This question is regarding the exercise 2.3.1. from Weibel's Introduction to Homological Algebra.
Let $d\mid m$ and there is a prime $p$ such that $p\mid d$ and $p\mid\frac{m}{d}$. Then I want to conclude that $\mathbb{Z}/d\mathbb{Z}$ is not injective over $\mathbb{Z}/m\mathbb{Z}$.
I know I have to use Baer's criterion. By the assumption d and $\frac{m}{d}$ are not relatively prime. So there is no injective map from $\mathbb{Z}/d\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$. Further Iam not able to. Please help me.
No need to use Baer's criterion here. Recall that a module $I$ is injective iff for every injective morphism $f: M\to N$ and every morphism $g:M\to I$, there is a morphism $h:N\to I$ such that $h\circ f = g$.
Now consider the unique injective morphism $f$ from $\mathbb{Z}/p \mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$, and let $g$ be the one from $\mathbb{Z}/p \mathbb{Z}$ to $\mathbb{Z}/d \mathbb{Z}$. Note that, for any morphism $h:\mathbb{Z}/m\mathbb{Z} \to \mathbb{Z}/d\mathbb{Z}$, the kernel of $h$ contains all multiples of $d$; note, moreover, that the image of $f$ is the set of all multiples of $\frac{m}{p}$. Since $p$ divides $\frac{m}{d}$, we get that $d$ divides $\frac{m}{p}$, so the image of $f$ is contained in the kernel of $h$.
Therefore, for any such $h$, we get that $h\circ f=0 \neq g$. Thus $\mathbb{Z}/d \mathbb{Z}$ cannot be injective.