**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.17**

Solution:

(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, $\varphi(u)^{-1} = \varphi(u^{-1})$.