Complete Intersection Hom Injective Dimension

Abstract

We introduce and investigate a new injective version of the complete intersection dimension of Avramov, Gasharov, and Peeva. It is like the complete intersection injective dimension of Sahandi, Sharif, and Yassemi in that it is built using quasi-deformations. Ours is different, however, in that we use a Hom functor in place of a tensor product. We show that (a) this invariant characterizes the complete intersection property for local rings, (b) it fits between the classical injective dimension and the G-injective dimension of Enochs and Jenda, (c) it provides modules with Bass numbers that are bounded by polynomials, and (d) it improves a theorem of Peskine, Szpiro, and Roberts (Bass’ conjecture).

Appendix A: Derived Functors

The first result in this appendix is for use in Theorem 3.6 and Proposition 3.8.

Lemma A.1

Let $R,S,\widetilde {R},\widetilde {S}$ be commutative noetherian rings (not necessarily local) and consider the following commutative diagram of ring homomorphisms

such that $\widetilde {S} \cong S\otimes _{R} \widetilde {R}$ and ${\operatorname {Tor}_{i}^{R}}(S,\widetilde {R}) = 0$ for all i > 0. Let $Y\in \mathcal {D}_{-}(S)$. Then

$$ \operatorname{id}_{\widetilde{R}}(\mathbf{R}\!\operatorname{Hom}_{S}(\widetilde{S},Y))\leq \operatorname{id}_{R}(Y). $$

Proof

In $\mathcal {D}(R)$, we have isomorphisms

$$ \begin{array}{@{}rcl@{}} \text{\textbf{R}Hom}_{S}({\widetilde{S}}, Y) &\simeq&\text{\textbf{R}Hom}_{S}(S\otimes^{\mathbf{L}}_{R}\widetilde{R}, Y)\\ &\simeq&\text{\textbf{R}Hom}_{R}(\widetilde{R},\text{\textbf{R}Hom}_{S}(S, Y))\\ &\simeq&\text{\textbf{R}Hom}_{R}(\widetilde{R},Y). \end{array} $$

It suffices to show that these isomorphisms respect the $\widetilde R$-structure; then the result follows directly since $\operatorname {id}_{\widetilde R}(\mathbf {R}\!\operatorname {Hom}_{R}(\widetilde {R},Y))\leq \operatorname {id}_{R}(Y)$.

In order to respect the $\widetilde R$-structure, fix an S-injective resolution $Y\xrightarrow \simeq J$ to compute $\text {\textbf {Rhom}}_{S}({\widetilde {S}}, Y)$. Arguing as in the preceding display, this yields

$$ \begin{array}{@{}rcl@{}} \text{\textbf{Rhom}}_{S}({\widetilde{S}}, Y) \simeq\operatorname{Hom}[S]{\widetilde S}J \cong\operatorname{Hom}{\widetilde{R}}{J}. \end{array} $$

It remains to show that the complex $\operatorname {Hom}{\widetilde {R}}{J}$ represents $\mathbf {R}\!\operatorname {Hom}_{S}(\widetilde {R},Y)$ in $\mathcal {D}(\widetilde R)$. For this, it suffices to show that the injective S-modules J_i are $\text {\textbf {R}Hom}_{R}(\widetilde {R}, -)$ acyclic, i.e., that $\operatorname {Ext}_{R}^{i}(\widetilde {R},J_{j})=0$ for all i ≥ 1 and all j. Computing as in the above displays, we find that $\operatorname {Ext}_{R}^{i}(\widetilde {R},J_{j})\cong \text {Ext}_{S}^{i}(\widetilde {S}, J)_{j}=0$ for all such i and j, where the vanishing comes from the fact that J_j is S-injective. □

Remark A.2

One can also prove Lemma A.1 using differential graded algebra resolutions, but the current proof is more direct.

The next result is proved like Lemma A.1. It is not needed for the results of this paper; however, it is used implicitly in [30, Theorem F].

Lemma A.3

Let $R,S,\widetilde {R},\widetilde {S}$ be commutative noetherian rings (not necessarily local) and consider the following commutative diagram of ring homomorphisms

such that $\widetilde {S} \cong S\otimes _{R} \widetilde {R}$ and ${\operatorname {Tor}_{i}^{R}}(S,\widetilde {R}) = 0$ for all i > 0. Let $Y\in \mathcal {D}_{+}(S)$. Then

$$ \operatorname{fd}_{\widetilde{R}}(\widetilde{S} \otimes^{\mathbf{L}}_{S} Y)\leq \operatorname{fd}_{R}(Y). $$

The following result is a slight improvement on [22, Theorem 3.13]. Our proof is similar to that of op. cit., but we include it here for the sake of completeness.

Proposition A.4

Let R be a commutative noetherian ring (not necessarily local) with $d = \dim (R) < \infty$. Let R^′ be a faithfully flat R-algebra. Then R is in the thick subcategory T of $\mathcal {D}(R)$ generated by Add(R^′).

Proof

• Claim 1: For all projective R-modules P and all n ≥ 1 the tensor product P ⊗_R(R^′)^⊗n is in T. In particular, for all n ≥ 1, we have (R^′)^⊗n ∈ T.

Proof of Claim 1. We argue by induction on n. For the base case n = 1, note that P is a summand of R^(B) for some set B. By definition we have R^(B) ⊗_RR^′≅(R^′)^(B) ∈ Add(R^′). Thus the summand P ⊗_RR^′ is also in $\operatorname {Add}(R') \subseteq T$.

Induction step: Assume that n ≥ 1 and that P ⊗_R(R^′)^⊗n ∈ T for all P. Fact 2.1 implies that pd _R(R^′) ≤ d. This provides a bounded projective resolution

$$ 0\to P_{d}\to\cdots\to P_{0}\to R'\to 0 $$

(A.4.2)

where each projective R-module P_i is a summand of a free R-module $R^{(B_{i})}$ with basis B_i. Apply $-\otimes _{R}(P\otimes _{R} (R^{\prime })^{\otimes n})$ to the resolution (A.4.2). As $(R^{\prime })^{\otimes n}$ is flat, so is $P\otimes _{R} (R')^{\otimes n}$. This yields an exact sequence

$$ 0\to P_{d} \otimes_{R} P \otimes_{R} (R')^{\otimes n} \to\cdots\to P_{0} \otimes_{R} P\otimes (R')^{\otimes n} \to P\otimes_{R} (R')^{\otimes (n+1)}\to 0. $$

Our induction hypothesis implies P_i ⊗_RP ⊗_R(R^′)^⊗n ∈ T for all i. As T is thick, the above exact sequence implies that P ⊗_R(R^′)^⊗(n+ 1) ∈ T. This establishes Claim 1.

Set M = R^′/R, which is flat over R since R^′ is faithfully flat. Next set I = Σ^− 1M so there is a natural exact triangle in $\mathcal {D}(R)$

$$ I\xrightarrow{\phi} R\to R'\to. $$

(A.4.3)

• Claim 2: For all m, n ≥ 1 we have (R^′)^⊗m ⊗_RI^⊗n ∈ T. In particular, R^′⊗_RI^⊗n ∈ T for all n ≥ 1.

Proof of Claim 2. We argue by induction on n. For the base case n = 1, apply the functor (R^′)^⊗m ⊗_R − to the triangle (A.4.3) and use the flatness of (R^′)^⊗m to get the exact triangle

$$ {(R')^{\otimes m}}\otimes_{R}I\to (R')^{\otimes m}\to(R')^{\otimes (m+1)}\to. $$

Since (R^′)^⊗m and (R^′)^⊗(m+ 1) are in T by Claim 1, so is (R^′)^⊗m ⊗_RI.

The induction step is similar to the base case. Assume that n ≥ 1 and that (R^′)^⊗m ⊗_RI^⊗n ∈ T for all m ≥ 1. Apply ((R^′)^⊗m ⊗_RI^⊗n) − to the triangle (A.4.3) and use the flatness of (R^′)^⊗m ⊗_RM^⊗n to get the exact triangle

$$ (R')^{\otimes m}\otimes_{R}I^{\otimes (n+1)}\to (R')^{\otimes m}\otimes_{R}I^{\otimes n}\to(R')^{\otimes (m+1)}\otimes_{R}I^{\otimes n}\to. $$

Since (R^′)^⊗m ⊗_RI^⊗n and (R^′)^⊗(m+ 1) ⊗_RI^⊗n are in T, so is (R^′)^⊗m ⊗_RI^⊗(n+ 1). This establishes Claim 2.

Recall the morphism ϕ from Eq. A.4.3. For each $n\in \mathbb {N}$, consider the natural morphism $I^{\otimes n}\xrightarrow {\phi ^{\otimes n}}R^{\otimes n}\simeq R$ and the induced exact triangle

$$ I^{\otimes n}\xrightarrow{\phi^{\otimes n}} R\to C(n)\to. $$

(A.4.4)

• Claim 3: For all m ≥ 0 and all n ≥ 1 we have I^⊗m ⊗_RC(n) ∈ T. In particular, C(n) ∈ T for all n ≥ 1.

Proof of Claim 3. We argue by induction on n. For the base case n = 1, compare the triangles (A.4.3) and (A.4.4) to conclude that ${I^{\otimes 0}}\otimes _{R} C(1) \simeq C(1) \simeq R'\in T$. For m ≥ 1 it follows that ${I^{\otimes m}}\otimes _{R}{C(1)}\simeq {I^{\otimes m}}\otimes _{R}{R'}\in T$ by Claim 2.

Induction step: Assume that n ≥ 1 and that I^⊗m ⊗_RC(n) ∈ T for all m ≥ 0. The morphism ϕ^⊗(n+ 1) decomposes as the composition of the next morphisms

$$I^{\otimes (n+1)}\xrightarrow{\phi^{\otimes n}\otimes I}R\otimes_{R} I\xrightarrow\cong I\xrightarrow{\phi}R.$$

Apply ⊗_R − I to the triangle (A.4.4) to produce the next exact triangle

$$ I^{\otimes (n+1)}\xrightarrow{\phi^{\otimes n}\otimes I} R \otimes_{R} I\to {C(n)}\otimes_{R} I\to. $$

The Octahedral Axiom applied to the morphisms ϕ^⊗n ⊗ I and ϕ (and their composition ϕ^⊗(n+ 1)) yields the next exact triangle.

$$ {C(n)}\otimes_{R} I \to C(n+1) \to C(1) \to $$

Apply −⊗_RI^⊗m to this triangle to obtain the next one.

$$C(n)\otimes_{R}I^{\otimes (m+1)} \to C(n+1)\otimes_{R}I^{\otimes m} \to C(1)\otimes_{R}I^{\otimes m} \to $$

Since C(n) ⊗_RI^⊗(m+ 1), C(1) ⊗_RI^⊗m ∈ T, we have C(n + 1) ⊗_RI^⊗m ∈ T. This establishes Claim 3.

Now we complete the proof. The module M = R^′/R is flat, hence so is M^⊗(d+ 1). Thus, we have pd _R(M^⊗(d+ 1)) ≤ d and so $\text {Ext}^{d+1}_{R}(M^{\otimes (d+1)}, R)=0$. It follows that

$$\text{Ext}^{0}_{R}(I^{\otimes (d+1)}, R)\cong \text{Ext}^{0}_{R}(\mathsf{\Sigma}^{-d-1}M^{\otimes (d+1)}, R)\cong\text{Ext}^{d+1}_{R}(M^{\otimes (d+1)}, R)=0.$$

It follows that the homotopy class of the morphism ϕ^⊗(d+ 1) is in Ext 0I^⊗(d+ 1)R = 0, thus ϕ^⊗(d+ 1) is nullhomotopic. It follows that the codomain R is a retract of C(d + 1). Claim 3 implies that C(d + 1) is in T, which is closed under retracts. Therefore we have R ∈ T, as desired. □

Our point for including Proposition A.4 is to obtain the next two results for use in Remark 3.2.

Proposition A.5

Continue with the assumptions of Proposition A.4 and let $X\in \mathcal {D}(R)$ be such that $\text {\textbf {R}Hom}_{R}(R^{\prime },X)\in \mathcal {D}_{*}(R)$, where ∗∈{+,−, b}. Then for all Z ∈ T we have $\text {\textbf {R}Hom}_{R}(Z,X)\in \mathcal {D}_{*}(R)$.

Proof

By Remark 2.2 we have Z ∈ T_n for some n ≥ 1. Argue by induction on n.

Base case: n = 1. In this case, Z is a summand of (R^′)^(A) for some A. The condition $\text {\textbf {R}Hom}_{R}(R^{\prime },X)\in \mathcal {D}_{*}(R)$ implies that

$$\text{\textbf{R}Hom}_{R}((R^{\prime})^{(A)}, X)\simeq\text{\textbf{R}Hom}_{R}(R^{\prime},X))^{A}\in\mathcal{D}_{*}(R).$$

It follows that the summand R Hom_R(Z, X) of $\text {\textbf {R}Hom}_{R}((R^{\prime })^{(A)}, X)$ is also in $\mathcal {D}_{*}(R)$.

Inductive step. Assume that n ≥ 1 and for all Z^′∈ T_n we have $\text {\textbf {R}Hom}_{R}(Z^{\prime },X)\in \mathcal {D}_{*}(R)$. Let Z ∈ T_n+ 1. Then Z is a retract of an object $Y\in \mathcal {D}(R)$ such that there is an exact triangle $Y'\to Y\to Y^{\prime \prime }\to$ in $\mathcal {D}(R)$ with Y^′∈ T₁ and $Y^{\prime \prime }\in T_{n}$. Our base case and induction hypothesis imply that $\text {\textbf {R}Hom}_{R}(Y^{\prime },X),\text {\textbf {R}Hom}_{R}(Y^{\prime \prime },X)\in \mathcal {D}_{*}(R)$. A long exact sequence argument shows that $\text {\textbf {R}Hom}_{R}(Y,X)\in \mathcal {D}_{*}(R)$. It follows that the retract R Hom_R(Z, X) must also be in $\mathcal {D}_{*}(R)$. □

Corollary A.6

Continue with the assumptions of Proposition A.4. Let $X\in \mathcal {D}(R)$ be such that $\text {\textbf {R}Hom}_{R}(R^{\prime },X)\in \mathcal {D}_{*}(R)$, where ∗∈{+,−, b}. Then $X\in \mathcal {D}_{*}(R)$.

Proof

Proposition A.4 implies that R ∈ T, so we have $X\simeq \text {\textbf {R}Hom}_{R}(R,X)\in \mathcal {D}_{*}(R)$ by Proposition A.5. □