Hom group $Hom_{\mathbb{Z}/m}(I, \mathbb{Z}/m)$ of an Ideal $I$

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?

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.