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This is probably easy, but I'm not seeing things. I just read the definition of an injective module on Wikipedia and found the claim that the $\mathbb{Z}$-module $\mathbb{Q}$ is an example of an injective module.

So, suppose we have a submodule $M^{\prime}\subseteq M$ of a $\mathbb{Z}$-module $M$.

Given a $\mathbb{Z}$-module homomorphism $f\colon M^{\prime}\to\mathbb{Q}$, we should be able to extend it to $M$.

How can one do that? Thanks.

Shoutre
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  • hint: the field of fractions of an integral domain $R$ as an $R$-module is always injective. Use the Baer criterion. – ArtW Jan 14 '17 at 11:55
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    Since $\Bbb Z$ is a PID, the injective $\Bbb Z$-modules are exactly the divisible modules. – Watson Jan 14 '17 at 12:40

1 Answers1

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This is an easy application of Baer’s criterion. A homomorphism $f\colon n\mathbb{Z}\to\mathbb{Q}$ extends to a homomorphism $g\colon\mathbb{Z}\to\mathbb{Q}$; just take $y\in\mathbb{Q}$ such that $ny=f(n)$ and define $g(z)=zy$.

egreg
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