This is related to Weibel, Homological Algebra, Chpt 4, Sec 2, Vista 4.2.7. Assume $R$ commutative.
If $R$ is a $0-$dimensional local ring with maximal ideal $m$, then $R$ is quasi-Frobenius iff $\operatorname{ann}(m)\cong R/m$.
$\textbf{Q:}$ How do I prove above assertion? My first thought is to use show $R$ is a Frobenius $R/m=k$ algebra.(i.e. $R\cong \operatorname{Hom}_k(R,k)$) If this is the case, then it follows that $R$ is self injective by hom-tensor adjunction with hom preserving injective objects. Since $R$ is local 0-dimensional, it follows that $R$ is artinian which makes filtration $0=m^{n+1}\subset m^n\subset\dots m\subset R$ finite length and it gives the splitting of $R\cong k\oplus_{i\leq n}\frac{m^i}{m^{i+1}}$ as a $k$-vector space. However, I have to note that first fact $k$ comes from identification of $k=R/m$ which contains $1$. Thus, I could not realize first factor $k$ as $\operatorname{Ann}(m)$. I do not know whether I am even on the right track.
$\textbf{Q':}$ The book correction part indicates if $R$ is finite dimensional over field $k$, then $R$ is a Frobenius algebra over $k$ iff $R$ is quasi-Frobenius. It is clear that if $R$ is a Frobenius algebra over $k$, then $R$ is self injective which implies $R$ is quasi-Frobenius. How do I get reversed implication?
$\textbf{Q'':}$ The book says $R$ is Gorenstein if $\operatorname{id}(R)$ injective dimension is finite. Since in the commutative case, I have $\operatorname{id}(R)$ identified with Krull dimension. So $R$ s has finite krull dimension which means $R$ may not be $0-$dimensional. "If $R$ is a $0$-dimensional local ring with maximal ideal $m$, then $R$ is quasi-Frobenius iff $\operatorname{ann}_R(m)\cong R/m$. This recognition criterion is at the heart of current research into the Gorenstein rings that arise in algebraic geometry." Why this criterion is at the heart of Gorenstein rings arising in algebraic geometry?