Proving that ker is $0$ or $ p \mathbb Z$

Question

Let $k$ be a field. Let $\varphi$ be the ring homomorphism $\varphi : \mathbb Z \to k $ defined by $\varphi (n.1) = n \varphi (1) = n.1 = n.$ Prove that $$ \ker(\varphi) = \begin{cases} (0), \\ p \mathbb Z, \quad p \ \rm prime. \end{cases}$$

Could someone give me a hint of how to tackle this please?

Answer 1

For homomorphisms of rings $f:A\rightarrow B$, where $B$ is an integral domain, the kernel is a prime ideal (this is easy to check). Now, the only prime ideals of $\mathbb{Z}$ are $(0)$ and the ideals generated by a prime number (remember that $\mathbb{Z}$ is a PID).