Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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
add comment

2 Answers

up vote 3 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.

share|improve this answer
add comment

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.

share|improve this answer
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. –  Bruno Stonek Dec 10 '11 at 17:21
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.