Ring homomorphisms from a field are injective

Statement

Let $K$ be a field, let $S$ be a nonzero ring and let $f \colon K \to S$ be a ring homomorphism. Then $f$ is injective.

Remarks

The proof: if $a \neq 0$ and $f(a) = 0$, then $1 = f(1) = f(a) f(a^{-1}) = 0$ in $S$. This contradicts $S \neq 0$. The hypothesis $S \neq 0$ is necessary, because the map from $K$ to the zero ring is a ring homomorphism. A consequence: a homomorphism of fields makes the first field a copy of a subfield of the second.

Explanations

L1 human

A field has no ideal other than $0$ and the whole field. The kernel of $f$ is an ideal, and it does not contain $1$. So the kernel is $0$, and $f$ is injective.