Lemma 15.69.6. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal contained in the Jacobson radical of $R$. Let $K \in D^+(R)$ have finite cohomology modules. Then the following are equivalent

1. $K$ has finite injective dimension, and

2. there exists a $b$ such that $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(R/J, K) = 0$ for $i > b$ and any ideal $J \supset I$.

Proof. The implication (1) $\Rightarrow$ (2) is immediate. Assume (2). Say $H^ i(K) = 0$ for $i < a$. Then $\mathop{\mathrm{Ext}}\nolimits ^ i(M, K) = 0$ for $i < a$ and all $R$-modules $M$. Thus it suffices to show that $\mathop{\mathrm{Ext}}\nolimits ^ i(M, K) = 0$ for $i > b$ any finite $R$-module $M$, see Lemma 15.69.2. By Algebra, Lemma 10.62.1 the module $M$ has a finite filtration whose successive quotients are of the form $R/\mathfrak p$ where $\mathfrak p$ is a prime ideal. If $0 \to M_1 \to M \to M_2 \to 0$ is a short exact sequence and $\mathop{\mathrm{Ext}}\nolimits ^ i(M_ j, K) = 0$ for $i > b$ and $j = 1, 2$, then $\mathop{\mathrm{Ext}}\nolimits ^ i(M, K) = 0$ for $i > b$. Thus we may assume $M = R/\mathfrak p$. If $I \subset \mathfrak p$, then the vanishing follows from the assumption. If not, then choose $f \in I$, $f \not\in \mathfrak p$. Consider the short exact sequence

$0 \to R/\mathfrak p \xrightarrow {f} R/\mathfrak p \to R/(\mathfrak p, f) \to 0$

The $R$-module $R/(\mathfrak p, f)$ has a filtration whose successive quotients are $R/\mathfrak q$ with $(\mathfrak p, f) \subset \mathfrak q$. Thus by Noetherian induction and the argument above we may assume the vanishing holds for $R/(\mathfrak p, f)$. On the other hand, the modules $E^ i = \mathop{\mathrm{Ext}}\nolimits ^ i(R/\mathfrak p, K)$ are finite by our assumption on $K$ (bounded below with finite cohomology modules), the spectral sequence (15.67.0.1), and Algebra, Lemma 10.71.9. Thus $E^ i$ for $i > b$ is a finite $R$-module such that $E^ i/fE^ i = 0$. We conclude by Nakayama's lemma (Algebra, Lemma 10.20.1) that $E^ i$ is zero. $\square$

Comment #4278 by Bogdan on

I think $f\in R, f\notin \mathfrak p$ should read $f\in I, f\notin \mathfrak p$.

It seems (unless I am missing something) that the claim "We have the desired vanishing for $R/(𝔭,f)$ by assumption" requires a justification since $(\mathfrak p, f)$ does not necessarily contain $I$.

I guess the proof can be fixed by reducing to the case of noetherian local ring $R$ and ideal $I=\mathfrak \m$ and then immitating the proof of Lemma $4$ on page $141$ of the book "Commutative Ring Theory" by Matsumura.

Comment #4443 by on

Good catch! I fixed it by using Noetherian induction. The proof could be fleshed out more, but at least now it isn't nonsense. Thanks very much. Fix is here.

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