Ring homomorphism between a ring R with an identity and an integral domain S is either 0 or maps $1_R$ to $1_S$

Question

I have been reading my Rings & Module course notes and the lecturer has written a proposition with certain properties of ring homomorphisms such as $\phi(0_R) = 0_S$ and some other without any proof. One of them is stated as follows:

If $S$ is an integral domain, $R$ a ring with an identity and $\phi : R \to S$ is a homomorphism, then either $\phi \equiv 0$ or $\phi(1_R) = 1_S$.

I'm sure the lecturer must have left it because its something straight forward but I can't seem to get anywhere with this one. Any help is appreciated.

Answer 1

Let $\phi;R\to S$ be a ring homomorphism, then for any $r\in R$, $$\phi(r)=\phi(r1_R)=\phi(r)\phi(1_R)$$ or $$\phi(r)-\phi(r)\phi(1_R)=0\Rightarrow \phi(r)(1_S-\phi(1_R))=0 $$ since $S$ is a domain, then either $\phi(r)=0$ or $1_S-\phi(1_R)=0$, i.e. $\phi\equiv0$ or $\phi(1_r)=1_S$.