Characteristic of a ring: $\ker(\varphi) = (n)$.

Question

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".

Answer 1

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.

Answer 2

Note that the kernel of that ring homomorphism is an ideal of $\mathbb{Z}$. In particular, it must be a subgroup of the cyclic group $\mathbb{Z}$. But subgroups of cyclic groups are again cyclic, so there is $n \in \mathbb{Z}$ such that

$$\ker \varphi = n \mathbb{Z}$$

Example: There is a unique ring morphism $\mathbb{Z} \to \mathbb{R}$, which is the inclusion map. This is injective, and thus the kernel is $(0)$. Thus $\operatorname{Char}(\mathbb{R})=0$.