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  "body": "\n## Proof\n\nWe first prove a lemma by induction on the degree. All fields are commutative, and homomorphisms are [[ring homomorphisms|0GGE0Z0JQ1]] that send $1$ to $1$.\n\nLemma. Let $E/F$ be a [[finite extension|0NRYPM9Y27]], let $M$ be a field and let $\\rho \\colon F \\to M$ be a homomorphism. Then at most $[E : F]$ homomorphisms $\\tau \\colon E \\to M$ satisfy $\\tau(c) = \\rho(c)$ for all $c \\in F$.\n\n1. Let $d = [E : F]$. If $d = 1$, then $E = F$, and the only such $\\tau$ is $\\rho$.\n2. Let $d \u003e 1$, and assume the lemma for all finite extensions of degree less than $d$. Choose $a \\in E$ with $a \\notin F$.\n3. The extension $E/F$ is finite, so $a$ is [[algebraic|0DFYTNZ6AM]] over $F$. Let $g = \\sum_k g_k x^k$ be its [[minimal polynomial|0GK1WB9R98]], of degree $r$. Then $[F(a) : F] = r$ by the [[degree of a simple extension|0N5QW9K7MH]].\n4. Since $a \\notin F$, we have $F(a) \\neq F$, so $r \\ge 2$. By the [[tower law|0A7KJ68XHT]], $[E : F(a)] = d/r \u003c d$.\n5. Let $g^\\rho = \\sum_k \\rho(g_k) x^k \\in M[x]$. It has degree $r$, because $\\rho(1) = 1$ and a homomorphism of fields is injective.\n6. Let $\\tau$ extend $\\rho$. Applying $\\tau$ to $g(a) = 0$ gives $g^\\rho(\\tau(a)) = 0$. So $\\tau(a)$ is one of the [[at most $r$ roots|0KJVQV3E0F]] of $g^\\rho$ in $M$.\n7. By the [[structure of simple extensions|0ZN39QDTP9]], every element of $F(a)$ is $h(a)$ for some $h \\in F[x]$. Then $\\tau(h(a)) = h^\\rho(\\tau(a))$.\n8. So the restriction of $\\tau$ to $F(a)$ is determined by $\\tau(a)$. Hence there are at most $r$ possible restrictions $\\sigma \\colon F(a) \\to M$.\n9. For each such $\\sigma$, the induction hypothesis for $E/F(a)$ and $\\sigma$ gives at most $[E : F(a)]$ homomorphisms $\\tau$ that extend $\\sigma$.\n10. Every $\\tau$ extending $\\rho$ extends its own restriction $\\sigma$. So there are at most $r \\cdot [E : F(a)] = [E : F]$ such $\\tau$, by the tower law. This proves the lemma.\n\nNow let $L/K$ be a finite extension.\n\n11. Every element $\\sigma$ of [[$\\operatorname{Aut}(L/K)$|0XVR72M51M]] is a homomorphism $L \\to L$ with $\\sigma(c) = c$ for $c \\in K$.\n12. Apply the lemma with $E = L$, $F = K$, $M = L$ and $\\rho$ the inclusion $K \\to L$. It gives at most $[L : K]$ such homomorphisms.\n13. So $\\operatorname{Aut}(L/K)$ is finite, and $\\abs{\\operatorname{Aut}(L/K)} \\le [L : K]$.\n\n## Idea\n\nAn automorphism fixing $K$ must send each generator to a root of its minimal polynomial, and such a polynomial has few roots. Adjoining one element at a time, the number of choices at each stage is at most the degree of that stage. The tower law multiplies these bounds into $[L : K]$.\n",
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