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  "body": "\n## Proof\n\n1. 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 $S$.\n2. Let $a \\in \\ker f$, the [[kernel|0AAHE7GEW6]] of $f$. Suppose that $a \\neq 0$.\n3. Since $K$ is a field, there is $b \\in K$ with $a b = 1$.\n4. Then $1 = f(1) = f(a b) = f(a) f(b) = 0 \\cdot f(b) = 0$ in $S$.\n5. This contradicts step 1. So $a = 0$, and $\\ker f \\subseteq \\set{0}$.\n6. Also $f(0) = 0$, since $f(0) = f(0) + f(0)$. So $\\ker f = \\set{0}$.\n7. A ring homomorphism with kernel $\\set{0}$ [[is injective|0MMCB8RXGA]]. So $f$ is injective.\n\n## Idea\n\nIf 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.\n",
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