathfrak {p}}$ , we obtain the first equality below

$$\begin{align*}\mathrm{n}_{R_{\mathfrak{p}}}(R_{\mathfrak{p}}/J)={\mathrm{pd}}_{R_{\mathfrak{p}}}(R_{\mathfrak{p}}/J)={\mathrm{depth}}(R_{\mathfrak{p}}), \end{align*}$$

where the second equality is by our choice of J. Combining with (1), we conclude that $ n_R( {\mathrm {mod}}(R)) \geq \sup \{ {\mathrm {depth}} (R_{\mathfrak {p}}) \mid \mathfrak {p} \in {\mathrm {Spec}}(R) \} \geq \dim (R) - 1, $ where the second inequality is by [Reference Christensen, Foxby and Frankild15, Lemma 1.4].

Proof First, we prove the $ {\mathrm {ca}}_R(M)\subseteq {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(M)$ . Let $r\in {\mathrm {ca}}_R^1(M)$ . Note that $ {\mathrm {ca}}_R^1(M)= {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)$ (see Lemma 3.8). Thus, the multiplication map $r\colon M\rightarrow M$ factors through a projective R-module. In particular, the multiplication map $r\colon M\rightarrow M$ in the singularity category $\mathsf {D}_{\mathsf {sg}}(R)$ is zero. This yields that $r\in {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(M)$ , and hence $ {\mathrm {ca}}_R^1(M)\subseteq {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(M)$ . For each $n>1$ ,

$$ \begin{align*} {\mathrm{ca}}_R^n(M) ={\mathrm{ca}}_R^1(\Omega^{n-1}_R(M))\subseteq {\mathrm{ann}}_{\mathsf{D}_{\mathsf{sg}}(R)}(\Omega^{n-1}_R(M)), \end{align*} $$

where the equality is by Lemma 3.8, and the inclusion follows from the argument above. By the isomorphism $\Omega ^{n-1}_R(M)[n-1]\cong M$ , we have

$$ \begin{align*}{\mathrm{ann}}_{\mathsf{D}_{\mathsf{sg}}(R)}(\Omega^{n-1}_R(M))={\mathrm{ann}}_{\mathsf{D}_{\mathsf{sg}}(R)}(M).\end{align*} $$

Thus, $ {\mathrm {ca}}_R^n(M)\subseteq {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(M)$ for all $n>1$ , and hence $ {\mathrm {ca}}_R(M)\subseteq {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(M).$

Now, we prove the converse direction. Assume $a\in {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(M)$ . That is, the multiplication map $a\colon M\rightarrow M$ in $\mathsf {D}_{\mathsf {sg}}(R)$ is zero. This implies that the multiplication map $a\colon M\rightarrow M$ in $\mathsf {D}^f(R)$ factors through a perfect complex P. That is, in $\mathsf {D}^f(R)$ , there is a commutative diagram

Since P is perfect, there exists $n>0$ such that

$$ \begin{align*}{\mathrm{Hom}}_{\mathsf{D}^f(R)}(P,N[i])=0 \end{align*} $$

for all $i\geq n$ and $N\in {\mathrm {mod}}(R)$ . Applying $ {\mathrm {Hom}}_{\mathsf {D}^f(R)}(-,N[i])$ on the above commutative diagram, we get that the multiplication map

$$ \begin{align*}a\colon {\mathrm{Hom}}_{\mathsf{D}^f(R)}(M,N[i])\rightarrow {\mathrm{Hom}}_{\mathsf{D}^f(R)}(M,N[i]) \end{align*} $$

is zero for all $i\geq n$ and $N\in {\mathrm {mod}}(R)$ . Combining with the isomorphism

$$ \begin{align*}{\mathrm{Ext}}^i_R(M,N)\cong {\mathrm{Hom}}_{\mathsf{D}^f(R)}(M,N[i]),\end{align*} $$

we conclude that $a\cdot {\mathrm {Ext}}^i_R(M,N)=0$ for all $i\geq n$ and $N\in {\mathrm {mod}}(R)$ . That is, $a\in {\mathrm {ca}}_R^n(M)$ , and hence $a\in {\mathrm {ca}}_R(M)$ . It follows that $ {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(M)\subseteq {\mathrm {ca}}_R(M)$ .

Proof Let X be a complex in $\mathsf {D}_{\mathsf {sg}}(R)$ . By choosing a projective resolution of X, we may assume that X is a bounded above complex of finitely generated projective R-modules with finitely many nonzero cohomologies. Then by taking brutal truncation, we conclude that $X[n]$ is isomorphic to a finitely generated R-module in $\mathsf {D}_{\mathsf {sg}}(R)$ for $n\ll 0$ . Assume $X[n]\cong M$ for some $n\ll 0$ and $M\in {\mathrm {mod}}(R)$ . Note that $ {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(X)= {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(M)$ . This yields the second equality below:

$$ \begin{align*} {\mathrm{ann}}_R\mathsf{D}_{\mathsf{sg}}(R)&=\bigcap_{X\in \mathsf{D}_{\mathsf{sg}}(R)}{\mathrm{ann}}_{\mathsf{D}_{\mathsf{sg}}(R)}(X)\\ &=\bigcap_{M\in {\mathrm{mod}}(R)}{\mathrm{ann}}_{\mathsf{D}_{\mathsf{sg}}(R)}(M)\\ &=\bigcap_{M\in {\mathrm{mod}}(R)}{\mathrm{ca}}_R(M), \end{align*} $$

where the third equality follows from Proposition 4.2.

Proof It follows from Proposition 3.10 that $V( {\mathrm {ca}}_R(M))= {\mathrm {IPD}}(M)$ . It is clear that

$$ \begin{align*}{\mathrm{IPD}}(M)\subseteq {\mathrm{Sing}}(R).\end{align*} $$

By [Reference Iyengar and Takahashi32, Lemma 2.6], the equality holds if M is a generator of $\mathsf {D}_{\mathsf {sg}}(R)$ .

Proof Let $s,t$ be nonnegative integers and G be a finitely generated R-module. Assume $\Omega ^s_R( {\mathrm {mod}}(R))\subseteq |G|_t^R$ . Since $|G|_t^R\subseteq [G]_t^R$ , $\Omega _{R}^{s}( {\mathrm {mod}}(R))\subseteq {\lbrack G\rbrack }_{t}^{R}$ (cf. 4.5). Let $\mathfrak {p}$ be a prime ideal of R. We claim that $\Omega _{R_{\mathfrak {p}}}^{s}( {\mathrm {mod}} (R_{\mathfrak {p}}))\subseteq {\lbrack G_{\mathfrak {p}}\rbrack }_{t}^{R_{\mathfrak {p}}}.$ Combining with this, 4.5 yields that $\dim (R_{\mathfrak {p}})\leq s$ , and hence $\dim (R) \leq s$ as $\mathfrak {p}$ is arbitrary.

Now, we proceed to prove the claim. Let X be an object in $\Omega _{R_{\mathfrak {p}}}^{s}( {\mathrm {mod}}(R_{\mathfrak {p}}))$ . Since the functor $(-)_{\mathfrak {p}}\colon {\mathrm {mod}}(R)\to {\mathrm {mod}}(R_{\mathfrak {p}})$ is dense, there exists $M\in {\mathrm {mod}}(R)$ such that $X\cong \Omega _{R_{\mathfrak {p}}}^{s}(M_{\mathfrak {p}})$ . On the other hand, there exist $n,m\geq 0$ such that $\Omega _{R_{\mathfrak {p}}}^{s}(M_{\mathfrak {p}})\oplus R_{\mathfrak {p}}^{\oplus n}\cong \left (\Omega _{R}^{s}(M)\right )_{\mathfrak {p}}\oplus R_{\mathfrak {p}}^{\oplus m}$ . Hence,

$$ \begin{align*}X\oplus R_{\mathfrak{p}}^{\oplus n}\cong \left(\Omega_{R}^{s}(M)\right)_{\mathfrak{p}} \oplus R_{\mathfrak{p}}^{\oplus m}\cong \left(\Omega_{R}^{s}(M) \oplus R^{\oplus m}\right)_{\mathfrak{p}}\in \left(\Omega_{R}^{s}({\mathrm{mod}}(R))\right)_{\mathfrak{p}}\subseteq {\lbrack G_{\mathfrak{p}}\rbrack}_{t}^{R_{\mathfrak{p}}};\end{align*} $$

the inclusion here uses $ \left (\Omega _{R}^{s}( {\mathrm {mod}}(R))\right )_{\mathfrak {p}}\subseteq ( [G]_{t}^{R})_{\mathfrak {p}}\subseteq {\lbrack G_{\mathfrak {p}}\rbrack }_{t}^{R_{\mathfrak {p}}}.$ As ${\lbrack G_{\mathfrak {p}}\rbrack }_{t}^{R_{\mathfrak {p}}}$ is closed under direct summands, one has $X\in {\lbrack G_{\mathfrak {p}}\rbrack }_{t}^{R_{\mathfrak {p}}}$ .

Suppose that R is a quasi-excellent ring of finite Krull dimension d and either admits a dualizing complex or is a local ring. Combining [Reference Iyengar and Takahashi32, Theorem 5.2] and [Reference Kimura37, Corollary 2.6], there exist $t\geq 0$ and $G\in {\mathrm {mod}}(R)$ such that $\Omega ^d_R( {\mathrm {mod}}(R))\subseteq |G|_t^R$ . This implies that the converse of the inequality stated in the proposition holds.

Proof (1) Assume G is a generator of $\mathsf {D}_{\mathsf {sg}}(R)$ . With the same argument in the proof of Corollary 4.3, there exists $n\in \mathbb {Z}$ such that $G[n]\cong M$ in $\mathsf {D}_{\mathsf {sg}}(R)$ for some $M\in {\mathrm {mod}}(R)$ . Then M is also a generator of $\mathsf {D}_{\mathsf {sg}}(R)$ . Combining with Lemma 4.4 and Proposition 4.2,

$$ \begin{align*}{\mathrm{Sing}}(R)=V({\mathrm{ca}}_R(M))=V({\mathrm{ann}}_{\mathsf{D}_{\mathsf{sg}}(R)}(M)). \end{align*} $$

The desired result follows as $V( {\mathrm {ca}}_R(M))= {\mathrm {IPD}}(M)$ (see Proposition 3.10).

(2) By Corollary 4.3, $ {\mathrm {ann}}_R\mathsf {D}_{\mathsf {sg}}(R)= {\mathrm {ca}}_R( {\mathrm {mod}}(R))$ . It remains to prove the first equality. Assume $\mathsf {D}_{\mathsf {sg}}(R)= {\mathrm {\mathsf {thick}}}^s_{\mathsf {D}_{\mathsf {sg}}(R)}(G)$ for some $G\in \mathsf {D}_{\mathsf {sg}}(R)$ and $s\geq 0$ . By definition, $ {\mathrm {ann}}_R\mathsf {D}_{\mathsf {sg}}(R)\subseteq {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(G)$ . Combining with the assumption, it follows from [Reference Esentepe24, Lemma 2.1] that

$$ \begin{align*}({\mathrm{ann}}_{\mathsf{D}_{\mathsf{sg}}(R)}(G))^s\subseteq {\mathrm{ann}}_R\mathsf{D}_{\mathsf{sg}}(R).\end{align*} $$

Thus, $ {\mathrm {ann}}_R\mathsf {D}_{\mathsf {sg}}(R)$ is equal to $ {\mathrm {ann}}_{\mathsf {D}_{\mathsf {sg}}(R)}(G)$ up to radical. This yields that

$$ \begin{align*}V({\mathrm{ann}}_R\mathsf{D}_{\mathsf{sg}}(R))=V({\mathrm{ann}}_{\mathsf{D}_{\mathsf{sg}}(R)}(G)). \end{align*} $$

The desired result now follows immediately from (1); note that $G[n]\cong M$ in $\mathsf {D}_{\mathsf {sg}}(R)$ for some $M\in {\mathrm {mod}}(R)$ and $n\in \mathbb {Z}$ .

(3) The strong generation of $ {\mathrm {mod}}(R)$ implies the strong generation of $\mathsf {D}_{\mathsf {sg}}(R)$ (see 2.13). Combining with (2), it remains to show the first equality, and this follows immediately by combining Proposition 4.6 and [Reference Iyengar and Takahashi32, Theorem 1.1].

(4) The first statement follows from Corollaries 3.5 and 4.3. The second statement follows from (3).

Proof By Proposition 4.8, $ {\mathrm {ann}}_{R}\mathsf {D}_{\mathsf {sg}}(R)$ defines the singular locus of R. It follows from this that $ {\mathrm {ann}}_{R}\mathsf {D}_{\mathsf {sg}}(R)\neq 0$ . If not, the zero ideal of R, denoted $(0)$ , is in $ {\mathrm {Sing}}(R)$ . This contradicts that $R_{(0)}$ is a field. Hence, $ {\mathrm {ann}}_{R}\mathsf {D}_{\mathsf {sg}}(R)\neq 0$ .

Proof This follows immediately from Proposition 4.8 and Example 4.12.

Proof $(1)\Rightarrow (2)$ : By Proposition 4.8, one has $ {\mathrm {Sing}} (R)=V( {\mathrm {ann}}_{R}\mathsf {D}_{\mathsf {sg}}(R))$ . Since R has isolated singularities, $ {\mathrm {Sing}}(R)$ consists of maximal ideals. Hence, $V( {\mathrm {ann}}_{R}\mathsf {D}_{\mathsf {sg}}(R))$ consists of maximal ideals. This implies $(2)$ .

$(2)\Rightarrow (3)$ : This follows from Corollary 4.3.

$(3)\Rightarrow (4)$ : If $R/ {\mathrm {ca}}_R( {\mathrm {mod}}(R))=0$ , then R is regular (see Lemma 3.5). Thus, for each $M\in {\mathrm {mod}}(R)$ , $\Omega ^i_R(M)$ is finitely generated projective for some $i\geq 0$ , and hence $\Omega ^i_R(M)\in |R|_1$ . This implies that $ {\mathrm {mod}}(R)$ has a point-wise strong generator. Next, we assume that $R/ {\mathrm {ca}}( {\mathrm {mod}}(R))$ is Artinian.

Set $I := {\mathrm {ca}}_R( {\mathrm {mod}}(R))$ and assume ${\boldsymbol {x}}=x_1,\ldots ,x_n$ is a generating set of I. Let M be a finitely generated R-module. Then

$$ \begin{align*}I={\mathrm{ca}}_{R}({\mathrm{mod}}(R))\subseteq{\mathrm{ca}}_{R}(M)={\mathrm{ca}}_{R}^{i+1}(M)={\mathrm{ca}}_{R}^{1}(\Omega_{R}^{i}(M))\end{align*} $$

for some $i\geq 0$ , where the second equality is by Lemma 3.8. This implies that

$$ \begin{align*}I\subseteq {\mathrm{ann}}_R{\mathrm{Ext}}^1_R(\Omega^i_R(M),\Omega^{i+1}_R(M)). \end{align*} $$

Combining with this, [Reference Bahlekeh, Hakimian, Salarian and Takahashi5, Theorem 3.1(1)] shows that there exists $L\in {\mathrm {mod}}(R/I)$ such that $ \Omega ^i_R(M)\in |\displaystyle \bigoplus _{j=1}^n\Omega _R^j(L)|^R_{n+1}. $ Since $R/I$ is Artinian, $L\in |N|^{R/I}_{\ell \ell (R/I)}$ , where $\ell \ell (R/I)$ is the Loewy length of $R/I$ and N is the quotient of $R/I$ by its Jacobson radical. Restriction scalars along the map $R\rightarrow R/I$ , one has $L\in |N|^R_{\ell \ell (R/I)}$ , and hence $\Omega ^j_R(L)\subseteq |\Omega ^j_R(N)|^R_{\ell \ell (R/I)}$ for each $j\geq 0$ . It follows from this that $\displaystyle \bigoplus _{j=1}^n\Omega ^j_R(L)\in |\displaystyle \bigoplus _{j=1}^n\Omega ^j_R(N)|^R_{\ell \ell (R/I)}$ . By [Reference Dey, Lank and Takahashi19, Lemma 3.8], $\Omega ^i_R(M)\in |\displaystyle \bigoplus _{j=1}^n\Omega ^j_R(N)|^R_{(n+1)\ell \ell (R/I)}$ . Hence, $\displaystyle \bigoplus _{j=1}^n\Omega ^j_R(N)$ is a point-wise strong generator of $ {\mathrm {mod}}(R)$ .

$(4)\Rightarrow (1)$ : This can be deduced by [Reference Bahlekeh, Salarian, Takahashi and Toosi6, Proposition 5.3].

Proof (1) We claim that $ \mathrm{n}_R( {\mathrm {mod}}(R))<\infty $ . By Proposition 3.16, this is equivalent to showing that $\sup \{ \mathrm{n}_{R_{\mathfrak {m}}}( {\mathrm {mod}}(R_{\mathfrak {m}})\mid \mathfrak {m}\in {\mathrm {Max}}(R)\}<\infty $ . If $\mathfrak {m}$ is a maximal ideal of height $0$ , then by hypothesis $\mathfrak {m}^2R_{\mathfrak {m}}=0$ , and hence $ \mathrm{n}_{R_{\mathfrak {m}}}( {\mathrm {mod}}(R_{\mathfrak {m}})) \leq 1$ by Remark 3.15(3). If $\mathfrak {m}$ is a maximal ideal height $1$ , then, if we can show that $ {\mathrm {\Omega }} {\mathrm {CM}}^{\times }(R_{\mathfrak {m}})$ is closed under canonical dual, we can conclude by Lemma 3.11 that $ \mathrm{n}_{R_{\mathfrak {m}}}( {\mathrm {mod}} (R_{\mathfrak {m}})) \leq~1$ . So we now verify that for any one-dimensional almost Gorenstein local ring R, $ {\mathrm {\Omega }} {\mathrm {CM}}^{\times }(R)$ is closed under canonical dual. Indeed, if R is almost Gorenstein, then $\mathfrak {m}^{\dagger }\in \mathrm { Ul}_{\omega }(R)$ by [Reference Dao, Maitra and Sridhar16, Proposition 7.2 and Lemma 4.15]. Notice that for any $M\in {\mathrm {\Omega }} {\mathrm {CM}}^{\times }(R)$ , $M\cong {\mathrm {Hom}}_R(M^*, R)= {\mathrm {Hom}}_R(M^*,\mathfrak {m})\cong {\mathrm {Hom}}_R(\mathfrak {m}^{\dagger }, (M^*)^{\dagger })$ , where the last module belongs to $\mathrm {Ul}_{\omega }(R)$ by [Reference Dao, Maitra and Sridhar16, Lemma 4.15]. Thus, by [Reference Dao, Maitra and Sridhar16, Corollary 4.27], we get $ {\mathrm {Hom}}_R(M,\omega )\cong M^*\in {\mathrm {\Omega }} {\mathrm {CM}}^{\times }(R)$ as we wanted. Therefore, $ \mathrm{n}_R( {\mathrm {mod}}(R))\leq 1$ . Combining this with Remark 3.15(2), we have $ {\mathrm {\mathfrak {ca}}}(R)= {\mathrm {ca}}_R( {\mathrm {mod}}(R))$ . Since $\mathsf {D}_{\mathsf {sg}}(R)$ has a strong generator, it follows from Proposition 4.8 that $ {\mathrm {Sing}}(R)=V( {\mathrm {\mathfrak {ca}}}(R))$ .

(2) By the proof of (1), $ \mathrm{n}_R( {\mathrm {mod}}(R))<\infty $ . Remark 4.17 now shows that $ {\mathrm {mod}}(R)$ has a strong generator.

Proof For each prime ideal $\mathfrak {p}$ of R, $R/\mathfrak {p}$ is a domain and $\dim R/\mathfrak {p}\leq 1$ . Thus, the equivalence $(1)\iff (2)$ follows from Corollary 4.20. The equivalence $(2)\iff (3)$ is a direct consequence of Corollary 4.3.

Proof (1) This follows from Lemma 3.5.

(2) By (1), it remains to prove $ {\mathrm {Sing}}(R) \subseteq V(\sqrt { {\mathrm {ca}}}_{R}( {\mathrm {mod}}(R)))$ . Let $\mathfrak {p}$ be a prime ideal of R such that $\mathfrak {p}\notin V(\sqrt { {\mathrm {ca}}}_{R}( {\mathrm {mod}}(R)))$ . Since the inclusion relation $V(\sqrt { {\mathrm {ca}}_{R}(R/\mathfrak {p})})\subseteq V(\sqrt { {\mathrm {ca}}}_{R}( {\mathrm {mod}}(R)))$ holds, the prime ideal $\mathfrak {p}$ does not belong to $V(\sqrt { {\mathrm {ca}}_{R}(R/\mathfrak {p})})=V( {\mathrm {ca}}_{R}(R/\mathfrak {p}))$ . This implies that $\mathfrak {p}\notin {\mathrm {IPD}}(R/\mathfrak {p})$ (see Proposition 3.10). Hence, the $R_{\mathfrak {p}}$ -module $R_{\mathfrak {p}}/\mathfrak {p} R_{\mathfrak {p}}$ has finite projective dimension. This implies that $\mathfrak {p}$ does not belong to $ {\mathrm {Sing}} R$ , and hence $ {\mathrm {Sing}}(R) \subseteq V(\sqrt { {\mathrm {ca}}}_{R}( {\mathrm {mod}}(R)))$ .

Proof For the forward direction, assume that $\mathsf {D}_{\mathsf {sg}}(R/\mathfrak {p})$ has a generator for each prime ideal $\mathfrak {p}$ of R. To obtain the conclusion, it suffices to show $\sqrt { {\mathrm {ca}}}_R( {\mathrm {mod}}(R)) \neq 0$ under the assumption that R is a domain. By assumption and Proposition 4.8(1), there exists $M\in {\mathrm {mod}}(R)$ such that M is a generator of $\mathsf {D}_{\mathsf {sg}}(R)$ and $ {\mathrm {Sing}}(R)=V( {\mathrm {ca}}_R(M))$ . With the same argument in the proof of Corollary 4.10, $ {\mathrm {ca}}_R(M)\neq 0$ , and hence there exists $f\neq 0$ in $ {\mathrm {ca}}^n_R(M)$ for some $n\geq 0$ . Combining this with Lemma 3.8(1), we get that the projective dimension of $M_f$ over $R_f$ is finite. Note that $M_f$ is a generator of $\mathsf {D}_{\mathsf {sg}}(R_f)$ as M is a generator of $\mathsf {D}_{\mathsf {sg}}(R)$ . Hence, $R_{f}$ is a regular ring. Let X be a finitely generated R-module. Since the $R_{f}$ -module $X_{f}$ has finite projective dimension, there exists $m\geq 0$ such that

$$ \begin{align*}{\mathrm{Ext}}_{R}^{m}(X, \Omega_{R}^{m} (X))_f \cong {\mathrm{Ext}}_{R_{f}}^{m}(X_{f}, (\Omega_{R}^{m} (X))_{f}) = 0. \end{align*} $$

It follows that $f^{r}\cdot {\mathrm {Ext}}_{R}^{m}(X, \Omega _{R}^{m} (X)) = 0$ for some $r \geq 0$ . This yields that $f\in \sqrt { {\mathrm {ca}}_{R}^{m}(X)} \subseteq \sqrt { {\mathrm {ca}}_{R}(X)}$ . Thus,

$$ \begin{align*}f \in \displaystyle\bigcap_{X \in {\mathrm{mod}}(R)} \sqrt{{\mathrm{ca}}_R(X)} = \sqrt{{\mathrm{ca}}}_{R}({\mathrm{mod}}(R)). \end{align*} $$

For the backward direction, assume that $\sqrt { {\mathrm {ca}}}_{R/\mathfrak {p}}( {\mathrm {mod}}(R/\mathfrak {p}))\neq 0$ for each prime ideal $\mathfrak {p}$ of R. Then $0 \in {\mathrm {Spec}}(R/\mathfrak {p})\setminus V(\sqrt { {\mathrm {ca}}}_{R/\mathfrak {p}}( {\mathrm {mod}}(R/\mathfrak {p}))) \subseteq {Reg}(R/\mathfrak {p})$ in $ {\mathrm {Spec}} (R/\mathfrak {p})$ , where $ {\mathrm {Reg}}(R/\mathfrak {p})= {\mathrm {Spec}}(R/\mathfrak {p})\setminus {\mathrm {Sing}}(R/\mathfrak {p})$ ; the inclusion is from Lemma 4.23. This implies that $ {\mathrm {Reg}}(R/\mathfrak {p})$ contains a nonempty open subset. By virtue of [Reference Iyengar and Takahashi33, Theorem 1.1], $\mathsf {D}_{\mathsf {sg}}(R/\mathfrak {p})$ has a generator for each prime ideal $\mathfrak {p}$ of R.

Proof Note that $ {\mathrm {id}}_{R_{\mathfrak {p}}}(M_{\mathfrak {p}})<n$ if and only if $ {\mathrm {Ext}}^n_{R_{\mathfrak {p}}}( {\mathrm {mod}}(R_{\mathfrak {p}}),M_{\mathfrak {p}})=0$ ; this can be proved by using Baer’s criterion for injectivity. This yields the first equality below:

$$ \begin{align*} \{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n\}& =\{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{Ext}}_{R_{\mathfrak{p}}}^n({\mathrm{mod}}(R_{\mathfrak{p}}), M_{\mathfrak{p}})\neq 0\}\\ &=\{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{Ext}}_{R}^n({\mathrm{mod}}(R), M)_{\mathfrak{p}}\neq 0\}\\ &={\mathrm{Supp}}_R{\mathrm{Ext}}^n_R({\mathrm{mod}}(R),M), \end{align*} $$

where the second one follows from 3.7 and the fact that $ {\mathrm {mod}}(R)\rightarrow {\mathrm {mod}}(R_{\mathfrak {p}})$ is dense. This finishes the proof.

Proof We only need to prove $ {\mathrm {Sing}}(R)\subseteq {\mathrm {IID}}(G)$ , that is, $ {\mathrm {Spec}}(R)\setminus {\mathrm {IID}}(G)\subseteq {\mathrm {Spec}}(R)\setminus {\mathrm {Sing}}(R)$ . By hypothesis, there exists $s\geq 0$ such that $ {\mathrm {\Omega }}^s_R( {\mathrm {mod}}(R))\subseteq \bigcup \limits _{t\ge 0}|G|_t$ . Pick $\mathfrak {p}\in {\mathrm {Spec}}(R)\setminus {\mathrm {IID}}(G)$ . That is, $ {\mathrm {id}}_{R_{\mathfrak {p}}} (G_{\mathfrak {p}})<\infty $ . Note that $( {\mathrm {\Omega }}^s_R (R/\mathfrak {p}))_{\mathfrak {p}}\in \bigcup \limits _{t\ge 0}|G_{\mathfrak {p}}|_t$ . Let $ {\mathrm {\Omega }}^s_{R_{\mathfrak {p}}}$ denote a minimal syzygy, then

$$ \begin{align*}R_{\mathfrak{p}}^\mathrm{n}\oplus{\mathrm{\Omega}}^s_{R_{\mathfrak{p}}}(k(\mathfrak{p}))\cong ({\mathrm{\Omega}}^s_R(R/\mathfrak{p}))_{\mathfrak{p}}\end{align*} $$

for some $n\geq 0$ , where $k(\mathfrak {p})$ is the residue field of $R_{\mathfrak {p}}$ . It follows that $ {\mathrm {\Omega }}^s_{R_{\mathfrak {p}}}(k(\mathfrak {p}))\in \bigcup \limits _{t\geq 0}|G_{\mathfrak {p}}|_t$ . Since $ {\mathrm {id}}_{R_{\mathfrak {p}}}(G_{\mathfrak {p}})<\infty $ , each module in $\bigcup \limits _{t\geq 0}|G_{\mathfrak {p}}|_t$ has finite injective dimension over $R_{\mathfrak {p}}$ . In particular, $ {\mathrm {id}}_{R_{\mathfrak {p}}}( {\mathrm {\Omega }}^s_{R_{\mathfrak {p}}} (k(\mathfrak {p})))<\infty $ . Combining this with [Reference Ghosh, Gupta and Puthenpurakal28, Theorem 3.7], we conclude that $R_{\mathfrak {p}}$ is regular, that is, $\mathfrak {p} \in {\mathrm {Spec}}(R)\setminus {\mathrm {Sing}}(R)$ .

Proof By Lemma 5.2, we have the first equality of the following:

$$ \begin{align*} \{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n\} & ={\mathrm{Supp}}_R{\mathrm{Ext}}^n_R({\mathrm{mod}}(R), M)\\ &=\bigcup_{N\in {\mathrm{mod}}(R)}{\mathrm{Supp}}_R{\mathrm{Ext}}^n_R(N,M)\\ & \subseteq\bigcup_{N\in {\mathrm{mod}}(R)}V({\mathrm{ann}}_R\ {\mathrm{Ext}}^n_R(N,M))\\ & \subseteq V(\bigcap_{N\in {\mathrm{mod}}(R)}{\mathrm{ann}}_R\ {\mathrm{Ext}}^n_R(N,M)). \end{align*} $$

Combining with $ {\mathrm {coca}}_R^n(M)=\displaystyle \bigcap \limits _{N\in {\mathrm {mod}}(R)} {\mathrm {ann}}_R\ {\mathrm {Ext}}^n_R(N,M)$ , the first statement of the lemma follows.

By the first statement, we get the inclusion below:

$$ \begin{align*} {\mathrm{IID}}(M)&=\bigcap_{n>0}\{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n\}\\ &\subseteq \bigcap_{n>0}V({\mathrm{coca}}_R^n(M)). \end{align*} $$

Combining with $\displaystyle \bigcap _{n>0}V( {\mathrm {coca}}_R^n(M)) =V(\bigcup \limits _{n>0} {\mathrm {coca}}_R^n(M)) =V( {\mathrm {coca}}_R(M))$ , we conclude that $ {\mathrm {IID}}(M)\subseteq V( {\mathrm {coca}}_R(M))$ , and hence $ {\mathrm {IID}}(M)\subseteq V(\sqrt { {\mathrm {coca}}_R(M)})$ . This yields the second inclusion below:

$$ \begin{align*} {\mathrm{Sing}}(R)& \subseteq \bigcup_{M\in {\mathrm{mod}}(R)}{\mathrm{IID}}(M) \\ &\subseteq \bigcup_{M\in {\mathrm{mod}}(R)} V(\sqrt{{\mathrm{coca}}_R(M)})\\ &\subseteq V(\bigcap_{M\in {\mathrm{mod}}(R)} \sqrt{{\mathrm{coca}}_R(M)})\\ &\subseteq V(\bigcap_{M\in {\mathrm{mod}}(R)} {\mathrm{coca}}_R(M)). \end{align*} $$

This implies the last statement.

Proof By 4.5, $\dim (R)$ is finite. The last equality holds because $ {\mathrm {\mathfrak {ca}}}^{n+d+1}(R)= \bigcap \limits _{M\in {\mathrm {mod}}(R)} {\mathrm {coca}}^{n+d+1}_R(M)$ . Next, we prove the other equalities. By hypothesis, there exists $G\in {\mathrm {mod}}(R)$ and $t\geq 0$ such that $\Omega _R^n ( {\mathrm {mod}}(R)) \subseteq [G]_t^R$ . It follows that $ {\mathrm {\Omega }}^{n+d}_R( {\mathrm {mod}} (R))\subseteq [H]_t^R$ , where $H:= {\mathrm {\Omega }}^d_R(G)$ . For each $M\in {\mathrm {mod}}(R)$ , we have

$$ \begin{align*} ({\mathrm{ann}}_R {\mathrm{Ext}}^1_R(H, \Omega_R^1(H)))^t& \subseteq ({\mathrm{ann}}_R {\mathrm{Ext}}^{\geq 1}_R(H,M))^t\\ & \subseteq {\mathrm{ann}}_R{\mathrm{Ext}}_R^{\geq 1}([H]_t^R,M)\\ & \subseteq {\mathrm{ann}}_R {\mathrm{Ext}}_R^{\ge n+d+1}({\mathrm{mod}}(R), M)\\ & \subseteq {\mathrm{coca}}(M), \end{align*} $$

where the first inclusion is by Lemma 3.8, the second one is due to [Reference Dao and Takahashi18, Lemma 5.3(1)], and the third one is by the inclusion $ {\mathrm {\Omega }}^{n+d}_R( {\mathrm {mod}} (R))\subseteq [H]_t^R$ . Thus,

$$ \begin{align*}({\mathrm{ann}}_R {\mathrm{Ext}}^1_R(H, \Omega_R^1(H)))^t\subseteq \bigcap_{M\in {\mathrm{mod}}(R)}{\mathrm{coca}}^{n+d+1}_R(M)\subseteq \bigcap_{M\in {\mathrm{mod}}(R)} {\mathrm{coca}}_R(M).\end{align*} $$

This yields the second and third inclusions below:

$$ \begin{align*} {\mathrm{Sing}}(R)& \subseteq V(\bigcap_{M\in {\mathrm{mod}}(R)} {\mathrm{coca}}_R(M))\\ & \subseteq V(\bigcap_{M\in {\mathrm{mod}}(R)}{\mathrm{coca}}^{n+d+1}_R(M))\\ & \subseteq V({\mathrm{ann}}_R {\mathrm{Ext}}^1_R(H, \Omega^1_R(H)))\\ &={\mathrm{IPD}}(H), \end{align*} $$

where the first inclusion is by Lemma 5.5, and the last equality can be deduced by combining the assumption $d\geq \dim (R)$ with Lemma 3.8 and Proposition 3.10. The desired equalities now follow by combining with $ {\mathrm {IPD}}(H)\subseteq {\mathrm {Sing}}(R)$ .

Proof (1) Assume $ {\mathrm {mod}}(R)$ has an extension generator G. That is, there exists $s\geq 0$ satisfying: for each $X\in {\mathrm {mod}}(R)$ , there exists $t\geq 0$ such that $\Omega ^s_R(X)\subseteq |G|_{t}$ . For each $n>0$ , Lemma 5.2 yields the first equality below:

$$ \begin{align*} \{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n+s\}& ={\mathrm{Supp}}_R{\mathrm{Ext}}^{n+s}_R({\mathrm{mod}}(R), M)\\ & ={\mathrm{Supp}}_R{\mathrm{Ext}}^\mathrm{n}_R(\Omega^s_R({\mathrm{mod}}(R)), M)\\ & =\bigcup_{N\in {\mathrm{mod}}(R)}{\mathrm{Supp}}_R{\mathrm{Ext}}^n_R(\Omega_R^s(N),M). \end{align*} $$

For each $N\in {\mathrm {mod}}(R)$ , it follows from the assumption that $\Omega ^s_R(N)\in |G|_{t}$ for some $t\geq 0$ . This yields that $ {\mathrm {Supp}}_R {\mathrm {Ext}}^n_R(\Omega ^s_R(N),M)\subseteq {\mathrm {Supp}}_R {\mathrm {Ext}}_R^n(G,M). $ Hence,

$$ \begin{align*}\{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n+s\}\subseteq {\mathrm{Supp}}_R {\mathrm{Ext}}_R^n(G,M). \end{align*} $$

Taking intersections throughout all $n>0$ , $ {\mathrm {IID}}(M)\subseteq \displaystyle \bigcap \limits _{n>0} {\mathrm {Supp}}_R {\mathrm {Ext}}^n_R(G,M). $ On the other hand, by Lemma 5.2,

$$ \begin{align*}\bigcap_{n>0}{\mathrm{Supp}}_R {\mathrm{Ext}}_R^n(G,M)\subseteq \bigcap_{n>0}\{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n\}={\mathrm{IID}}(M).\end{align*} $$

Thus,

$$ \begin{align*}{\mathrm{IID}}(M)=\bigcap\limits_{n>0}{\mathrm{Supp}}_R{\mathrm{Ext}}^n_R(G,M). \end{align*} $$

The desired result follows as $ {\mathrm {Supp}}_R {\mathrm {Ext}}^n_R(G,M)=V( {\mathrm {ann}}_R {\mathrm {Ext}}^n_R(G,M))$ is closed.

(2) By Lemma 5.5, it remains to prove $V( {\mathrm {coca}}_R(M))\subseteq {\mathrm {IID}}(M)$ . Let $n>0$ be an integer. Since $ {\mathrm {mod}}(R)$ has a strong generator, there exist $s,t\geq 0$ and $G\in {\mathrm {mod}}(R)$ such that $\Omega ^s_R( {\mathrm {mod}}(R))\subseteq |G|_t$ (see 5.7). Hence,

$$ \begin{align*}{\mathrm{coca}}_R^{n+s}(M)={\mathrm{ann}}_R{\mathrm{Ext}}^n_R(\Omega^s_R({\mathrm{mod}}(R)),M)\supseteq {\mathrm{ann}}_R{\mathrm{Ext}}^n_R(|G|_t, M). \end{align*} $$

It is routine to check that $ {\mathrm {ann}}_R {\mathrm {Ext}}^n_R(|G|_t,M)\supseteq ( {\mathrm {ann}}_R {\mathrm {Ext}}_R^n(G,M))^t $ . Thus,

$$ \begin{align*} V({\mathrm{coca}}_R^{n+s}(M))&\subseteq V({\mathrm{ann}}_R{\mathrm{Ext}}_R^n(G,M))\\ & \subseteq \{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n\}, \end{align*} $$

where the second inclusion is due to $V( {\mathrm {ann}}_R {\mathrm {Ext}}^n_R(G,M))= {\mathrm {Supp}}_R {\mathrm {Ext}}^n_R(G,M)$ and 3.7; the equality here holds as $ {\mathrm {Ext}}_R^n(G,M)$ is finitely generated. Combining with $ {\mathrm {coca}}_R^{n+s}(M)\subseteq {\mathrm {coca}}_R(M)$ , we get

$$ \begin{align*}V({\mathrm{coca}}_R(M))\subseteq \{\mathfrak{p}\in{\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n\}. \end{align*} $$

Since $n>0$ is arbitrary,

$$ \begin{align*}V({\mathrm{coca}}_R(M))\subseteq \bigcap_{n>0}\{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n\}={\mathrm{IID}}(M). \end{align*} $$

This completes the proof.

Proof By Example 4.12, $ {\mathrm {mod}}(R)$ has a strong generator. The desired result now follows from Theorem 5.9.

Proof (1) This follows from Proposition 4.6 and [Reference Bruns and Herzog12, Theorem 3.1.17].

(2) By Theorem 5.9, $ {\mathrm {IID}}(M)=V( {\mathrm {coca}}_R(M))$ . It is clear that

$$ \begin{align*}V({\mathrm{coca}}_R(M))\subseteq V({\mathrm{coca}}_R^{d+s+1}(M)).\end{align*} $$

Assume this is not equal. Combining with $ {\mathrm {IID}}(M)=V( {\mathrm {coca}}_R(M))$ , there exists a prime ideal $\mathfrak {p}$ of R containing $ {\mathrm {coca}}_R^{d+s+1}(M)$ and $\mathfrak {p}\notin {\mathrm {IID}}(M)$ . Thus, $\mathfrak {p}$ contains $ {\mathrm {ann}}_R {\mathrm {Ext}}^{d+s+1}_R( {\mathrm {mod}}(R),M)$ . By the argument in the proof of Theorem 5.9(2),

$$ \begin{align*}V({\mathrm{coca}}_R^{n+s}(M)) \subseteq \{\mathfrak{p}\in {\mathrm{Spec}}(R)\mid {\mathrm{id}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})\geq n\} \end{align*} $$

for each $n>0$ . Thus, $ {\mathrm {id}}_{R_{\mathfrak {p}}}(M_{\mathfrak {p}})\geq d+1$ . This contradicts with $\mathfrak {p}\notin {\mathrm {IID}}(M)$ ; by assumption, $ {\mathrm {id}}_{R_{\mathfrak {p}}}(M_{\mathfrak {p}})\leq d$ if $\mathfrak {p}\notin {\mathrm {IID}}(M)$ . This completes the proof.

Proof By 5.14, $ {\mathrm {Sing}}(R)$ is closed. Assume $ {\mathrm {Sing}}(R)=V(x_1,\ldots ,x_n)$ for some elements $x_1,\ldots ,x_n$ in R. Let M be a finitely generated R-module. For each $\mathfrak {p}\notin {\mathrm {Sing}}(R)$ , we have

$$ \begin{align*}{\mathrm{Ext}}^1_R(\Omega^d_R(M),\Omega^{d+1}_R(M))_{\mathfrak{p}}=0. \end{align*} $$

It follows that the non-free locus of $\Omega ^d_R(M)$ , given by $\{\mathfrak {p}\in {\mathrm {Spec}}(R)\mid \Omega ^d_R(M)_{\mathfrak {p}} \text { is not free}\}$ , is contained in $ {\mathrm {Sing}}(R)$ . By [Reference Bahlekeh, Hakimian, Salarian and Takahashi5, Theorem 3.1 (2)], $\Omega ^d_R(M)\in \displaystyle \bigcup _{t\geq 0}\Big |\bigoplus _{i=0}^n \bigoplus _{\mathfrak {p}\in {\mathrm {Sing}}(R)} \Omega ^i_R(R/\mathfrak {p})\Big |_t$ .

Proof We prove it by induction on $d = \dim R$ . Note that when $d = 0$ , $R/\mathfrak {p}$ is Artinian for every $\mathfrak {p}$ , and hence $ {\mathrm {mod}}(R/\mathfrak {p})$ has an extension generator. So we assume $d> 0$ .

First, we consider the case when R is a domain. By [Reference Iyengar and Takahashi33, Theorem 1.1], $ {\mathrm {Sing}}(R)=V(I)$ for some ideal I. As R is a domain, I is a nonzero ideal, so pick $0 \neq x \in I$ . Then $R_x$ is regular. As every prime ideal of $R/(x)$ is of the form $\mathfrak {p}/(x)$ for some $\mathfrak {p} \in {\mathrm {Spec}}(R)$ , and $(R/(x))/(\mathfrak {p}/(x)) \cong R/\mathfrak {p}$ , thus $\mathsf {D}_{\mathsf {sg}} ((R/(x))/{\mathfrak {q}})$ has a classical generator for each $\mathfrak {q} \in {\mathrm {Spec}}(R/(x))$ . Since R is a domain, $n := \dim (R/(x)) < \dim (R) = d$ . Thus, by induction hypothesis, $\Omega ^n_{R/(x)}( {\mathrm {mod}}(R/(x)) \subseteq \displaystyle \bigcup _{t\geq 0} |G|_t^{R/(x)}$ for some $G \in {\mathrm {mod}}(R/(x))$ . By [Reference Araya, Iima and Takahashi1, Lemma 4.5(2)], there exists $H \in {mod}(R)$ such that $\Omega ^d_R( {\mathrm {mod}}(R)) \subseteq \displaystyle \bigcup _{t\geq 0} |H|_t^R$ . This completes the case where R is a domain.

The proof of the general case now proceeds similarly to part (b) of the proof in [Reference Araya, Iima and Takahashi1, Proposition 4.6(3)]; note that each $R/\mathfrak {p}_i$ in loc. cit. is a domain and satisfies the same hypothesis on the singularity category.

Proof By 3.13, $ \mathrm{Rfd}(G)$ is finite. Set $ \mathrm{Rfd}(G)=n$ . By [Reference Christensen, Foxby and Frankild15, Definition 2.1 and Theorem 2.4], we conclude that the category $\mathcal {C} := \{M \in {\mathrm {mod}} (R) | \mathrm{Rfd}(M) \leq n\}$ is a resolving subcategory of $ {\mathrm {mod}}(R)$ containing G, hence it contains $ {\mathrm {res}}(G)$ . It follows from the hypothesis that $\Omega _R^s(M)\in \mathcal {C}$ for each $M \in {\mathrm {mod}}(R)$ . Thus, for each $M \in {\mathrm {mod}}(R)$ , $ \mathrm{Rfd}(\Omega _R^s (M)) \leq n$ . Again, by [Reference Christensen, Foxby and Frankild15, Definition 2.1 and Theorem 2.4], we conclude that $ \mathrm{Rfd}(M) \leq s+n$ for each $M \in {\mathrm {mod}}(R)$ . Since $ {\mathrm {depth}} (R_{\mathfrak {p}})\leq \mathrm{Rfd}(R/\mathfrak {p})$ , we get $\sup \{ {\mathrm {depth}}(R_{\mathfrak {p}}) \mid \mathfrak {p} \in {\mathrm {Spec}}(R)\} \leq s+n$ . Since prime ideals of localization are localization of prime ideals, [Reference Christensen, Foxby and Frankild15, Lemma 1.4] implies that $\dim (R) -1 \leq s+n$ .

Proof The second statement and the implication $(1)\Rightarrow (2)$ follows from Proposition 5.19. The implication $(2)\Rightarrow (3)$ follows from the definition.

$(3)\Rightarrow (1)$ : Assume $ {\mathrm {mod}}(R/\mathfrak {p})$ has a resolving generator for each $\mathfrak {p}\in {\mathrm {Spec}}(R)$ . By Proposition 5.20, $R/\mathfrak {p}$ has finite Krull dimension for each prime ideal $\mathfrak {p}$ of R. Since the set of minimal prime ideals of R is finite, we conclude that R has finite Krull dimension.

Fix a prime ideal $\mathfrak {p}$ of R, assume G is a resolving generator of $ {\mathrm {mod}}(R/\mathfrak {p})$ . Note that the modules in $ {\mathrm {res}}(G)$ are contained in $ {\mathrm {\mathsf {thick}}}_{\mathsf {D}_{\mathsf {sg}}(R/\mathfrak {p})}(G)$ . Since each complex in $\mathsf {D}_{\mathsf {sg}}(R/\mathfrak {p})$ is isomorphic to some shift of a module M in $ {\mathrm {mod}}(R/\mathfrak {p})$ , we conclude that G is a generator of $\mathsf {D}_{\mathsf {sg}}(R/\mathfrak {p})$ . This finishes the proof.

Proof (1) For each $X\in {\mathrm {mod}}(R)$ , since $\Omega ^d_R(X)\in {\mathrm {CM}}(R)$ , we get

$$ \begin{align*}{\mathrm{ca}}^1_R(M^{\dagger})\subseteq {\mathrm{ann}}_R {\mathrm{Ext}}^1_R(M^{\dagger}, (\Omega_R^d (X))^{\dagger})={\mathrm{ann}}_R {\mathrm{Ext}}^1_R(\Omega_R^d(X), M)={\mathrm{ann}}_R {\mathrm{Ext}}^{d+1}_R(X,M).\end{align*} $$

Since this is true for any $X\in {\mathrm {mod}}(R)$ , we get $ {\mathrm {ca}}_R^1(M)\subseteq {\mathrm {ann}}_R {\mathrm {Ext}}^{d+1}_R( {\mathrm {mod}}(R), M)= {\mathrm {coca}}_R^{d+1}(M)$ .

Assume R is Gorenstein in codimension $d-1$ . Note that $(\Omega _R^1(M^{\dagger }))^{\dagger }\in {\mathrm {CM}}(R)$ . By [Reference Evans and Griffith27, Theorem 3.8], $(\Omega _R^1(M^{\dagger }))^{\dagger }\cong \Omega ^d_R(Y)$ for some $Y\in {\mathrm {mod}}(R)$ . This yields the first equality below:

$$ \begin{align*}{\mathrm{coca}}_R^{d+1}(M)\subseteq {\mathrm{ann}}_R {\mathrm{Ext}}^{d+1}_R(Y, M)&={\mathrm{ann}}_R {\mathrm{Ext}}^1_R((\Omega_R^1(M^{\dagger}))^{\dagger},M)={\mathrm{ann}}_R {\mathrm{Ext}}^1_R(M^{\dagger}, \Omega_R^1(M^{\dagger}))\\ &={\mathrm{ca}}_R^1(M^\dagger),\end{align*} $$

where the last equality is by Lemma 3.8. This proves the converse direction.

(2) For each $n>0$ , assume $r\in {\mathrm {coca}}_R^n(M)$ . This yields

$$ \begin{align*}0=r\cdot {\mathrm{Ext}}^n_R({\mathrm{mod}}(R), M)=r\cdot {\mathrm{Ext}}^1_R(\Omega_R^{n-1}({\mathrm{mod}}(R)),M). \end{align*} $$

Since $M\in {\mathrm {CM}}(R)$ and R is strongly Gorenstein (see 2.1), M is an infinite syzygy. In particular, there is a short exact sequence $ \xi \colon 0\rightarrow M\rightarrow P\rightarrow C\rightarrow 0, $ where P is a finitely generated projective and $C\in \Omega _R^{n-1}( {\mathrm {mod}}(R))$ . By above, $r\cdot \xi =0$ . In particular, $r\colon M\rightarrow M$ factors through P. This implies that $r\in {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)$ , and therefore $ {\mathrm {coca}}_R^n(M)\subseteq {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)$ for each $n>0$ . It follows from this that $ {\mathrm {coca}}_R(M)\subseteq {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)$ .

By above, $ {\mathrm {coca}}_R^{d+1}(M)\subseteq {\mathrm {coca}}_R(M)\subseteq {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)= {\mathrm {ca}}_R(M)$ , where the equality here follows from Lemma 3.8 and Proposition 3.14. Next, we prove the inclusion $ {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)\subseteq {\mathrm {coca}}_R^{d+1}(M)$ . Assume $a\in {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)$ . Namely, there is a factorization

where Q is finitely generated projective. Note that $ {\mathrm {id}}_R(Q)\leq d$ (see 2.1). Hence, $ {\mathrm {Ext}}^{d+1}_R(X,Q)=0$ for each $X\in {\mathrm {mod}}(R)$ . Applying $ {\mathrm {Ext}}^{d+1}_R(X,-)$ to the above diagram, we conclude that $a\colon {\mathrm {Ext}}^{d+1}_R(X,M)\rightarrow {\mathrm {Ext}}^{d+1}_R(X,M)$ is zero for each $X\in {\mathrm {mod}}(R)$ . That is, $a\in {\mathrm {coca}}_R^{d+1}(M)$ , and hence $ {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)\subseteq {\mathrm {coca}}_R^{d+1}(M)$ . Finally, $ {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)= {\mathrm {ca}}_R(M)$ follows from Proposition 3.14(1).

(3) For the forward direction, assume that R is Gorenstein with finite Krull dimension. For each $M\in {\mathrm {mod}}(R)$ , if $ \mathrm{Rfd}(M)=0$ , then M is maximal Cohen–Macaulay. By (2), this yields that $ {\mathrm {ca}}_R(M)= {\mathrm {coca}}_R(M)$ .

For the backward direction, assume that $ {\mathrm {ca}}_R(M)= {\mathrm {coca}}_R(M)$ for each $M\in {\mathrm {mod}}(R)$ with $ \mathrm{Rfd}_R(M)=0$ . In particular, since $ \mathrm{Rfd}_R(R)=0$ , we have $ {\mathrm {ca}}_R(R)= {\mathrm {coca}}_R(R)$ . It follows that $ {\mathrm {coca}}_R(R)=R$ as $ {\mathrm {ca}}_R(R)=R$ . Hence, there exists $n\geq 0$ such that $ {\mathrm {coca}}^n_R(R)=R$ . In particular, $ {\mathrm {Ext}}^j_R(R/\mathfrak {p}, R)=0$ for all $j \geq n$ and $\mathfrak {p}\in {\mathrm {Spec}}(R)$ . Localizing at $\mathfrak {p}$ , we conclude that $ {\mathrm {id}}_{R_{\mathfrak {p}}}(R_{\mathfrak {p}})\leq n$ for each $\mathfrak {p}\in {\mathrm {Spec}}(R)$ , and hence $ {\mathrm {id}}_R(R)\leq n$ . This shows that R is Gorenstein with finite Krull dimension (see 2.1).

Proof (1) By Lemma 3.8, it remains to prove $ {\mathrm {coca}}_R(M)\subseteq {\mathrm {ann}}_R\ \underline { {\mathrm {End}}}_R(M)$ . This follows from the same argument as in the proof of Lemma 5.22(2). If, in addition, R is strongly Gorenstein, then M is in $ {\mathrm {CM}}(R)$ . Hence, by Lemma 5.22(2), we have $ {\mathrm {coca}}_R(M)= {\mathrm {ca}}_R(M)$ .

(2) The statement follows from Proposition 3.10, Theorem 5.9, and (1).