Let $(R, \mathfrak m)$ be a local ring and in the same time a finite dimensional algebra over the complex numbers. How one can prove that if $\operatorname{Ann}_R(m)$ has dimension one then $R$ is an injective $R$-module?
Gorenstein ring of dimension zero
3
$\begingroup$
commutative-algebra
injective-module
gorenstein
1 Answers
1
Since $\mathbb{C}$ is algebraically closed, Hilbert's Nullstellensatz implies that $R/m \cong \mathbb{C}$ It is not too hard to show $\text{Ann}_{R}(m) \cong \text{Hom}_R(R/m, R) $ If dim$_{\mathbb{C}}$Hom$_R(R/m, R)=1$, then, since dim$(R)=0$, the result you want follows from Theorem 18.1 in Matsumura's "Commutative Ring Theory".
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0I should specify here that, when I say dim(R) above, I mean Krull dimension. – 2013-01-07