This is the problem I want to solve :

Show that for any finitely generated projective module $P$ over a ring $R $, $\mathrm{Hom}(P,M)$ is isomorphic with $\mathrm{Hom}(P,R)\otimes M $.

This is what I've done:

I define a map : $Hom (P,R) × M \rightarrow Hom (P,M) $, with the law: $(\phi,m)\rightarrow \phi_m $, $\phi_m (x):=\phi(x) m $.

This map is obviously bilinear, so we have a map $Hom (P,R) \otimes M \rightarrow Hom (P,M) $, with the law $\phi\otimes m \rightarrow \phi_m $, $\phi_m (x):=\phi(x) m $.

Now I have to construct the inverse homomorphism to end the proof.

Because $P $ is finitely generated, we can assume $P=R^n $. I define a map:

$Hom (P,M) \rightarrow Hom (P,R)\otimes M $ that sends $f $ to $f'\otimes f (1) $, when $1$ is the identity element of $ P=R^n $. I don't know how to define $f'$.

Is there any hint?

If you don't agree with these, do you have another idea?


  • $\begingroup$ You can't just assume that $P = R^n$; in general, finitely generated projective modules need not be free. $\endgroup$
    – lokodiz
    Oct 28, 2016 at 5:47
  • $\begingroup$ A special case is here: math.stackexchange.com/questions/679584 $\endgroup$
    – Watson
    Sep 26, 2018 at 14:42

2 Answers 2


Hint: This is where you need to use finite generation and projectiveness of $P$. Since $P$ is finitely generated, there is a surjective $R$-map $p : R^n \to P$ for some $n > 0$. Projectiveness of $P$ yields a section $s : P \to R^n$ of $p$ (so that $p\circ s = \operatorname{id}_P$). For $1\le i\le n$, let $s_i$ be the composition $P \xrightarrow{s} R^n \xrightarrow{\pi_i} R$, where $\pi_i$ is the projection onto the $i$th coordinate. Now we define an $R$-map $$\operatorname{Hom}(P,M) \to \operatorname{Hom}(P,R) \otimes M$$ by sending a $g\in \operatorname{Hom}(P,M)$ to $\sum s_i \otimes g(p(e_i))$, where $e_1,\ldots, e_n$ is the standard basis of $R^n$.

  • $\begingroup$ thank you a lot!but how do you know that $\pi_i$'s are well defined? $\endgroup$
    – user115608
    Oct 28, 2016 at 12:51
  • $\begingroup$ @user115608 actually they weren't well-defined unless the sum was direct. I've made an edit to my answer. $\endgroup$
    – kobe
    Oct 28, 2016 at 16:20

Correct me if anything worng. If we consider left modules. Then $P=Re$ for some idempotent $e \in R$. Hence $Hom_R(P,M)\cong eM$ as (eRe,R)-modules. Also $Hom_R(P,R) \otimes_R M\cong eR\otimes_R M \cong eM$. So $Hom_R(P,M)\cong Hom_R(P,R)\otimes_RM$.


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