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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?

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    $\begingroup$ Is $M$ an arbitrary $R$-module? Where does $P$ show up in $\operatorname{hom}_R(M,R)$? $\endgroup$
    – Ben West
    Feb 14, 2017 at 6:45
  • $\begingroup$ Sorry, I edited the post. $\endgroup$
    – t3nch3r
    Feb 14, 2017 at 6:52

1 Answer 1

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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.

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  • $\begingroup$ Can you comment a little on $\operatorname{hom}_R(R^n,R)\simeq\bigoplus_{i=1}^n\hom_R(R,R)\simeq R^n.$ and $\operatorname{hom}_R(P\oplus\ker\pi,R)\simeq\operatorname{hom}_R(P,R)\oplus\operatorname{hom}_R(\ker\pi,R)$ Sorry, I am new to working with Hom. $\endgroup$
    – t3nch3r
    Feb 14, 2017 at 7:10
  • $\begingroup$ @t3nch3r If $R$ is commutative, and $A_i$, $B$ are $R$-modules, we have $hom_R(\bigoplus_{i\in I}A_i,B)\simeq\prod_{i\in I}hom_R(A_i,B)$ via $f\mapsto (f\circ\alpha_i)$, where $\alpha_i\colon A_i\to\bigoplus_i A_i$ is the inclusion map. If the index set is finite, the direct product is just the direct sum as we've used above. In the above situation, we're just using the fact that $R^n=\bigoplus_{i=1}^n R$. Details can be found as Thm 2.31 in Rotman's Homological Algebra, for instance. This is the same thing used to split the hom over $P\oplus\ker\pi$. $\endgroup$
    – Ben West
    Feb 14, 2017 at 7:14
  • $\begingroup$ Secondly, the standard isomorphism is if $M$ is an $R$-module, $hom_R(R,M)\simeq M$ via $f\mapsto f(1)$. So $hom_R(R,R)\simeq R$ in particular. $\endgroup$
    – Ben West
    Feb 14, 2017 at 7:15
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    $\begingroup$ awesome, yes I just proved the second one for myself. I will look at details for the first. Thank you. I am practicing for masters exam, and I am finding out that my knowledge of modules is lacking. $\endgroup$
    – t3nch3r
    Feb 14, 2017 at 7:16

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