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?
Are free modules injective?
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$\begingroup$
abstract-algebra
modules
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0Rings for which free modules are injective are called quasi-frobenius rings (noetherian self-injective rings). – 2011-12-11
2 Answers
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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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0@Manos: I edited in a reference for the stated result. I recommend you learn it, it is very useful. – 2011-12-10