Galois Extensions

(a) An embedding $\sigma: F(\alpha) \to \overline{F}$ extending $\iota$ is determined by $\sigma(\alpha)$, since every element of $F(\alpha)$ is a polynomial in $\alpha$ with coefficients in $F$. Since $\sigma$ fixes $F$ and is a ring homomorphism, applying $\sigma$ to $m_\alpha(\alpha) = 0$ gives $m_\alpha(\sigma(\alpha)) = 0$ (coefficients of $m_\alpha$ lie in $F$, hence are fixed). So $\sigma(\alpha)$ must be a root of $m_\alpha$ in $\overline F$. Conversely, for any root $\beta \in \overline F$ of $m_\alpha$, the map $F[x]/(m_\alpha) \to \overline F$, $x \mapsto \beta$, is a well-defined field homomorphism (since $m_\alpha(\beta)=0$, this factors through the quotient by the ideal $(m_\alpha)$), and composing with the isomorphism $F(\alpha) \cong F[x]/(m_\alpha)$ gives an embedding sending $\alpha \mapsto \beta$. This is a bijection between embeddings and roots, proving (a).

(b) Induct on $[K:F]$. If $[K:F]=1$, $K=F$, $n=1=[K:F]$, trivial. Otherwise pick $\alpha \in K \setminus F$, so $F \subsetneq F(\alpha) \subseteq K$. By (a), $n(F(\alpha)/F) =$ (number of distinct roots of $m_\alpha$) $\le \deg m_\alpha = [F(\alpha):F]$, with equality iff $m_\alpha$ is separable (has no repeated root in $\overline F$; note $m_\alpha$ irreducible over $F$, and separability of $m_\alpha$ doesn't depend on which algebraically closed field we count roots in, since a repeated root of $m_\alpha$ is detected by $\gcd(m_\alpha,m_\alpha')\ne 1$ over $F$ itself). By induction on the tower $F(\alpha) \subseteq K$, $n(K/F(\alpha)) \le [K:F(\alpha)]$. Each of the $n(F(\alpha)/F)$ embeddings of $F(\alpha)$ extends to some embedding of $\overline F$ (since $\overline F$ is algebraically closed and $K/F(\alpha)$ is algebraic, an isomorphism-extension theorem guarantees this), and for each such extension the number of further extensions to $K$ is $n(K/F(\alpha))$ by definition, independent of the chosen extension because any two embeddings of $F(\alpha)$ into $\overline F$ over $F$ differ by an automorphism of $\overline F$ over $F$ (both being generated by roots of the same $m_\alpha$), and post-composing with such an automorphism permutes the extensions to $K$ bijectively. Hence

by the tower law. Equality holds iff both factors are equalities: $m_\alpha$ separable for every choice of generator $\alpha$ of a simple subextension in the chain, and by induction this happens for the whole tower iff every element of $K$ has a separable minimal polynomial over the relevant intermediate field. Chasing through the induction (or invoking that separability of $K/F$ means every $\alpha \in K$ has separable minimal polynomial over $F$, and using that minimal polynomials over intermediate fields divide those over $F$, hence share the separability property, since $F$-separability of $m_\alpha \in F[x]$ forces its irreducible factor over any extension field to remain separable) shows equality in the whole chain holds if and only if $K/F$ is separable. This proves (b).

(c) This is exactly the multiplicative bookkeeping described in the equality case of the argument in (b): every embedding of $E$ over $F$ extends to $\overline F$, and the extensions of that embedding to $K$ number $n(K/E)$ regardless of which extension to $\overline F$ was used, by the same automorphism-of-$\overline F$ argument. Summing over the $n(E/F)$ embeddings of $E$ gives $n(K/F) = n(K/E)\cdot n(E/F)$. $\blacksquare$

Proof of the Theorem (equivalence of (1)–(4)). We prove $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$, and separately $(1)\wedge(2) \Rightarrow (4) \Rightarrow (1)$, closing the cycle. Throughout, fix an algebraic closure $\overline F$ of $F$ containing $K$ (possible since $K/F$ is finite, hence algebraic).

$(1) \Rightarrow (2)$. Assume $K/F$ is normal and separable. By the Lemma (b), since $K/F$ is separable, $n(K/F) = [K:F]$, where $n(K/F)$ counts embeddings $K \hookrightarrow \overline F$ over $F$. We claim every such embedding is in fact an automorphism of $K$, i.e. its image is $K$ itself; this converts the embedding count into an automorphism count. Let $\sigma: K \to \overline F$ be an $F$-embedding. For any $\alpha \in K$ with minimal polynomial $m_\alpha \in F[x]$, the argument in Lemma (a) shows $\sigma(\alpha)$ is a root of $m_\alpha$ in $\overline F$. Since $K/F$ is normal, $m_\alpha$ splits completely in $K$ (this is the defining property of a normal extension applied to $\alpha \in K$: the minimal polynomial of every element of $K$ splits in $K$). Hence every root of $m_\alpha$ in $\overline F$ already lies in $K$, so $\sigma(\alpha) \in K$. As $\alpha$ was arbitrary, $\sigma(K) \subseteq K$. Since $\sigma$ is an injective $F$-linear map (it is a field homomorphism, automatically injective) on the finite-dimensional $F$-vector space $K$, and $\sigma(K) \subseteq K$, dimension-counting forces $\sigma(K) = K$: an injective linear self-map of a finite-dimensional vector space is surjective. Thus $\sigma$ restricts to an automorphism of $K$ fixing $F$, i.e. $\sigma \in \mathrm{Aut}(K/F)$. Conversely every element of $\mathrm{Aut}(K/F)$ is in particular an $F$-embedding $K \to \overline F$ (compose with inclusion $K \hookrightarrow \overline F$), and distinct automorphisms give distinct embeddings. So the embeddings $K \to \overline F$ over $F$ are in bijection with $\mathrm{Aut}(K/F)$, giving

$(2) \Rightarrow (3)$. Assume $|\mathrm{Aut}(K/F)| = [K:F]$. First we show $K/F$ is separable. By Lemma (b), $n(K/F) \le [K:F]$ always, with equality iff separable. But $|\mathrm{Aut}(K/F)| \le n(K/F)$ always (every automorphism of $K$ over $F$, followed by inclusion into $\overline F$, is one of the $n(K/F)$ embeddings, and distinct automorphisms give distinct embeddings — this uses no normality, just injectivity of the map "automorphism $\mapsto$ embedding"). Hence

forcing $n(K/F) = [K:F]$, so by Lemma (b), $K/F$ is separable. Next we show $K/F$ is normal, and identify a defining separable polynomial. Since $K/F$ is finite, write $K = F(\alpha_1,\dots,\alpha_r)$ (any finite extension is finitely generated by algebraic elements — e.g. an $F$-basis of $K$). Let $m_i \in F[x]$ be the minimal polynomial of $\alpha_i$, and set $f = m_1 m_2 \cdots m_r$ after removing duplicate irreducible factors so that $f$ has no repeated irreducible factors; note each $m_i$ is separable (as a factor of the polynomial defining a separable extension — indeed $\alpha_i \in K$ and $K/F$ separable means every element, in particular $\alpha_i$, has separable minimal polynomial) and distinct irreducible polynomials share no roots, so $f$ is a separable polynomial (no repeated roots in $\overline F$). We claim $K$ is the splitting field of $f$ over $F$. Since each $\alpha_i \in K$, and we must show all roots of each $m_i$ lie in $K$ (splitting), consider any automorphism $\sigma \in \mathrm{Aut}(K/F)$; it permutes the roots of $m_i$ that lie in $K$ among themselves (as in the proof of (a) $\Rightarrow$ (b), $\sigma(\alpha_i)$ is again a root of $m_i$, and $\sigma(\alpha_i) \in K$ automatically since $\sigma$ maps $K \to K$). We must further show all roots of $m_i$ (not just those already known in $K$) lie in $K$. Consider the orbit $O_i = \{\sigma(\alpha_i) : \sigma \in \mathrm{Aut}(K/F)\} \subseteq K$; this is a finite subset of roots of $m_i$ lying in $K$. We show $O_i$ is all the roots of $m_i$ in $\overline F$, by an embedding-counting argument: extend $F \subseteq F(\alpha_i)$ and use that $|\mathrm{Aut}(K/F)| = [K:F] = [K:F(\alpha_i)]\cdot[F(\alpha_i):F]$. Every root $\beta$ of $m_i$ in $\overline F$ gives an $F$-embedding $F(\alpha_i) \to \overline F$, $\alpha_i \mapsto \beta$ (Lemma (a)); since $K/F$ has $n(K/F) = [K:F]$ embeddings into $\overline F$ total, and by the multiplicativity of Lemma (c) these decompose as extensions of the $[F(\alpha_i):F]$-many embeddings of $F(\alpha_i)$ (one for each root $\beta$ of $m_i$, since $m_i$ separable means all $[F(\alpha_i):F] = \deg m_i$ roots are distinct and give distinct embeddings) — each extending in exactly $[K:F(\alpha_i)]$ ways. In particular for every root $\beta$ of $m_i$, there is at least one embedding $\tau: K \to \overline F$ over $F$ with $\tau(\alpha_i) = \beta$. But we showed above (in $(1)\Rightarrow(2)$'s argument, which only used normality — so we cannot yet invoke it circularly here) — instead, argue directly: $|\mathrm{Aut}(K/F)| = n(K/F)$ was already established in this direction (both equal $[K:F]$), meaning every $F$-embedding $K \to \overline F$ has image $K$ itself (else the images would be embeddings not coming from automorphisms, but we've shown the count of automorphisms already saturates the total embedding count $n(K/F)$, so no embedding can have image other than $K$, since automorphisms are automatically embeddings and there is no room for extra ones). Hence $\tau(K) = K$, so $\tau(\alpha_i) = \beta \in K$. Thus every root $\beta$ of $m_i$ lies in $K$: $m_i$ splits in $K$. As this holds for each $i=1,\dots,r$, $f = \prod m_i$ splits in $K$, and $K$ is generated over $F$ by the $\alpha_i$'s, which are among the roots of $f$; hence $K$ is generated over $F$ by roots of $f$ and $f$ splits in $K$, so $K$ is precisely the splitting field of $f$ over $F$. This proves (3), with $f$ separable as shown.

$(3) \Rightarrow (1)$. Assume $K$ is the splitting field over $F$ of a separable polynomial $f \in F[x]$. That $K/F$ is normal is a standard fact about splitting fields (a splitting field of any polynomial is a normal extension of the base field — every splitting field is normal because any $F$-embedding of $K$ into $\overline{F}$ must send roots of $f$ to roots of $f$, hence stays within the subfield generated by those roots, which is $K$ itself, forcing every embedding to be an automorphism; equivalently, minimal polynomials of elements of $K$, being factors of powers of $f$ up to the structure of $K$, split in $K$). We take this standard characterization of splitting fields as previously established under Normal Extensions and do not reprove it here. It remains to prove $K/F$ is separable, i.e. every $\alpha \in K$ has separable minimal polynomial over $F$. Since $K$ is generated over $F$ by the roots $\beta_1,\dots,\beta_n$ of $f$ (with $n = \deg f$, all distinct since $f$ is separable), it suffices to show that any subfield generated over $F$ by roots of a separable polynomial is separable over $F$; this again is a standard closure property of separable extensions (separability is preserved under composita and subextensions generated by separable elements), which we invoke from Separable Extensions. Hence $K/F$ is separable, completing $(1)$. This closes the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$.

$(2) \Rightarrow (4)$. Assume $|\mathrm{Aut}(K/F)| = [K:F]$; by the cycle just proved this is equivalent to (1), so $K/F$ is normal and separable. Let $F' = K^{\mathrm{Aut}(K/F)}$, the fixed field; clearly $F \subseteq F'$ since every $\sigma \in \mathrm{Aut}(K/F)$ fixes $F$ by definition of $\mathrm{Aut}(K/F)$. We must show $F' \subseteq F$, i.e. $F' = F$. Regard $K$ as an extension of the intermediate field $F'$; every automorphism in $\mathrm{Aut}(K/F)$ fixes $F'$ pointwise by definition of fixed field, so $\mathrm{Aut}(K/F) \subseteq \mathrm{Aut}(K/F')$. Also every automorphism fixing $F'$ certainly fixes the smaller field $F$, so $\mathrm{Aut}(K/F') \subseteq \mathrm{Aut}(K/F)$. Hence $\mathrm{Aut}(K/F') = \mathrm{Aut}(K/F)$, of size $[K:F]$. Now $K/F'$ is normal (since $K/F$ is normal: every $\alpha \in K$ has minimal polynomial over $F$ splitting in $K$; its minimal polynomial over the larger field $F'$ divides this splitting polynomial's factor, hence also splits in $K$ — a standard fact that normality persists over intermediate fields) and separable (a subextension of a separable extension is separable, since minimal polynomials over $F'$ divide those over $F$, and divisors of separable polynomials are separable). So by $(1)\Rightarrow(2)$ already proved, $|\mathrm{Aut}(K/F')| = [K:F']$. Combining with $|\mathrm{Aut}(K/F')| = |\mathrm{Aut}(K/F)| = [K:F]$ gives

Since $F \subseteq F' \subseteq K$ and $[K:F] = [K:F']\cdot[F':F]$ (tower law) with $[K:F']=[K:F]$ finite and nonzero, we get $[F':F] = 1$, i.e. $F' = F$. This proves (4).

$(4) \Rightarrow (1)$. Assume $K^{\mathrm{Aut}(K/F)} = F$. We show $K/F$ is separable, then normal. Separability: Let $\alpha \in K$ be arbitrary, and let $O = \{\sigma(\alpha) : \sigma \in \mathrm{Aut}(K/F)\}$ be its finite orbit under $\mathrm{Aut}(K/F)$ (finite because $\mathrm{Aut}(K/F)$ is a finite group — it injects into the set of embeddings $K \to \overline F$, which is finite of size $\le [K:F]$ by Lemma (b), as $K/F$ is finite). List the distinct elements of $O$ as $\beta_1=\alpha, \beta_2,\dots,\beta_s$, and form

Every $\sigma \in \mathrm{Aut}(K/F)$ permutes $\{\beta_1,\dots,\beta_s\}$ (since $\sigma$ maps the orbit of $\alpha$ to the orbit of $\sigma(\alpha)$, which is the same orbit, as $\mathrm{Aut}(K/F)$ is a group acting on itself's orbits — more precisely, composing $\sigma$ with any $\tau$ with $\tau(\alpha)=\beta_j$ gives $\sigma\tau(\alpha) = \sigma(\beta_j)$, still in the orbit of $\alpha$). Hence $\sigma$ permutes the roots $\beta_j$, so it permutes the linear factors of $g$, and therefore fixes each coefficient of $g$ (symmetric functions of the $\beta_j$ are permutation-invariant). This holds for every $\sigma \in \mathrm{Aut}(K/F)$, so every coefficient of $g$ lies in $K^{\mathrm{Aut}(K/F)} = F$ by hypothesis (4). Thus $g \in F[x]$, $g(\alpha) = 0$, and $g$ has distinct roots $\beta_1,\dots,\beta_s$ by construction (they were listed as distinct). Since the minimal polynomial $m_\alpha$ of $\alpha$ over $F$ divides $g$ (as $g \in F[x]$ vanishes at $\alpha$), and $g$ has no repeated roots, $m_\alpha$ has no repeated roots either. Hence $\alpha$ is separable over $F$. As $\alpha \in K$ was arbitrary, $K/F$ is separable. Normality: With the same notation, we show $m_\alpha = g$, which shows $m_\alpha$ splits in $K$ (its roots are the $\beta_j \in K$), giving normality since $\alpha$ was arbitrary. We know $m_\alpha \mid g$. Conversely, the identity map $\mathrm{id} \in \mathrm{Aut}(K/F)$ shows $\alpha = \beta_1$ is a root of $m_\alpha$; and for each $j$, since $\beta_j = \sigma(\alpha)$ for some $\sigma \in \mathrm{Aut}(K/F)$, applying $\sigma$ to $m_\alpha(\alpha)=0$ (with $m_\alpha \in F[x]$ fixed coefficient-wise by $\sigma$) gives $m_\alpha(\sigma(\alpha)) = m_\alpha(\beta_j) = 0$. So every $\beta_j$ is a root of $m_\alpha$, meaning $g = \prod(x-\beta_j)$ divides $m_\alpha$ (as $m_\alpha$ has all $s$ distinct roots $\beta_j$, each simple since... to be careful: $g \mid m_\alpha$ requires the $\beta_j$ to be roots of $m_\alpha$ with at least their multiplicity in $g$, which is $1$ each; since all $\beta_j$ are distinct roots of $m_\alpha$, $\prod(x-\beta_j) = g$ divides $m_\alpha$). Combined with $m_\alpha \mid g$ and both monic, $m_\alpha = g$. Since $g$ splits completely in $K$ by construction (its roots are $\beta_1,\dots,\beta_s \in K$), $m_\alpha$ splits in $K$. As $\alpha \in K$ was arbitrary, every element of $K$ has minimal polynomial splitting in $K$: $K/F$ is normal. This proves $(4) \Rightarrow (1)$, closing all implications and establishing the full equivalence of (1), (2), (3), (4). $\blacksquare$