Galois Theory — Fundamentals

Identity: $\mathrm{id}_K$ fixes $F$ pointwise. Closure: if $\sigma,\tau$ both fix $F$ pointwise, so does $\sigma\circ\tau$: $(\sigma\tau)(a)=\sigma(\tau(a))=\sigma(a)=a$ for $a\in F$. Inverses: if $\sigma$ fixes $a\in F$, applying $\sigma^{-1}$ to $\sigma(a)=a$ gives $a=\sigma^{-1}(a)$, so $\sigma^{-1}$ also fixes $F$ pointwise. Associativity is inherited from function composition.

Write $f=\sum a_ix^i$ ($a_i\in F$). Since $\sigma$ is a ring homomorphism fixing $F$: $f(\sigma(\alpha)) = \sum a_i\sigma(\alpha)^i = \sum \sigma(a_i)\sigma(\alpha)^i = \sigma\Big(\sum a_i\alpha^i\Big) = \sigma(f(\alpha)) = \sigma(0)=0$.

By Theorem 20.2, $\sigma$ permutes $\{\alpha_1,\ldots,\alpha_n\}$ among themselves (a root maps to a root of the same polynomial, and the roots are exactly this finite set). Since $K=F(\alpha_1,\ldots,\alpha_n)$, every element of $K$ is an $F$-polynomial expression in the $\alpha_i$'s; since $\sigma$ fixes $F$ and is determined on each $\alpha_i$, it is determined on every such expression, hence on all of $K$. The map $\sigma \mapsto (\text{induced permutation of the }\alpha_i)$ is therefore an injective function into $S_n$, and it's a homomorphism since composing automorphisms composes the induced permutations.

The induction has to be run on a stronger statement than the theorem itself, in the same shape as Theorem 19.2 — counting extensions of an arbitrary base isomorphism, not just automorphisms over a fixed base field. (Counting only automorphisms would leave no way to handle the intermediate stages, where $\alpha$ gets sent to a different root and the base map is no longer the identity.)

Claim. Let $\sigma: F\to F'$ be a field isomorphism, $f\in F[x]$ separable, $K$ a splitting field of $f$ over $F$, and $K'$ a splitting field of $\sigma(f)$ over $F'$. Then the number of isomorphisms $K\to K'$ extending $\sigma$ is exactly $[K:F]$.

Theorem 20.1 is the special case $F'=F$, $\sigma=\mathrm{id}_F$, $K'=K$: an isomorphism $K\to K$ extending $\mathrm{id}_F$ is precisely an element of $\mathrm{Gal}(K/F)$.

Proof of the Claim, by induction on $[K:F]$.

If $[K:F]=1$, then $K=F$, so $f$ splits into linear factors over $F$; applying $\sigma$ coefficientwise to that factorization shows $\sigma(f)$ splits over $F'$, so $K'=F'$ and $\sigma$ itself is the only extension: exactly $1=[K:F]$.

If $[K:F]>1$, pick a root $\alpha\in K\setminus F$ of $f$, with minimal polynomial $p=m_\alpha$ over $F$ of degree $d=[F(\alpha):F]>1$ (Theorem 18.4). Then $p$ is an irreducible factor of $f$, so $\sigma(p)$ is an irreducible factor of $\sigma(f)$ (irreducibility transfers across $\sigma$, as in Theorem 19.2's proof), and $\sigma(p)$ has $d$ distinct roots in $K'$ — distinct because $f$ is separable and $\sigma$ carries $f$'s factorization over to $\sigma(f)$'s, preserving the absence of repeated roots.

Counting the possible images of $\alpha$. Any extension $\tau: K\to K'$ of $\sigma$ must send $\alpha$ to a root of $\sigma(p)$ in $K'$ (apply $\tau$ to $p(\alpha)=0$ coefficientwise, exactly as in Theorem 20.2), so there are at most $d$ choices for $\tau(\alpha)$. Conversely, for each of the $d$ roots $\alpha'$ of $\sigma(p)$ in $K'$, Theorem 18.4's identifications $F(\alpha)\cong F[x]/(p)$ and $F'(\alpha')\cong F'[x]/(\sigma(p))$ compose with $\sigma$ to give exactly one isomorphism $\sigma_{\alpha'}: F(\alpha)\to F'(\alpha')$ extending $\sigma$ with $\alpha\mapsto\alpha'$ — unique because the basis $\{1,\alpha,\ldots,\alpha^{d-1}\}$ means the image of $\alpha$ determines the map completely.

Counting the extensions of each $\sigma_{\alpha'}$. Fix such an $\alpha'$. Then $K$ is a splitting field of $f$ over $F(\alpha)$ (adjoining $f$'s roots to the bigger base still gives $K$), and $K'$ is a splitting field of $\sigma_{\alpha'}(f)=\sigma(f)$ over $F'(\alpha')$, with $[K:F(\alpha)] = [K:F]/d < [K:F]$ (Tower Law). The induction hypothesis, applied to the base isomorphism $\sigma_{\alpha'}$, gives exactly $[K:F(\alpha)]$ isomorphisms $K\to K'$ extending $\sigma_{\alpha'}$.

Combining. Every extension $\tau$ of $\sigma$ restricts on $F(\alpha)$ to $\sigma_{\alpha'}$ for exactly one root $\alpha'=\tau(\alpha)$, so the extensions of $\sigma$ are partitioned according to that choice — $d$ classes, each of size $[K:F(\alpha)]$:

by the Tower Law. This proves the Claim, and with it $\lvert\mathrm{Gal}(K/F)\rvert = [K:F]$.

$F\subseteq K^H$: every $\sigma\in H\subseteq\mathrm{Gal}(K/F)$ fixes $F$ pointwise by definition, so $F\subseteq K^H$. Field: if $a,b\in K^H$, then for any $\sigma\in H$, $\sigma(a-b)=\sigma(a)-\sigma(b)=a-b$ (ring homomorphism) so $a-b\in K^H$; similarly $\sigma(ab)=\sigma(a)\sigma(b)=ab$, so $ab\in K^H$; and if $a\neq0$, $\sigma(a^{-1})=\sigma(a)^{-1}=a^{-1}$ (Theorem 13.2), so $a^{-1}\in K^H$. By the subfield analog of Theorem 11.6, $K^H$ is a field.

$\sigma(E)=E$ for every $\sigma\in G:=\mathrm{Gal}(K/F)$. Since $E=F(\beta_1,\ldots,\beta_m)$ for the roots $\beta_i$ of $g$, and $\sigma$ permutes these roots among themselves (Theorem 20.2), $\sigma$ maps $E$ (generated by the $\beta_i$ over $F$, with $\sigma$ fixing $F$) onto itself.

Normality. For $\sigma\in G$, $\tau\in\mathrm{Gal}(K/E)$, and $e\in E$: since $\sigma(E)=E$, $\sigma^{-1}(e)\in E$, so $\tau(\sigma^{-1}(e))=\sigma^{-1}(e)$ (as $\tau$ fixes $E$ pointwise). Then $(\sigma\tau\sigma^{-1})(e) = \sigma(\sigma^{-1}(e)) = e$. So $\sigma\tau\sigma^{-1}$ fixes $E$ pointwise for every $e\in E$, i.e. $\sigma\tau\sigma^{-1}\in\mathrm{Gal}(K/E)$. Hence $\mathrm{Gal}(K/E)\trianglelefteq G$.

The quotient isomorphism. Since $\sigma(E)=E$ for every $\sigma\in G$, define $\rho: G\to\mathrm{Gal}(E/F)$ by $\rho(\sigma) = \sigma|_E$ (the restriction, a genuine automorphism of $E$ fixing $F$, by the previous paragraph). $\rho$ is a homomorphism: $\rho(\sigma_1\sigma_2) = (\sigma_1\sigma_2)|_E = \sigma_1|_E\circ\sigma_2|_E = \rho(\sigma_1)\rho(\sigma_2)$.

$\ker\rho = \{\sigma\in G : \sigma|_E=\mathrm{id}_E\} = \mathrm{Gal}(K/E)$, exactly.

Surjective: given any $\tau_0\in\mathrm{Gal}(E/F)$, apply Theorem 19.2 (Isomorphism Extension Theorem) with the base isomorphism $\tau_0: E\to E$ and the polynomial $f$ — the separable polynomial whose splitting field over $F$ is $K$ (not $g$: since $E$ is already the splitting field of $g$ over $F$, $g$ splits completely inside $E$, so the splitting field of $g$ over $E$ is $E$ itself, not $K$). Two facts make Theorem 19.2 applicable: $K$ is the splitting field of $f$ over $E$ as well as over $F$ (adjoining $f$'s roots to the larger base $E$ still produces exactly $K$), and $\tau_0(f)=f$, since $f$'s coefficients lie in $F$ and $\tau_0$ fixes $F$ pointwise. So $\tau_0$ extends to an isomorphism $\tau: K\to K$, i.e. an automorphism of $K$. It fixes $F$ pointwise because it restricts to $\tau_0$ on $E\supseteq F$ and $\tau_0$ fixes $F$. Hence $\tau\in G$ and $\rho(\tau)=\tau|_E=\tau_0$.

By the First Isomorphism Theorem (Theorem 6.5): $G/\ker\rho \cong \mathrm{im}\,\rho$, i.e. $G/\mathrm{Gal}(K/E) \cong \mathrm{Gal}(E/F)$.