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Let $M$ be a finite module over a Noetherian ring $A$. Let $0 \rightarrow M \rightarrow I^1 \rightarrow I^2 \rightarrow \cdots $ be a minimal injective resolution of $M$.

Question: Is it true that each $I^k$ is a finite $A$-module? If yes, how can we see that?

Manos
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2 Answers2

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This is not true. Consider $\Bbb{Z}$ as a module over itself and the injective resolution

$$0 \to \Bbb{Z} \to \Bbb{Q} \to \Bbb{Q}/\Bbb{Z} \to 0.$$

The resolution is minimal because $\Bbb{Z}$ is not injective and so we can't have a resolution of length $0$. It is well-known that $\Bbb{Q}$ is not a f.g. Abelian group, so it remains to see that $\Bbb{Q}/\Bbb{Z}$ is not a f.g. Abelian group. Now every element of $\Bbb{Q}/\Bbb{Z}$ is torsion and so if it were f.g. we must have $\Bbb{Q}/\Bbb{Z}$ being a finite Abelian group, a contradiction.

Alternatively, if $\Bbb{Q}/\Bbb{Z}$ were f.g. this would imply $\Bbb{Q}$ is f.g. (since $\Bbb{Z}$ is trivially f.g.), a contradiction.

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The answer is negative.

Take $R$ local with $\dim R\ge 1$ and $M=R/m$. Then the injective hull $E(R/m)$ is not finitely generated, otherwise since $E$ is a faithful $R$-module we deduce that $R$ must be artinian, a contradiction.

  • How do we deduce that $R$ is Artinian? Which result are you using? – Manos Dec 05 '13 at 23:58
  • @Manos If a ring has a finitely generated faithful artinian module (and $E(R/m)$ is artinian!), then the ring itself is artinian. –  Dec 06 '13 at 06:25