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$\DeclareMathOperator{Hom}{Hom}$Let $R$ be a ring with $1$, $D$ a divisible $\mathbb{Z}$-module. Denote $X=\Hom_{\mathbb{Z}}(R,D)$ as the set of homomorphisms from $R$ to $D$ when viewing $R$ and $D$ as $\mathbb{Z}$-modules. We can associate an $R$-module structure on $X$ by defining $(rf)(x) = f(xr)$ for $r\in R, f\in X$. I want to prove that $X$ is an injective $R$-module.

To accomplish this, it suffices to prove:

Given an injective $g: M\to N$, with $M,N$ being $R$-modules, the induced map $$g^{\ast}: \Hom_R(N,\Hom_{\mathbb{Z}}(R,D)) \to \Hom_R(M,\Hom_{\mathbb{Z}}(R,D))$$ is surjective.

I have made some progress in proving the claim, but I don't know utilize the assumption that $g$ is injective and $D$ is divisible. There is also a hint, which I don't know how to use either: namely $$\Hom_{\mathbb{Z}}(Y,D) \cong \Hom_{R}(Y,\Hom_{\mathbb{Z}}(R,D))$$ for any $R$-module $Y$, the map is given by: $f\mapsto \hat{f}$, with $\hat{f}(y)(r) = f(ry)$. Proving this does not require divisibility of $D$.

Any hint or help is appreciated.

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1 Answer 1

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The hint is apposite; composing $g^*$ with these isomorphisms gives a $\Bbb Z$-module map $$\text{Hom}_{\Bbb Z}(N,D)\to\text{Hom}_{\Bbb Z}(M,D).$$ This is the map induced by the injective homomorphism $g:M\to N$. This map is surjective, as $D$ is injective in the category of $\Bbb Z$-modules. Therefore $g^*$ is surjective.

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  • $\begingroup$ Thank you for your help. I never thought it would be so simple. :) $\endgroup$
    – pisco
    Oct 17, 2017 at 17:45

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