# $M$ is a projective module iff $D(M)$ is an injective module

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.

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.