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  "body": "\n## Statement\n\nLet $K$ be a [[field|0QA8CHH8MG]], let $S$ be a nonzero [[ring|0894J980RK]] and let $f \\colon K \\to S$ be a [[ring homomorphism|0GGE0Z0JQ1]].\nThen $f$ is injective.\n\n## Remarks\n\nThe proof: if $a \\neq 0$ and $f(a) = 0$, then $1 = f(1) = f(a) f(a^{-1}) = 0$ in $S$. This contradicts $S \\neq 0$.\nThe hypothesis $S \\neq 0$ is necessary, because the map from $K$ to the [[zero ring|008TBCYDPX]] is a ring homomorphism.\nA consequence: a homomorphism of fields makes the first field a copy of a subfield of the second.\n\n## Explanations\n\n### L1 human\n\nA field has no ideal other than $0$ and the whole field.\nThe kernel of $f$ is an ideal, and it does not contain $1$.\nSo the kernel is $0$, and $f$ is injective.\n",
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