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Suppose that $R$ is a local Noetherian ring. Suppose that there exists a non-zero finitely generated injective module $M$. How can I prove that $R$ is Artinian?

It is easy if $R$ is Cohen-Macaulay, because we know that if there exists a nonzero finitely generated module $M$ of finite injective dimension then $\mathrm{id}\;M=\mathrm{depth}\;R$. So in our case we get $\mathrm{depth}\;R=0$, so if the ring is Cohen-Macaulay we can deduce that it is Artinian. But this should be true in general, any idea of how to prove it?

(I was told that it can be proved with Matlis duality, but since this is an exercise in Bruns-Herzog that comes before Matlis duality there should be a way to prove it without using it)

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Chris
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2 Answers2

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Let $Q$ be a finitely generated nonzero injective $R$-module. Assume that $\dim R>0$.

There exists a prime ideal $\mathfrak{p}\neq\mathfrak{m}$ such that $\operatorname{Hom}_R(R/\mathfrak{p},Q)\neq 0$. In order to prove this take $\mathfrak{q}\in\text{Ass}(Q)$ and consider two cases: $\mathfrak{q}\neq\mathfrak{m}$ (and we are done) or $\mathfrak{q}=\mathfrak{m}$ (and now any prime ideal $\mathfrak{p}\neq\mathfrak{m}$ is okay).

Let $a\in\mathfrak{m}$, $a\notin\mathfrak{p}$. Then we have an exact sequence $0\longrightarrow R/\mathfrak{p}\stackrel{a\cdot}\longrightarrow R/\mathfrak{p}$ and by using the injectivity of $Q$ any $f\in\operatorname{Hom}_R(R/\mathfrak{p},Q)$ has an "extension" $g\in\operatorname{Hom}_R(R/\mathfrak{p},Q)$. This shows that $\operatorname{Hom}_R(R/\mathfrak{p},Q)=a\operatorname{Hom}_R(R/\mathfrak{p},Q)$ and Nakayama implies that $ \operatorname{Hom}_R(R/\mathfrak{p},Q)=0$, a contradiction.

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    Nicely done. This is an excellent proof. – Dylan C. Beck Aug 13 '21 at 20:36
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    Beautiful proof. It took me a while to see one thing, I will reword it just for reference: the exactness of $0\rightarrow R/\mathfrak{p}\stackrel{a\cdot_}{\rightarrow} R/\mathfrak{p}$ implies, by the fact that $Q$ is injective, that $\operatorname{Hom}R(R/\mathfrak{p},Q)\stackrel{a\cdot\}{\rightarrow} \operatorname{Hom}_R(R/\mathfrak{p},Q) \rightarrow 0$ is exact, i.e. $a\cdot\operatorname{Hom}_R(R/\mathfrak{p},Q)=\operatorname{Hom}_R(R/\mathfrak{p},Q)$. Thus, NAK forces the module to be zero. – Joel Castillo Rey Jun 22 '24 at 18:44
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Although this is an old question, for completeness I feel the need to prove this using Matlis Duality. Hence let $E$ be injective and finitely generated. We have that $\textrm{dim}_R E \leq \textrm{id}_R E=0$, so this is an Artinian $R$ module with $\textrm{supp} M = \{ \mathfrak{m} \}$ (our maximal ideal).

Employing $3.2.8$ in the book, $E \cong \oplus_n E(k)$ where $k$ denotes our residue field, and since $E$ is finitely generated, so is $E(k)$. Applying the Matlis dual, $T(E(k))$ must then be an Artinian module (by the correspondence asserted by the Matlis dual). However, recall that $T(R) \cong E(k)$, and hence,

$$T(E(k)) \cong T(T(R)) \cong R$$

So that $R$ is Artinian as an $R$-module, and hence as a ring, completing the proof.

Rellek
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