## Compute Hom(ZZ/(30), ZZ/(21)) over ZZ

Exhibit all of the $\mathbb{Z}$-module homomorphisms from $\mathbb{Z}/(30)$ to $\mathbb{Z}/(21)$.

By this previous exercise, we have $\mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(30), \mathbb{Z}/(21)) \cong \mathsf{Ann}_{\mathbb{Z}/(21)}((30))$. We claim that $\mathsf{Ann}_{\mathbb{Z}/(21)} = (7)/(21)$. To see this, suppose first that $\overline{k} \in \mathsf{Ann}_{\mathbb{Z}/(21)}((30))$. In particular, we have $30 \cdot \overline{k} = \overline{30k} = 0$ mod 21. Then $21|30k$, and so $7|10k$. Now $10k \equiv 0$ mod 7, and hence $k \equiv 0$ mod 7. So $\overline{k} \in (7)/(21)$. Conversely, note that $30 \cdot \overline{7k} = \overline{210k} = 0$. Thus $\overline{7k} \in \mathsf{Ann}_{\mathbb{Z}/(21)}((30))$, and we have $\mathsf{Ann}_{\mathbb{Z}/(21)}((30)) = (7)/(21) \cong \mathbb{Z}/(3)$.

So there are precisely three $\mathbb{Z}$-module homomorphisms $\mathbb{Z}/(30) \rightarrow \mathbb{Z}/(21)$. As group homomorphisms, each of these is uniquely determined by the image of $\overline{1}$. These homomorphisms are given by $\overline{1} \mapsto \overline{0}$, $\overline{1} \mapsto \overline{7}$, and $\overline{1} \mapsto \overline{14}$.