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  "body": "\n## Proof\n\nWe count the homomorphisms $L \\to L$ that fix $K$, adjoining one generator at a time. Homomorphisms are [[ring homomorphisms|0GGE0Z0JQ1]] that send $1$ to $1$.\n\n1. Let $L/K$ be a finite [[Galois extension|0DS2F7GK05]], so $L/K$ is [[normal|0CBRFQ7DVZ]] and [[separable|0N019F6RE9]]. Let $n = [L : K]$.\n\nLemma. Let $K \\subseteq E \\subseteq L$ be a field, let $a \\in L$ and $E' = E(a)$. Let $\\rho \\colon E \\to L$ be a homomorphism with $\\rho(c) = c$ for $c \\in K$. Then exactly $[E' : E]$ homomorphisms $\\tau \\colon E' \\to L$ restrict to $\\rho$ on $E$.\n\n2. Let $f$ be the [[minimal polynomial|0GK1WB9R98]] of $a$ over $K$, of degree $N$. Let $m$ be the minimal polynomial of $a$ over $E$, of degree $d$. Then $d = [E' : E]$ by the [[degree of a simple extension|0N5QW9K7MH]].\n3. Since $f \\in E[x]$ and $f(a) = 0$, the polynomial $m$ [[divides|06K7SARZKR]] $f$. Write $f = m h$ with $h \\in E[x]$ of degree $N - d$.\n4. The polynomial $f$ is [[irreducible|0NFEG48RDR]] over $K$ and has the root $a$ in $L$. As $L/K$ is normal, $f = (x - b_1) \\cdots (x - b_N)$ with $b_1, \\dots, b_N \\in L$.\n5. As $a$ is separable over $K$, $f$ is a [[separable polynomial|03J8BS7FMV]]: $u f + v f' = 1$ for some $u, v \\in K[x]$.\n6. Suppose $b_i = b_j = b$ for some $i \\neq j$. Then $(x - b)^2$ divides $f$, so $b$ has [[multiplicity|03XVP8Q4GY]] at least $2$. So [[$f(b) = f'(b) = 0$|067F8YVYMH]], and evaluating step 5 at $b$ gives $1 = 0$. So $b_1, \\dots, b_N$ are distinct.\n7. For $g \\in E[x]$, let $g^\\rho \\in L[x]$ be obtained by applying $\\rho$ to the coefficients. The map $g \\mapsto g^\\rho$ respects sums and products. Since $\\rho$ is [[injective|09N7DZ7SDB]], it preserves degrees.\n8. Since $\\rho$ fixes the coefficients of $f$, step 3 gives $f = m^\\rho h^\\rho$ in $L[x]$, with $\\deg m^\\rho = d$ and $\\deg h^\\rho = N - d$.\n9. Each $b_i$ is a root of $m^\\rho$ or of $h^\\rho$, since $L$ is a field. The polynomial $h^\\rho$ has [[at most $N - d$ roots|0KJVQV3E0F]] in $L$.\n10. So at least $d$ of the distinct $b_i$ are roots of $m^\\rho$. As $m^\\rho$ has at most $d$ roots, it has exactly $d$ roots in $L$.\n11. Let $\\tau$ extend $\\rho$. Applying $\\tau$ to $m(a) = 0$ gives $m^\\rho(\\tau(a)) = 0$. So $\\tau(a)$ is one of the $d$ roots of $m^\\rho$.\n12. By the [[structure of simple extensions|0ZN39QDTP9]], every element of $E'$ is $g(a)$ for some $g \\in E[x]$. Then $\\tau(g(a)) = g^\\rho(\\tau(a))$, so $\\tau$ is determined by $\\tau(a)$.\n13. Let $b$ be a root of $m^\\rho$ in $L$. Define $\\tau(g(a)) = g^\\rho(b)$ for $g \\in E[x]$. If $g(a) = g_1(a)$, then $m$ divides $g - g_1$, so $m^\\rho$ divides $g^\\rho - g_1^\\rho$, and $g^\\rho(b) = g_1^\\rho(b)$.\n14. So $\\tau$ is well defined. It respects sums and products by step 7, it extends $\\rho$, and $\\tau(a) = b$.\n15. By steps 10 to 14, the extensions $\\tau$ correspond to the $d$ roots of $m^\\rho$. So there are exactly $d = [E' : E]$ of them. This proves the lemma.\n\nNow we count.\n\n16. Let $a_1, \\dots, a_s$ be a basis of $L$ over $K$. Then [[$L = K(a_1, \\dots, a_s)$|0MA5B744K3]]. Let $E_0 = K$ and $E_j = E_{j-1}(a_j)$, so $E_j = K(a_1, \\dots, a_j)$ and $E_s = L$.\n17. Claim: exactly $[E_j : K]$ homomorphisms $E_j \\to L$ fix $K$. For $j = 0$, the only one is the inclusion of $K$.\n18. Let $j \\ge 1$. Each such homomorphism on $E_j$ restricts to one on $E_{j-1}$. By the claim for $j - 1$ there are $[E_{j-1} : K]$ restrictions, and each has exactly $[E_j : E_{j-1}]$ extensions by the lemma.\n19. So there are $[E_j : E_{j-1}] \\cdot [E_{j-1} : K] = [E_j : K]$ of them, by the [[tower law|0A7KJ68XHT]]. With $j = s$, exactly $n$ homomorphisms $\\sigma \\colon L \\to L$ fix $K$.\n20. Each such $\\sigma$ is injective, and it is $K$-linear because $\\sigma(c y) = c \\sigma(y)$ for $c \\in K$. As $L$ has finite dimension $n$ over $K$, $\\sigma$ is [[surjective|0912EEEVCT]].\n21. So these $n$ maps are exactly the automorphisms of $L$ that fix $K$, that is, the elements of the [[Galois group|0XVR72M51M]]. Hence $\\abs{\\operatorname{Gal}(L/K)} = [L : K]$.\n\n## Idea\n\nAn automorphism is built by choosing the image of one generator at a time. At each stage the image must be a root of a minimal polynomial. Normality puts all these roots in $L$, and separability makes them distinct, so the number of choices is exactly the degree of that stage. The tower law multiplies these counts into $[L : K]$.\n",
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