How do I prove that $End(P_R)\cong End({_R}P^*)$, where P is f.g. projective module?

Question

I can't prove the 2d in Proposition 18.19 of Lam's book "lectures on modules and rings". Suppose P is a f.g. projective right R-module, and $S=\operatorname{End}(P)$ and $Q=\operatorname{Hom}(P,R)=P^*$. How do I define the isomorphism between S and End(Q)? I thought of $s\mapsto qs$ but it doesn't seems to work. Could anyone help?

Answer 1

Question: "I can't prove the 2d in Proposition 18.19 of Lam's book "lectures on modules and rings". Suppose P is a f.g. projective right R-module, and S=End(P) and Q=Hom(P,R)=P∗. How do I define the isomorphism between S and End(Q)? I thought of s↦qs but it doesn't seems to work. Could anyone help?"

Answer: There is for any projective $R$-module $P$ an isomorphism $End_R(P) \cong P^* \otimes_R P$ hence

$$S:=End_R(P) \cong P^* \otimes_R P \cong P^{**}\otimes_R P^* \cong End_R(P^*):=Q.$$

This is true since $P^{**}\cong P$ is a canonical isomorphism of $R$-modules.

In fact there is in general a ring isomorphism $S \cong Q^{op}$. You find a more detailed explanation below (the same proof holds for projective modules):

Dual space of vector space