What is $\operatorname{Hom}_R(P,R)$ isomorphic to when $P$ is projective?

Question

Let $R$ be a (possibly noncommutative) ring with $1$. Now, quite clearly we have $$\operatorname{Hom}_R(R^n,R)\cong R^n.$$

I am wondering if there is any similar result for $\operatorname{Hom}_R(P,R)$ where $P$ is a projective $R$-module.

Answer 1

If you want $\operatorname{Hom}_R(P,R)\simeq P$ for $P$ a finitely generated projective $R$-module, then forget it.

$\bullet$ If $R$ is an integral domain, and $I\subset R$ is a non-zero ideal, then $\operatorname{Hom}_R(I,R)\simeq I^{-1}$, where $I^{-1}=\{x\in Q(R):xI\subseteq R\}$. (Here $Q(R)$ stands for the field of fractions of $R$.)
$\bullet$ If $\operatorname{Hom}_R(I,R)\simeq I$, then $I^{-1}\simeq I$. This is equivalent to $\exists y\in Q(R)$, $y\ne0$ such that $yI^{-1}=I$.
$\bullet$ If moreover $I$ is projective (that is, invertible) multiply the above equation by $I$ and find $yR=I^2$, that is, $I^2$ is principal.

Now let $R=\mathbb Z[\sqrt{-14}]$, and $I=(3,1+\sqrt{-14})$. Since $R$ is Dedekind, $I$ is projective. Moreover, $I^2=(9,7+\sqrt{-14})$. The last step is to show that $I^2$ is not principal and this is left to the reader.

(Simpler examples are welcome.)

Remark. However we have the following result: Finitely generated projective modules are isomorphic to their double dual.