## Ring homomorphisms map an identity element to an identity or a zero divisor

Let $R$ and $S$ be nonzero rings with identies $1_R$ and $1_S$, respectively. Let $\varphi : R \rightarrow S$ be a nonzero ring homomorphism.

1. Prove that if $\varphi(1_R) \neq 1_S$, then $\varphi(1_R)$ is a zero divisor in $S$. Deduce that if $S$ is an integral domain then every ring homomorphism $R \rightarrow S$ sends the identity of $R$ to the identity of $S$.
2. Prove that if $\varphi(1_R) = 1_S$ then if $u \in R$ is a unit, then $\varphi(u) \in S$ is a unit and $\varphi(u)^{-1} = \varphi(u^{-1})$.

1. Suppose $\varphi(1_R) = r$, with $r \neq 1$. First, if $r = 0$, then $\varphi(x) = \varphi(x \cdot 1_R)$ $= \varphi(x) \varphi(1_R)$ $= \varphi(x) \cdot 0$ $= 0$, so that $\varphi = 0$, a contradiction. Thus $r \neq 0$. Now $\varphi(1_R) = \varphi(1_R \cdot 1_R)$ $= \varphi(1_R) \cdot \varphi(1_R)$ $= r^2$, so that $r^2 = r$, and we have $r(r-1) = 0$. Thus $r$ is a left zero divisor. Similarly, $(1-r)r = 0$, and $r$ is a right zero divisor. Hence $r$ is a zero divisor in $S$.

If $S$ is an integral domain, this yields a contradiction. Thus if $\varphi : R \rightarrow S$ is a nonzero ring homomorphism and $R$ and $S$ have identities, then $\varphi(1) = 1$. (I.e. $\varphi$ is a unital ring homomorphism.)

2. Suppose $u \in R$ is a unit. Now $\varphi(u) \varphi(u^{-1}) = \varphi(uu^{-1}) = \varphi(1) = 1$. Similarly, $\varphi(u^{-1}) \varphi(u) = 1$. Thus $\varphi(u)$ is a unit, and by the uniqueness of inverses, $\varphi(u)^{-1} = \varphi(u^{-1})$.