Gorenstein ring of dimension zero

Question

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?

Answer 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".