# Prove that if $\varphi(1_R) \ne 1_s$ then $\varphi(1_R)$ is a zero divisor in $S$.

Let $R$ and $S$ be nonzero rings with identity and denote their respective identities by $1_R$ and $1_S$. Let $\varphi: R \to S$ be a nonzero homomorphism of rings.

Prove that if $\varphi(1_R) \ne 1_s$ then $\varphi(1_R)$ is a zero divisor in $S$.

I've seen many variations of this problem elsewhere on MSE, but none in this particular form. I've got the following:

Since $\varphi$ is a homomorphism, it follows that $\varphi(1_R) = \varphi(1_R) \varphi(1_R)$; and so

$$\varphi(1_R)1_S = \varphi(1_R) \varphi(1_R) \implies \varphi(1_R)[1_S - \varphi(1_R)] = 0.$$

Since $\varphi(1_R) \ne 1_S$ by assumption, either $\varphi(1_R)$ is a zero divisor (in which case, we're done), or $\varphi(1_R) = 0$.

Here's where I'm stuck because I don't see why $\varphi(1_R)$ can't be equal to $0$. Indeed, the statement of the problem says that $\varphi$ is a nonzero homomorphism of rings, but is there some reason why just $1_R$ can't be in ker$(\varphi)$?

If $\phi(1_R)=0$, then we have for all $r\in R$ $$\phi(r)=\phi(1_Rr)=\phi(1_R)\phi(r)=0,$$ hence $\phi=0$, which was excluded.