I am trying to prove that if $R$ is a commutative ring with unity and $P$ is a finitely generated projective $R$-Module, then Hom$_R(P,R)$ is projective.
The exercise gives a hint that says 'prove the $P$ is a direct summand of a free module of finite rank'.
If $S$ is the generating set for $P$, I know that we have a unique homomorphism $\phi$ from $F(S)$ to $P$ that is the identity on $S$ given by the universal propety of free modules. And I read in Dummit and Foote that this gives the following short exact sequence (they reference the universal property):
$0\rightarrow \ker(\phi)\rightarrow F(S) \overset{\phi}{\rightarrow} P\rightarrow 0$
But how do I know $\phi$ is surgective?
Also, and equaly impotant, how does lemma get used to solve the original problem?
Best Answer
Since $P$ is finitely generated, there is a surjection $\pi\colon R^n\to P$ for some $n$. (Explicitly, if $P$ is generated by $\{a_1,\dots,a_n\}$, and $\{e_1,\dots,e_n\}$ a basis of $R^n$, you can define $\pi$ by $e_i\mapsto a_i$. It is surjective since the $a_i$ are a generating set.) Since $P$ is projective, we find $P\oplus\ker\pi\simeq R^n$, so $P$ is a direct summand of a free $R$-module of finite rank. Standard isomorphisms show, since $R$ is commutative, $$ \operatorname{hom}_R(R^n,R)\simeq\bigoplus_{i=1}^n\hom_R(R,R)\simeq R^n. $$
So we also have $$ R^n\simeq\operatorname{hom}_R(R^n,R)\simeq\operatorname{hom}_R(P\oplus\ker\pi,R)\simeq\operatorname{hom}_R(P,R)\oplus\operatorname{hom}_R(\ker\pi,R). $$
So $\operatorname{hom}_R(P,R)$ is a direct summand of a free $R$-module, hence projective.