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I'm trying to show that $R= \mathbb{Z}/m$ is an injective $R$-module using Baer's criterion (Weibel, Homological Algebra, exercise 2.3.1).

I want to find the homomorphism module $Hom_R(I, R)$, then show that every homomorphism is extendable to an $R$-linear map $R \rightarrow R$. Since $R$ is a PID, the ideal $I$ is of the form $(n)$ for some $n \in R$, and every $R$-linear map is given by $n \mapsto k$ for some $k$. My guess is that $n$ has to divide $k$ in $R$, and we simply send $1 \mapsto \frac{k}{n}$ for the extension. But, I can't prove my statement. Can someone help me find the Hom-module or is my idea wrong altogether?

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The ideals of $R$ are of the form $I = (n + m\mathbb{Z})$ with $n$ a divisor of $m$. Write $m = d n$ and you find that $n + m\mathbb{Z}$ is annihilated by $d + m \mathbb{Z}$ in $R$. Consequently, if $n + m\mathbb{Z}$ maps to $k \in R$ via a morphism $I \to R$, the element $k$ is also annihilated by $d + m \mathbb{Z}$, so if $k = f + m\mathbb{Z}$ for some $f$, then $m = d n$ divides $df$ and thus $n$ divides $f$. Write $f = g n$ and map $1 + m\mathbb{Z}$ to $g + m\mathbb{Z}$ as you envisioned.

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