Krull dimension of the injective hull of residue field

Question

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.

Answer 1

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 $\dim R/\operatorname{Ann}(M)$.)