I've been asked to prove that there is exactly one ring homomorphism from the ring of integers to any ring with unity. And I proved that let $R$ be a ring with unity and
$$
\begin{array}{rccl}
\varphi \colon & \mathbb{Z} & \longrightarrow & R\\
&z & \longmapsto & \varphi(z)=z \cdot 1_R,
\end{array}
$$
$\varphi$ is the unique posible homomorphism from $\mathbb{Z}$ to $R$.
The second part of the exercise ask: "Show that $\exists \,n \in \mathbb{N}$ shuch that $\ker(\varphi) = (n)$, where $(n) = \{n \cdot k \, \mid \, k \in \mathbb{Z} \}$ without using the definition of characteristic of a ring R".
Best Answer
The kernel of a ring homomorphism is an ideal. The only ideals in $\Bbb Z$ are of the form $(n)$.
That is, $\Bbb Z$ is a PID.