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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.

user26857
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Sam Williams
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    Let's say $P$ is a right $R$-module, so $\text{Hom}_R(P,R)$ is a left $R$-module. It is finitely generated projective if $P$ is finitely generated. Otherwise, it's a summand of a direct product of copies of $_RR$. Not projective, in general. – egreg Jun 26 '15 at 20:03
  • I think the right title is the following: "Is $Hom_R(P,R)$ isomorphic to $P$ when $P$ is projective?" – user26857 Jun 26 '15 at 21:13
  • @user26857: The answer to your question is false, e.g. $\mathrm{Hom}(\oplus^\omega \mathbb{Z},\mathbb{Z})\cong\prod^\omega\mathbb{Z}$. –  Jun 29 '15 at 15:06
  • @JasonPolak It's not my question (and I know the answer is wrong). I just wanted to made it clear what the question seems to be if follow the given example. (A field instead of $\mathbb Z$ makes things even easier.) – user26857 Jun 29 '15 at 15:17
  • Of course $Hom_R(P,R)$ is isomorphic to itself and in general little else can be said. I think the question should be more precise for a good answer. – Mohan Jun 29 '15 at 17:28
  • The example is a free finitely generated module. So you would expect the same result for a projective finitely generated module, which I think is true (if $R$ is commutative). – san Jun 29 '15 at 21:58

1 Answers1

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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.

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user26857
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  • Very nice example. In this case we have at least $I \simeq Hom(I,R)$ as $\Bbb{Z}$-modules, via $3\mapsto Inclusion$ and $1+\sqrt{-14}\mapsto \psi$, where $\psi(3):=1-\sqrt{-14}$ and $\psi(1+\sqrt{-14}):=5$. – san Jul 01 '15 at 02:36
  • Many thanks! To be honest, I expected $Hom_R(P,,R)\cong P$ to be false, but I was wondering if that doesn't work, is there there was anything similar which could be helpful? For example, your remark is very helpful. – Sam Williams Jul 02 '15 at 17:08