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Let $A$ be a commutative ring and consider the category of $A$-modules. Let $F$ be free $A$-module. Then the functor $Hom_A(F,\cdot)$ is exact. Is the functor $Hom_A(\cdot,F)$ also exact? Equivalently, is a free module injective?

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Just an easy comment: if $A$ is a field, then every $A$-module (i.e. $A$-vector space) is free, projective and injective. – Andrea Dec 10 '11 at 17:25
Rings for which free modules are injective are called quasi-frobenius rings (noetherian self-injective rings). – Jack Schmidt Dec 11 '11 at 18:04
up vote 4 down vote accepted

Here's how I remember this: working again over $\mathbf Z$, we have a short exact sequence \[ 0 \to \mathbf Z \stackrel2\longrightarrow \mathbf Z \to \mathbf Z/2\mathbf Z \to 0 \] and it's clear that this doesn't split.

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Of course, $\mathbb{Z}$ is free as a $\mathbb{Z}$-module: it has basis $\{1\}$.

A $\mathbb{Z}$-module is injective iff it is a divisible abelian group (see here). This is a well-known result that gives a very simple characterization of injective $\mathbb{Z}$-modules.

Hence $\mathbb{Z}$ is not an injective $\mathbb{Z}$-module, since 2 is not divisible by 3.

$\mathbb{Z}$ thus answers your question negatively.

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what is a divisible abelian group? – Manos Dec 10 '11 at 16:51
An abelian group $A$ (written additively) is divisible if for each $a\in A$ and each positive integer $n$ there is some $b\in A$ such that $nb=a$. – Keenan Kidwell Dec 10 '11 at 17:00
@Manos: I edited in a reference for the stated result. I recommend you learn it, it is very useful. – lentic catachresis Dec 10 '11 at 17:21

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