The Stacks project

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 (, 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. $\square$

Comments (2)

Comment #4278 by Bogdan on

I think should read .

It seems (unless I am missing something) that the claim "We have the desired vanishing for by assumption" requires a justification since does not necessarily contain .

I guess the proof can be fixed by reducing to the case of noetherian local ring and ideal and then immitating the proof of Lemma on page 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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