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  text: '[summary(Technical): Let $F$ and $G$ be [481 fields], and let $f: F \\to G$ be a [-field_homomorphism]. Then $f$ either is the constant $0$ map, or it is [4b7 injective].]\n\n[summary: A structure-preserving [3jy map] between two [481 fields] turns out either to be totally trivial (sending every element to $0$) or it preserves so much structure that its [3lh image] is an embedded copy of the [3js domain]. More succinctly, if $f$ is a field homomorphism, then $f$ is either the constant $0$ map, or it is [4b7 injective].]\n\nLet $F$ and $G$ be [481 fields], and let $f: F \\to G$ be a [-field_homomorphism]. Then one of the following is the case:\n\n - $f$ is the constant $0$ map: for every $x \\in F$, we have $f(x) = 0_G$.\n - $f$ is [4b7 injective].\n\n# Proof\n\nLet $f: F \\to G$ be non-constant.\nWe need to show that $f$ is injective; equivalently, for any pair $x,y$ of elements with $f(x) = f(y)$, we need to show that $x = y$.\n\nSuppose $f(x) = f(y)$.\nThen we have $f(x)-f(y) = 0_G$; so $f(x-y) = 0_G$ because $f$ is a field homomorphism and so respects the "subtraction" operation.\nHence in fact it is enough to show the following sub-result:\n\n> Suppose $f$ is non-constant. If $f(z) = 0_G$, then $z = 0_F$.\n\nOnce we have done this, we simply let $z = x-y$.\n\n## Proof of sub-result\n\nSuppose $f(z) = 0_G$ but that $z$ is not $0_F$, so we may find its multiplicative inverse $z^{-1}$.\n\nThen $f(z^{-1}) f(z) = f(z^{-1}) \\times 0_G = 0_G$; but $f$ is a homomorphism, so $f(z^{-1} \\times z) = 0_G$, and so $f(1_F) = 0_G$.\n\nBut this contradicts that the [-49z], because we may consider $f$ to be a [-47t] between the *multiplicative groups* $F \\setminus \\{ 0_F \\}$ and $G \\setminus \\{0_G\\}$, whereupon $1_F$ is the identity of $F \\setminus \\{0_F\\}$, and $1_G$ is the identity of $F \\setminus \\{0_G\\}$.\n\nOur assumption on $z$ was that $z \\not = 0_F$, so the contradiction means that if $f(z) = 0_G$ then $z = 0_F$.\nThis proves the sub-result and hence the main theorem.',
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