Proof that ring homomorphisms from a field are injective

Proof

1. Let $K$ be a field, let $S$ be a nonzero ring, and let $f \colon K \to S$ be a ring homomorphism. Since $S$ is nonzero, $1 \neq 0$ in $S$.
2. Let $a \in \ker f$, the kernel of $f$. Suppose that $a \neq 0$.
3. Since $K$ is a field, there is $b \in K$ with $a b = 1$.
4. Then $1 = f(1) = f(a b) = f(a) f(b) = 0 \cdot f(b) = 0$ in $S$.
5. This contradicts step 1. So $a = 0$, and $\ker f \subseteq \set{0}$.
6. Also $f(0) = 0$, since $f(0) = f(0) + f(0)$. So $\ker f = \set{0}$.
7. A ring homomorphism with kernel $\set{0}$ is injective. So $f$ is injective.