---
tag: 3RF6K0W06J
kind: proof
name: Proof that ring homomorphisms from a field are injective
msc: ["12E99"]
proves: 09N7DZ7SDB
uses: [0QA8CHH8MG, 0894J980RK, 0GGE0Z0JQ1, 0AAHE7GEW6, 0MMCB8RXGA]
sources:
  - url: https://proofwiki.org/wiki/Ring_Homomorphism_from_Field_is_Monomorphism_or_Zero_Homomorphism
---

## Proof

1. Let $K$ be a [[field|0QA8CHH8MG]], let $S$ be a nonzero [[ring|0894J980RK]], and let $f \colon K \to S$ be a [[ring homomorphism|0GGE0Z0JQ1]]. Since $S$ is nonzero, $1 \neq 0$ in $S$.
2. Let $a \in \ker f$, the [[kernel|0AAHE7GEW6]] 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|0MMCB8RXGA]]. So $f$ is injective.

## Idea

If a nonzero element $a$ went to $0$, then so would $a a^{-1} = 1$. But a ring homomorphism sends $1$ to $1$, and $1 \neq 0$ in $S$. So the kernel is zero, and the map is injective.