2. Preliminaries, examples, and remarks

In this section, we give new examples of modules that have finite reducing but infinite, projective dimension, see Examples 2.20, 2.13, and 2.15. We also record several important facts that are needed for our argument, see 2.8. We start by recalling the definition of reducing invariant of modules.

In the following, ${{\mathbb{I}}}$ denotes an invariant of modules over local rings, that is, a family of functions ${{\mathbb{I}}}_R$ from isomorphism classes of $R$ -modules to $\mathbb{Z} \cup \{ \pm \infty \}$ as $R$ varies over all local rings. Classical examples of such an invariant we use in this paper are the projective dimension ${{\mathbb{I}}}=\mathsf{pd}$ and the Gorenstein dimension ${{\mathbb{I}}}=\mathsf{G-dim}$ .

We will give various examples of modules that have finite reducing homological dimension, but first we record several remarks. The first one, which follows directly from Definition 2.1, is used throughout the paper, for example, in the proofs of Proposition 3.1, Proposition 3.9, and Theorem 3.11.

One of the motivations of our work in this paper comes from the fact that the reducing projective dimension of the residue field is finite over local rings with radical square zero:

The rings in Example 2.5 are zero-dimensional. Next is an example of a module—over a local ring of dimension two—that has finite reducing projective, but infinite projective, dimension.

In general, unlike the classical homological dimensions, the reducing projective dimension of a nonzero module over a local ring, if finite, may not be bounded by the depth of the ring in question, see, for example, Example 2.5. In fact, we do not know whether or not there is a uniform upper bound for reducing projective dimension of modules, even over Gorenstein rings. On the other hand, a consequence of what we know from [Reference Celikbas, Dey, Kobayashi and Matsui12] yields a satisfactory answer over complete intersection rings and also allows us to determine modules that have sufficiently large finite reducing projective dimension:

We proceed to give an example of a module that has reducing projective dimension equal two and subsequently determine a reducing sequence of it explicitly, see Example 2.10. First, we record some auxiliary results that are used for the argument of the example, as well as in Section 3.

The next observation is used for Example 2.10 and also in the proof of Proposition 3.9.

2.9. Let $R$ be a local ring and let $M$ be a nonzero $R$ -module. Assume there is a non zero-divisor $x\in \mathfrak{m}$ on $R$ such that $xM=0$ . Then there is a short exact sequence of $R/xR$ -modules:

(2.9.1) \begin{equation} 0 \to M \to \Omega _R M \otimes _R (R/xR) \to \Omega _{R/xR}(M) \to 0. \end{equation}

To establish the sequence in ( 2.9.1 ), we consider a syzygy sequence of $M$ , namely an exact sequence $0 \to \Omega _R M \to F \to M \to 0$ , where $F$ is a free $R$ -module. Then we obtain, by tensoring the syzygy sequence with $R/xR$ , an exact sequence of $R$ -modules of the form

(2.9.2) \begin{equation} 0 \to \textrm{Tor}_1^R(M,R/xR) \to \Omega _R M \otimes _R (R/xR) \to F\otimes _R(R/xR) \to M/xM \to 0. \end{equation}

Note that $\textrm{Tor}_1^R(M,R/xR) \cong M$ since $x$ is a non zero-divisor on $R$ and $xM=0$ . Hence, ( 2.9.2 ) yields the exact sequence in ( 2.9.1 ).

Example 2.10. Let $R=S/aS$ , where $(S, \mathfrak{n})$ is a one-dimensional local hypersurface ring and $a\in \mathfrak{n}^2$ is a nonzero divisor on $S$ . For example, we can set $R=\mathbb{C}[\![x,y]\!]/(x^2, y^2)$ , $S=\mathbb{C}[\![x,y]\!]/(x^2)$ , and $a=y^2$ . Then $R$ is an Artinian complete intersection ring of codimension two so that ${\mathsf{red-pd}}_R(k)=2$ , see 2.7. Now we proceed and find a reducing $\mathsf{pd}$ -sequence of $k$ over $R$ .

Let us first note, by making use of ( 2.9.1 ) with the $S$ -module $k$ over $S$ , we have a short exact sequence of $R$ -modules of the form:

(2.10.1) \begin{equation} 0 \to k \to R \otimes _{S} \mathfrak{n} \to \Omega _R k \to 0. \end{equation}

As we know ${\mathsf{red-pd}}_R(k)=2$ , it follows that ${\mathsf{red-pd}}_R(R \otimes _{S} \mathfrak{n})=1$ , see Remark 2.2 (ii). So it is enough to find a reducing $\mathsf{pd}$ -sequence of $R \otimes _{S} \mathfrak{n}$ over $R$ .

In general, if $T$ is a local ring and $X$ and $Y$ are (finitely generated) $T$ -modules such that $\mathsf{pd}_T(X)=1$ and $Y$ is torsion-free, then $\textrm{Tor}_i^T(X,Y)=0$ for al $i\geq 1$ , see, for example, [Reference Celikbas and Takahashi13, 2.7]. So, as $\mathsf{pd}_S(R)=1$ , and $\mathfrak{n}$ and $\Omega _S \mathfrak{n}$ are torsion-free $S$ -modules, we have

(2.10.2) \begin{equation} \textrm{Tor}_i^S(\Omega _S \mathfrak{n},R)=0=\textrm{Tor}_i^S(\mathfrak{n},R) \text{ for all } i\geq 1. \end{equation}

Note that, as $S$ is a one-dimensional hypersurface ring and $\mathfrak{n}$ is a maximal Cohen–Macaulay $S$ -module with no free summand, we have $\mathfrak{n} \cong \Omega ^2_S \mathfrak{n}$ . Moreover, since $\mathfrak{n}$ is generated by two elements, there is a short exact sequence of $S$ -modules of the form:

(2.10.3) \begin{equation} 0 \to \mathfrak{n} \to S^{\oplus 2} \to \Omega _S \mathfrak{n}\to 0. \end{equation}

We obtain, by using ( 2.10.2 ) and tensoring ( 2.10.3 ) with $R$ over $S$ , the exact sequence of $R$ -modules:

(2.10.4) \begin{equation} 0 \to R \otimes _S \mathfrak{n} \to R^{\oplus 2} \to R \otimes _S \Omega _S \mathfrak{n} \to 0. \end{equation}

Next note that $R\otimes _S\Omega _S \mathfrak{n} \cong \Omega _R (\mathfrak{n} \otimes _S R)$ , see 2.8(i) and (2.10.2). Therefore, (2.10.1) and (2.10.4) show that a reducing $\mathsf{pd}$ -sequence of $k$ over $R$ is $\{k, R \otimes _S \mathfrak{n}, R^{\oplus 2} \}$ .

Recall that an $\mathfrak{m}$ -primary ideal $I$ of a local ring $R$ is called Ulrich if $I/I^2$ is a free $R/I$ -module and $I^2 = \mathfrak{q}I$ for some parameter ideal $\mathfrak{q}$ of $R$ , where $\mathfrak{q}$ is a reduction of $I$ , see [Reference Goto, Ozeki, Takahashi, Watanabe and Yoshida17, 1.1] for details. Next we point out that Ulrich ideals, when exist, provide nontrivial examples of modules of finite reducing projective dimension.

Recall that a Cohen–Macaulay local ring $R$ is said to have minimal multiplicity if the codimension of $R$ is one less than the multiplicity of $R$ , see [Reference Bruns and Herzog11, 4.5.14]. In the following, we obtain an extension of the fact recorded in Example 2.5:

We prove Theorem 1.2 in the next section; we should note that Theorem 1.2, besides giving an affirmative answer for Question 1.1 for a large class of local rings, also yields an extension of the observation stated in Remark 2.12.

3. Proof of the main theorem

In this section, we give a proof of Theorem 1.2, see the paragraph following Proposition 3.9. Along the way, we obtain several new results and study further properties of reducing dimensions, especially properties under local homomorphisms of local rings.

Throughout, ${{\mathbb{I}}}$ denotes an invariant of modules over local rings, see the paragraph preceding Definition 2.1. Our first result yields a generalization of [Reference Araya, Celikbas, Cook and Kobayashi2, 2.9].

Proof. Let $N$ be an $R$ -module such that $\mathsf{red-}\mathbb{I}_R(N)=r\lt \infty$ . We proceed by induction on $r$ to prove $\mathsf{red-}\mathbb{I}_S(N\otimes _R S)\leq r$ . We may assume $N\otimes _R S\neq 0$ .

Assume $r=0$ . Then, by Definition 2.1, we have that ${{\mathbb{I}}}_R(N)\lt \infty$ . This implies, by the hypothesis, that ${{\mathbb{I}}}_S(N\otimes _R S)\leq {{\mathbb{I}}}_R(N)\lt \infty$ , and hence $\mathsf{red-}\mathbb{I}_S(N\otimes _R S)=0=r$ .

Next we assume $r\geq 1$ . Then, by Remark 2.2(i), there exists a short exact sequence of $R$ -modules $0\to N^{\oplus a} \to K \to \Omega ^c_R N^{\oplus b}\to 0$ , where $\mathsf{red-}\mathbb{I}_R(K)=r-1$ , and $a\geq 1$ , $b\geq 1$ , and $c\geq 0$ are integers. Note, since $\textrm{Tor}^R_{i}(N,S)=0$ for all $i\geq 1$ , we have the vanishing of $\textrm{Tor}^R_{i}(K,S)$ for all $i\geq 1$ . Hence, the induction hypothesis on $r$ yields

(3.1.1) \begin{equation} \mathsf{red-}\mathbb{I}_S(K\otimes _R S)\leq\mathsf{red-}\mathbb{I}_R(K)= r-1. \end{equation}

Once again we make use of the hypothesis that $\textrm{Tor}^R_{i}(N,S)$ vanishes for all $i\geq 1$ , tensor the short exact sequence $0\to N^{\oplus a} \to K \to \Omega ^c_R N^{\oplus b}\to 0$ with $S$ over $R$ , and obtain the short exact sequence of $S$ -modules $0\to (N\otimes _R S)^{\oplus a} \to K\otimes _R S \to \Omega ^c_S(N\otimes _R S)^{\oplus b}\to 0$ , see also 2.8(i). So, by Remark 2.2(ii), we deduce

(3.1.2) \begin{equation} \mathsf{red-}\mathbb{I}_S(N\otimes _R S)\le \mathsf{red-}\mathbb{I}_S(K\otimes _R S)+1. \end{equation}

Consequently, (3.1.1) and (3.1.2) show that $\mathsf{red-}\mathbb{I}_S(N\otimes _R S)\leq r=\mathsf{red-}\mathbb{I}_R(N)$ .

Proof. The first and the second conclusion, in view of Proposition 3.1, follow from 2.8(i) and 2.8(ii), respectively.

Proof. The result follows from Proposition 3.1 by using the ring homomorphism $R \to S=R/xR$ .

The following special case of Corollary 3.3 is worth recording separately:

Next we point out that the inequality in Corollary 3.2(ii) can be strict; we give an example of a local homomorphism of local rings $R \to S$ and an $R$ -module $N$ such that $S$ has finite flat dimension over $R$ , $\textrm{Tor}^R_{i}(N,S)=0$ for all $i\geq 1$ , and ${\mathsf{red-G-dim}}_S(N\otimes _R S)\lt \infty ={\mathsf{red-G-dim}}_R(N)$ .

We need the following result for the proof of Theorem 3.11.

The conclusion of Proposition 3.8 seems natural, but its proof is involved; therefore, we defer the proof of the lemma until the next section so it does not interfere with the flow of the paper.

Proof. We proceed by induction on $n$ . Assume first $n=1$ and set $x_1=x$ . Then, since $x$ is a non zero-divisor on $\Omega _R M$ , it follows from Corollary 3.3 that $\mathsf{red-}\mathbb{I}_{S} \left (\dfrac{\Omega _R M}{x\cdot \Omega _RM} \right ) \le \mathsf{red-}\mathbb{I}_R \left (\Omega _R M\right )$ .

Note, by (2.9.1), we have the following short exact sequence of $S$ -modules:

(3.9.1) \begin{equation} 0 \to M \to \dfrac{\Omega _R M}{x\cdot \Omega _RM} \to \Omega _{S}(M) \to 0. \end{equation}

Therefore, Remark 2.2(ii) applied to (3.9.1) gives $\mathsf{red-}\mathbb{I}_{S}(M)\leq \mathsf{red-}\mathbb{I}_{S} \left (\dfrac{\Omega _R M}{x\cdot \Omega _RM} \right ) +1$ . This establishes the base case $n=1$ of the induction.

Next we assume $n\geq 2$ and set $T=R/\underline{x'}R$ , where $\underline{x'}=x_1,\ldots,x_{n-1}$ . Then $x_n$ is a non zero-divisor on $T$ and $S=T/x_nT$ . So, by the case where $n=1$ , we have

(3.9.2) \begin{equation} \mathsf{red-}\mathbb{I}_{S}(M)\le \mathsf{red-}\mathbb{I}_T(\Omega _T M)+1. \end{equation}

We make use of the induction hypothesis with the module $\Omega _T M$ and conclude that:

(3.9.3) \begin{equation} \mathsf{red-}\mathbb{I}_T(\Omega _T M)\le \mathsf{red-}\mathbb{I}_R(\Omega ^{n-1}_{R}\Omega _{T}M)+n-1. \end{equation}

Hence, (3.9.2) and (3.9.3) yield:

(3.9.4) \begin{equation} \mathsf{red-}\mathbb{I}_{S}(M)\le \mathsf{red-}\mathbb{I}_R(\Omega ^{n-1}_{R}\Omega _{T}M)+n. \end{equation}

Next, note that $\Omega ^{n-1}_{R}\Omega _{T}M \cong \Omega ^{n}_RM\oplus F$ for some free $R$ -module $F$ , see 2.8(iii). Thus, the induction argument is now complete from (3.9.4) and Proposition 3.8.

We need the following observation for the proof of Theorem 3.11.

3.10. Let $R$ be a ring (not necessarily local and Noetherian) and let $0 \to L \to M \xrightarrow []{p} N\oplus F \to 0$ be a short exact sequence of $R$ -modules, where $F$ is free. Then, we have the following pullback diagram, that is, a commutative diagram with exact rows and columns:

This diagram yields a short exact sequence of $R$ -modules $0 \to L \to M' \to N \to 0$ , where $M\cong M'\oplus F$ .

The next theorem plays a key role for the proof of Theorem 1.2.

Proof. Let $N$ be a nonzero finitely generated $S$ -module such that $\mathsf{red-}\mathbb{I}_S (N)=r\lt \infty$ . We proceed by induction on $r$ to show that $\mathsf{red-}\mathbb{I}_R(\Omega ^n_R N)\le r$ .

Assume first $r=0$ . Then $\mathsf{red-}\mathbb{I}_S(N)=0$ , that is, ${{\mathbb{I}}}_S(N)\lt \infty$ . Hence, by the hypotheses, we conclude that ${{\mathbb{I}}}_{R}(\Omega _{R}{N})\le {{\mathbb{I}}}_{R}(N)\lt \infty$ . This shows that $\mathsf{red-}\mathbb{I}_R(\Omega ^n_R M)=0=r$ .

Next assume $r\geq 1$ . Then it follows from Definition 2.1 that there exists a short exact sequence of (finitely generated) $S$ -modules

(3.11.1) \begin{equation} 0\to N^{\oplus a} \to K \to \Omega ^c_S N^{\oplus b}\to 0, \end{equation}

where $a\geq 1$ , $b\geq 1$ , and $c\geq 0$ are integers, and $\mathsf{red-}\mathbb{I}_S(K)=\mathsf{red-}\mathbb{I}_S(N)-1=r-1$ . So, by the induction hypothesis, we have

(3.11.2) \begin{equation} \mathsf{red-}\mathbb{I}_R(\Omega ^n_R K)\le \mathsf{red-}\mathbb{I}_S (K)=r-1. \end{equation}

We obtain, by applying $\Omega ^n_R(\!-\!)$ to (3.11.1), the following exact sequence of $R$ -modules:

(3.11.3) \begin{equation} 0\to \Omega ^n_RN^{\oplus a} \to \Omega ^n_RK\oplus G \to \Omega ^{n}_R \Omega ^c_S N^{\oplus b} \to 0, \end{equation}

where $G$ is a free $R$ -module, see [Reference Dao and Takahashi16, 2.2(1)].

Note that $\Omega ^{n}_R \Omega ^c_S N^{\oplus b} \cong \Omega ^{n+c}_R N^{\oplus b}\oplus F$ for some free $R$ -module $F$ , see 2.8(iii). Then, by using 3.10, we obtain the following exact sequence of $R$ -modules:

(3.11.4) \begin{equation} 0\to \Omega ^n_RN^{\oplus a} \to K' \to \Omega ^{n+c}_R N^{\oplus b} \to 0, \end{equation}

where $\Omega ^n_RK\oplus G\cong K'\oplus F$ . Now we observe the following (in) equalities, which complete the proof:

(3.11.5) \begin{align} \mathsf{red-}\mathbb{I}_R(\Omega ^n_R N) \leq & \mathsf{red-}\mathbb{I}_R(K')+1 \nonumber \\ = & \mathsf{red-}\mathbb{I}_R(\Omega ^n_RK)+1 \nonumber \\ \leq &(r-1)+1=r. \end{align}

Here, in (3.11.5), the first inequality follows from (3.11.4) and Remark 2.2(ii), and the second one follows from (3.11.3); moreover, the first equality is due to Proposition 3.8.

Next is a consequence of Theorem 3.11; this result, for the case where ${{\mathbb{I}}}$ is the projective dimension, or the Gorenstein dimension of modules, can also be deduced from [Reference Araya and Celikbas1]. However, as mentioned in Remark 2.3, the definition given in [Reference Araya and Celikbas1] is slightly more restrictive than the definition of reducing dimensions we adopt in this paper, see Definition 2.1.

Proof. The corollary follows from Theorem 3.11 by setting $R=S$ .

We need a few more results to prove Theorem 1.2; in the following $\widehat{(\!-\!)}$ denotes the completion functor.

We can show, under the hypothesis of 3.14, that $\mathsf{red-}\mathbb{I}_{\widehat R}(\widehat N)= \mathsf{red-}\mathbb{I}_R(N)$ if $N$ is an $R$ -module, which is locally free on the punctured spectrum of $R$ , that is, if $N_{\mathfrak{p}}$ is free over $R_{\mathfrak{p}}$ for all non-maximal prime ideals $\mathfrak{p}$ of $R$ .

Proof. Let $N$ be an $R$ -module, which is locally free on punctured spectrum of $R$ . Then, in view of 3.14, it is enough to prove $\mathsf{red-}\mathbb{I}_R(N)\le \mathsf{red-}\mathbb{I}_{\widehat R}(\widehat N)$ . Notice we may assume $\mathsf{red-}\mathbb{I}_{\widehat R}(\widehat N)\lt \infty$ as otherwise there is nothing to prove. We set $r=\mathsf{red-}\mathbb{I}_{\widehat R}(\widehat N)$ and proceed by induction on $r$ .

If $r=0$ , then we have ${{\mathbb{I}}}_{\widehat R}(\widehat N)\lt \infty$ , see Definition 2.1. Hence, our assumption yields ${{\mathbb{I}}}_R (N)={{\mathbb{I}}}_{\widehat R}(\widehat N)$ , and so shows that $\mathsf{red-}\mathbb{I}_R (N)=0=\mathsf{red-}\mathbb{I}_{\widehat R}(\widehat N)$ .

Now let $r\ge 1$ . Then there exists a short exact sequence of $\widehat R$ -modules: $0\to \widehat{N^{\oplus a}} \to K \to \widehat{\Omega ^c_{ R} N^{\oplus b}}\to 0$ , where $\mathsf{red-}\mathbb{I}_{\widehat R}( K)=r-1$ , and $a\geq 1$ , $b\geq 1$ , and $c\geq 0$ are integers, see Remark 2.2(i). Note that, since $\textrm{Ext}^1_R(\Omega ^c_{ R} N^{\oplus b}, N^{\oplus a})$ is a finite length $R$ -module, in the following the natural maps are isomorphisms:

\begin{equation*}\textrm{Ext}^1_R(\Omega ^c_{ R} N^{\oplus b}, N^{\oplus a}) \cong \textrm{Ext}^1_R(\Omega ^c_{ R} N^{\oplus b}, N^{\oplus a})\otimes _R\widehat {R} \cong \textrm{Ext}^1_{\widehat {R}}(\widehat {\Omega ^c_{ R} N^{\oplus b}}, \widehat {N^{\oplus a}})\end{equation*}

As the composition of these two natural maps is given by the completion of short exact sequences, we deduce that there is a short exact sequence of $R$ -modules $0\to N^{\oplus a} \to X \to \Omega ^c_{ R} N^{\oplus b}\to 0$ , where $K\cong \widehat X$ , see also [Reference Leuschke and Wiegand22, 2.7(i)].

Note that an $R$ -module $T$ is locally free on the punctured spectrum of $R$ if and only if $\widehat{T}$ is locally free on the punctured spectrum of $\widehat{R}$ . Therefore, since $K\cong \widehat X$ and $K$ is locally free on the punctured spectrum of $\widehat{R}$ , we conclude that $X$ is locally free on punctured spectrum of $R$ . As $\mathsf{red-}\mathbb{I}_{\widehat R}(\widehat{X})=\mathsf{red-}\mathbb{I}_{\widehat R}(K)=r-1$ , we see by the induction hypothesis that $\mathsf{red-}\mathbb{I}_{R}(X)\le \mathsf{red-}\mathbb{I}_{\widehat R}(\widehat X)$ . Hence, in view of Remark 2.2(ii), the exact sequence $0\to N^{\oplus a} \to X \to \Omega ^c_R N^{\oplus b}\to 0$ implies that $\mathsf{red-}\mathbb{I}_R(N)\le \mathsf{red-}\mathbb{I}_R(X)+1\leq (r-1)+1=r$ . This completes the induction and the proof.

Proof. Note that $\mathsf{pd}_{\widehat R}(\widehat M)=\mathsf{pd}_R(M)$ and $\mathsf{G-dim}_{\widehat R}(\widehat M)=\mathsf{G-dim}_R(M)$ for each $R$ -module $M$ , see, for example, 2.8(i) and [Reference Christensen15, 5.10(c)]. Therefore, the result follows from Proposition 3.15.

We are now ready to prove Theorem 1.2 as advertised in the introduction.

Proof of Theorem 1.2. Note that we have

\begin{equation*} {\mathsf {red-pd}}_{\widehat {S}}(k)\leq {\mathsf {red-pd}}_R( \Omega ^n_{R} k)+n \leq {\mathsf{red-pd}}_R( k)+n \lt \infty. \end{equation*}

Here, in view of 3.7, the first inequality is due to Proposition 3.9 and the second one is due to Corollary 3.13; the third inequality is clear if $R$ is regular, and it holds by 2.14 in case $R$ is singular. Now, since $k$ is a complete $S$ -module that is locally free on the punctured spectrum of $S$ , it follows that ${\mathsf{red-pd}}_{S}(k)={\mathsf{red-pd}}_{\widehat S}(\widehat k)={\mathsf{red-pd}}_{\widehat S}(k)$ , see from Corollary 3.16. This completes the proof.

We finish this section with a further application of Proposition 3.9 and Corollary 3.13. Recall that a local ring $R$ is called G-regular [Reference Takahashi25] provided that $\mathsf{G-dim}_R(M)\lt \infty$ if and only if $\mathsf{pd}_R(M)\lt \infty$ for each $R$ -module $M$ . As, in general, over a local ring $R$ , we have that $\mathsf{G-dim}_R(M)=\mathsf{pd}_R(M)$ for each $R$ -module $M$ with $\mathsf{pd}_R(M)\lt \infty$ , it is clear that each regular local ring is G-regular. A nontrivial fact is that a non-Gorenstein Cohen–Macaulay local ring of minimal multiplicity is $G$ -regular, see [Reference Takahashi25, 5.1] for details. On the other hand, deformations of G-regular rings, that is, G-regular rings modulo ideals generated by regular sequences, need not be G-regular as we see next:

Although deformations of G-regular rings are not G-regular in general, we finish this section by making a related observation: a deformation of a G-regular ring is reducing G-regular, that is, a ring over which each module that has finite reducing Gorenstein dimension also has finite reducing projective dimension. First, we note that each G-regular ring is reducing G-regular.

Proposition 3.19. Let $R$ be a G-regular local ring and let $S=R/\underline{x}R$ , where $\underline{x}=x_1,\ldots, x_n \subseteq \mathfrak{m}$ is an $R$ -regular sequence. If $M$ is a (finitely generated) $S$ -module, then

\begin{equation*}{\mathsf {red-G-dim}}_S(M)\leq {\mathsf {red-pd}}_S(M) \leq {\mathsf {red-G-dim}}_S(M)+n.\end{equation*}

So, for each finitely generated $S$ -module $N$ , we have ${\mathsf{red-pd}}_S(N)\lt \infty$ if and only if ${\mathsf{red-G-dim}}_S(N)\lt \infty$ .

Proof. Let $M$ be a finitely generated $S$ -module. Then the first inequality follows by Definition 2.1. For the second inequality, note that we have

\begin{equation*} {\mathsf{red-pd}}_{S}(M)\le {\mathsf{red-pd}}_R(\Omega ^n_RM)+n={\mathsf{red-G-dim}}_R(\Omega ^n_R M)+n \leq {\mathsf{red-G-dim}}_S(M)+n. \end{equation*}

Here, the first inequality, in view of [Reference Bruns and Herzog11, 1.3.5], follows from Proposition 3.9, the second inequality follows from Theorem 3.11 and Remark 3.12, and the equality is due to 3.18.

4. Proof of Proposition 3.8

This section is devoted to a proof of Proposition 3.8. We start by noting that:

Proposition 3.8, in view of Remark 4.1, is subsumed by the following result:

Proof. To prove the claim, we assume $\mathsf{red-}\mathbb{I}_R(N)\le r$ for some integer $r\geq 0$ and proceed by induction on $r$ . Note, if $r=0$ , then by definition, ${{\mathbb{I}}}_R(N)\lt \infty$ , and then condition (iii) of 3.6 implies ${{\mathbb{I}}}_R(M)\lt \infty$ , and hence $\mathsf{red-}\mathbb{I}_R(M)=0\le\mathsf{red-}\mathbb{I}_R(M)$ . So we assume $r\geq 1$ .

It follows that there is a short exact sequence of $R$ -modules $0 \to N^{\oplus a} \xrightarrow []{\chi } K \to \Omega _R^n N^{\oplus b} \to 0$ , where $\mathsf{red-}\mathbb{I}_R(K)=\mathsf{red-}\mathbb{I}_R(N)-1\le r-1$ and $a,b\ge 1$ , $n\ge 0$ are integers, see Remark 2.2(ii). Now we consider the following commutative diagram (with exact rows and columns) obtained by taking the pushout of the maps $\chi$ and $\psi ^{\oplus a}$ , where $0\to F \to N \xrightarrow []{\psi } M \to 0$ is the exact sequence we consider.

Now, since $\mathsf{red-}\mathbb{I}_R(K)\le r-1$ , we use the induction hypothesis on $r$ with the short exact sequence $0\to F^{\oplus a} \to K \to L \to 0$ obtained in the above diagram and conclude that

(4.2.1) \begin{equation} \mathsf{red-}\mathbb{I}_R(L)\le \mathsf{red-}\mathbb{I}_R(K)=\mathsf{red-}\mathbb{I}_R(N)-1. \end{equation}

Case 1: Suppose $n\geq 1$ and consider the exact sequence $0\to M^{\oplus a} \to L \to \Omega _R^n N^{\oplus b} \to 0$ .

We take syzygies of the short exact sequence $0\to F \to N \to M \to 0$ , and see that, for each $v\geq 1$ , there exists a free $R$ -module $H_v$ such that $\Omega _R^v M \cong \Omega _R^v N \oplus H_v$ . So, by adding free summands to the exact sequence $0\to M^{\oplus a} \to L \to \Omega _R^n N^{\oplus b} \to 0$ , we obtain the exact sequence:

(4.2.2) \begin{equation} 0\to M^{\oplus a} \to L \oplus H_n^{\oplus b} \to \Omega _R^n N^{\oplus b} \oplus{H_n^{\oplus b}} \to 0. \end{equation}

The sequence (4.2.2), in view of the isomorphism $\Omega _R^n M \cong \Omega _R^n N \oplus H_n$ , gives the exact sequence:

(4.2.3) \begin{equation} 0\to M^{\oplus a} \to L \oplus H_n^{\oplus b} \to \Omega _R^n M^{\oplus b} \to 0. \end{equation}

Now we use (4.2.3) and conclude from Remarks 2.2(ii) and 4.1 that:

(4.2.4) \begin{equation} \mathsf{red-}\mathbb{I}_R(M)\le \mathsf{red-}\mathbb{I}_R(L \oplus H_n^{\oplus b} )+1 \leq \mathsf{red-}\mathbb{I}_R(L)+1. \end{equation}

Thus, since $\mathsf{red-}\mathbb{I}_R(K)+1=\mathsf{red-}\mathbb{I}_R(N)$ , the proof is complete in view of (4.2.1) and (4.2.4).

Case 2: Suppose $n=0$ and consider the exact sequence $0\to M^{\oplus a} \to L \to \Omega _R^n N^{\oplus b} \to 0$ .

Recall that, by the hypothesis and the previous pushout diagram, we have the short exact sequences $0\to M^{\oplus a} \to L \to N^{\oplus b} \to 0$ and $0\to F^{\oplus b} \to N^{\oplus b} \xrightarrow []{\psi ^{\oplus b}} M^{\oplus b} \to 0$ . Next we take the pullback of the maps $L \to N^{\oplus b}$ and $\psi ^{\oplus b}$ and obtain the following commutative diagram with exact rows and columns:

The top exact sequence in the above diagram shows that $Y\cong M^{\oplus a}\oplus F^{\oplus b}$ as $F^{\oplus b}$ is a free $R$ -module. So the left vertical exact sequence in the above diagram gives the short exact sequence:

(4.2.5) \begin{equation} 0 \to M^{\oplus a}\oplus F^{\oplus b} \to L \to M^{\oplus b} \to 0. \end{equation}

We add, if needed, free $R$ -modules to (4.2.5) and obtain an exact sequence for some integers $c, d\geq 0$ :

(4.2.6) \begin{equation} 0 \to (M \oplus F^{\oplus c})^{\oplus a} \to L \oplus F^{\oplus d} \to (M \oplus F^{\oplus c})^{\oplus b} \to 0. \end{equation}

Next we observe the following inequalities:

\begin{align*}\mathsf{red-}\mathbb{I}_R(M) \le \mathsf{red-}\mathbb{I}_R(M\oplus F^{\oplus c}) \leq \mathsf{red-}\mathbb{I}_R(L \oplus F^{\oplus d})+1 &\le \mathsf{red-}\mathbb{I}_R(L) + 1 \\ &\le \mathsf{red-}\mathbb{I}_R(K)+1 \\ &= \mathsf{red-}\mathbb{I}_R(N) \\ &\leq r. \end{align*}

Here, we obtain the second inequality by applying Remark 2.2(ii) to (4.2.6), the third inequality is due to Remark 4.1, the fourth inequality and the equality follow from (4.2.1), the last inequality is our assumption, and the first inequality can be deduced by the next claim:

Proof of the Claim: We note, due to Definition 2.1 and Remark 2.2(ii), that there exists a short exact sequence of $R$ -modules $0 \to (A\oplus H)^{\oplus a} \to K \xrightarrow []{p} \Omega _R^{s}(A\oplus H)^{\oplus b} \to 0$ , where $a,b\ge 1$ , $s\ge 0$ are integers and $\mathsf{red-}\mathbb{I}_R(K)=\mathsf{red-}\mathbb{I}_R(A\oplus H)-1\le r-1$ . Note also that $\Omega _R^s(A \oplus H)^{\oplus b}\cong \Omega _R^sA^{ \oplus b} \oplus G$ for some free $R$ -module $G$ . Thus, we have an exact sequence of the form:

(4.2.7) \begin{equation} 0 \to (A\oplus H)^{\oplus a} \to K \to \Omega _R^sA^{ \oplus b} \oplus G \to 0. \end{equation}

Now we use 3.10 with the sequence (4.2.7) and obtain the following exact sequence of $R$ -modules:

(4.2.8) \begin{equation} 0 \to (A \oplus H)^{ \oplus a} \xrightarrow []{\alpha } K' \to \Omega _R^{n}A^{\oplus b} \to 0, \end{equation}

where $K\cong K'\oplus G$ .

Next we consider the commutative diagram (with exact rows and columns) that is obtained by taking the pushout of the map $\alpha$ from (4.2.8) with the canonical surjection $(A\oplus H)^{\oplus a} \twoheadrightarrow A^{\oplus a}$ :

Note that, since $K\cong K'\oplus G$ , we have an exact sequence $0 \to G \to K \to K' \to 0$ . We use this exact sequence and, since $\mathsf{red-}\mathbb{I}_R(K)\le r-1$ , conclude by the induction hypothesis on $r$ that

(4.2.9) \begin{equation} \mathsf{red-}\mathbb{I}_R(K') \leq \mathsf{red-}\mathbb{I}_R(K) \leq r-1. \end{equation}

Next, we consider the exact sequence obtained from the above diagram: $0 \to H^{\oplus a} \to K' \to X \to 0$ . As $\mathsf{red-}\mathbb{I}_R(K')\le r-1$ by (4.2.9), we make use of the induction hypothesis on $r$ and conclude that

(4.2.10) \begin{equation} \mathsf{red-}\mathbb{I}_R(X)\le \mathsf{red-}\mathbb{I}_R(K') \leq \mathsf{red-}\mathbb{I}_R(K). \end{equation}

We also consider the exact sequence obtained from the above diagram: $0 \to A^{\oplus a} \to X \to \Omega _R^n A^{\oplus b} \to 0$ . We deduce from this short exact sequence and Remark 2.2(ii) that:

(4.2.11) \begin{equation} \mathsf{red-}\mathbb{I}_R(A) -1\le \mathsf{red-}\mathbb{I}_R(X). \end{equation}

Finally, as $\mathsf{red-}\mathbb{I}_R(K)=\mathsf{red-}\mathbb{I}_R(A\oplus H)-1$ , we see $\mathsf{red-}\mathbb{I}_R(A) \le \mathsf{red-}\mathbb{I}_R(A\oplus H)$ by (4.2.10) and (4.2.11). This completes the proof of the Claim and hence that of the proposition.