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Let $(R,\mathfrak{m})$ be a noetherian local ring, and $E=E_R(R/\mathfrak{m})$ the injective hull of $R/\mathfrak{m}$. What do we know about the Krull dimension of $E$? Thank you.

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@ Tobias. Thank you! –  tlquyen Jan 17 '13 at 6:45

2 Answers 2

We have $\operatorname{Ann}_RE(R/\mathfrak m)=(0)$.

Let $a\in R$ such that $aE=0$. If $\operatorname{Ann}(a)=R$, we are done. Otherwise, $\operatorname{Ann}(a)\subseteq\mathfrak m$. Set $x=\hat 1\in E.$ Obviously $r\in \operatorname{Ann}(a)\Rightarrow r\in\mathfrak m\Rightarrow rx=0\Rightarrow r\in\operatorname{Ann}(x)$, so $\operatorname{Ann}(a)\subseteq\operatorname{Ann}(x)$. Now define $f:Ra\to E$ by $f(ra)=rx$. Since $E$ is injective $f$ can be extended to $R$ and thus we get an element $y\in E$ such that $x=ay$. It follows that $x=0$, a contradiction.

Since $\operatorname{Ann}_RE_R(R/\mathfrak{m})=(0)$ we get $\dim E_R(R/\mathfrak{m})=\dim R$. (Here the Krull dimension of an $R$-module $M$ is considered as being $\dim R/\operatorname{Ann}(M)$.)

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Could you explain why $rx=0$ and how $f$ is extended to $R$. THanks. –  user58764 Jan 18 '13 at 15:13
$rx=0$ since $x=1+\mathfrak m$ and $r\in\mathfrak m$. $f$ extends to $R$ because $Ra\subseteq R$ and $E$ is injective. –  user26857 Jan 18 '13 at 18:24
Where did you use the Noetherianess of $R$? –  user58764 Jan 19 '13 at 5:31
@user58764 As you can see, I didn't use it. –  user26857 Jan 19 '13 at 9:20

If one considers the Krull dimension of a module $M$ as being the supremum of the lengths of all chains of prime ideals in $\operatorname{Supp}(M)$, then $\dim E_R(R/\mathfrak m)=0$ since $E_R(R/\mathfrak m)$ is an artinian $R$-module (see here) whose support equals $\{\mathfrak m\}$.

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