Assume that $A$ is a finite dimensional $k$-algebra, and $M$ is a left $A$-module. Let $D=\operatorname{Hom}_k(-,k)$. Problem: $M$ is a projective module iff $D(M)$ is an injective right $A$-module.
This problem is an exercise follows the 'injective module' chapter. The main result covered in this chapter is: every module is a submodule of injective modules. The hint is to consider the universal property, and that $D(D(M))=M$.
I tried to prove the "$\Rightarrow$" using Baer's criterion, but failed. Given $L \triangleleft A$, $L$ being an ideal of $A$, $f:L \to D(M)$ being an $A$-module homomorphism, I couldn't figure out a way to extend $f$ to $\tilde{f} : A \to D(M)$ with $\tilde{f}| _L = f$.
Any help is appreciated.
Best Answer
A little late, but here is an idea:
Suppose that $M$ is projective, and consider a short exact sequence (of right $A$-modules) starting with $D(M)$, say $$0 \longrightarrow D(M) \longrightarrow \bullet \longrightarrow * \longrightarrow 0.$$ Applying $D$ we get a new short exact sequence of left $A$-modules, $$0 \longrightarrow D(*) \longrightarrow D(\bullet) \longrightarrow M \longrightarrow 0,$$ which splits, by assumption. By applying $D$ again we see that the first sequence splits, so $D(M)$ is injective.
A similar proof works for the other direction.