KindiakMath

Galois Par-Excellence

The goal of this post is to prove the fundamental theorem of Galois theory.

Let \mathbb K \supseteq \mathbb F denote a finite extension. Given any finite subgroup G \leq \mathrm{Aut}(\mathbb K), define the following sets:

\begin{aligned} \mathbb F_{\text{ext}} &:= \{ \mathbb L \subseteq \mathbb K : \mathbb L \supseteq \mathbb F\ \text{is a field extension}\}, \\ G_{\text{sub}} &:= \{H \subseteq G : H \leq G \}. \end{aligned}

Elements of \mathbb F_{\mathrm{ext}} are called intermediate extensions.

Theorem 1. Define the maps \Phi : \mathbb F_{\text{ext}} \to G_{\text{sub}} and \Psi : G_{\text{sub}} \to \mathbb F_{\text{ext}} by

\Phi(\mathbb L) := \mathrm{Aut}_{\mathbb L}(\mathbb K),\quad \Psi(H) := \mathbb K^H.

Then \Phi, \Psi are inverses of each other, so that

\mathbb K^{\mathrm{Aut}_{\mathbb L}(\mathbb K)} = \mathbb L, \quad \mathrm{Aut}_{\mathbb K^H}(\mathbb K) = H.

Before proving this theorem, we are going to need some intermediate results.

Lemma 1. Any finite subgroup \{\sigma_1,\dots, \sigma_n\} \leq \mathrm{Aut}(\mathbb K) is linearly independent over \mathbb F.

Proof. We proceed by induction. The set \{\sigma_1\} is trivially linearly independent over \mathbb F. Now consider the subgroup

\{\sigma_1,\dots,\sigma_k,\sigma_{k+1}\}

and the equation \sum_{i=1}^{k+1} c_i \sigma_i = 0. Since the subgroup is not a singleton, \sigma_1 \neq \sigma_2 trivially. Hence, there exists u \in \mathbb K such that \sigma_1(u) \neq \sigma_2(u). Defining \tau_i := \sigma_i(u) \sigma_i and \omega_i := \sigma_2(u) \sigma_i, we have

\displaystyle \sum_{i=1}^{k+1} c_i \tau_i = 0, \quad \sum_{i=1}^{k+1} c_i \omega_i = 0.

Since \tau_2 = \omega_2, by denoting K_j := \{1,\dots,k+1\}\backslash \{j\} so that |K_j| = k < k+1,

\begin{aligned} \sum_{i \in K_2} c_i (\sigma_i(u)-\sigma_2(u))\sigma_i &= \sum_{i \in K_2} c_i (\tau_i-\omega_i) \\ &= \sum_{i=1}^{k+1} c_i (\tau_i-\omega_i) \\ &= \sum_{i=1}^{k+1} c_i \tau_i - \sum_{i=1}^{k+1} c_i \omega_i = 0. \end{aligned}

By the induction hypothesis, c_i(\sigma_i(u)-\sigma_2(u)) = 0 for each i. In particular, for i = 1, since \sigma_1(u) \neq \sigma_2(u),

\displaystyle c_1(\sigma_1(u)-\sigma_2(u)) = 0 \quad \Rightarrow \quad c_1 = 0.

Therefore, \sum_{i=2}^{k+1} c_i \sigma_i = 0. By the induction hypothesis, c_i = 0 for i = 1,\dots,k+1, as required.

Lemma 2. For any finite subgroup G \leq \mathrm{Aut}(\mathbb K),

[\mathbb K : \mathbb K^G] = |G|,\quad \mathrm{Aut}_{\mathbb K^G}(\mathbb K) = G.

Proof. Define the collection

\mathcal G := \{ \sigma : \mathbb K \to \overline{\mathbb K^G} : \sigma\text{ is a }\mathbb K^G\text{-homomorphism} \}.

It is not hard to check that

G \subseteq \mathrm{Aut}_{\mathbb K^G}(\mathbb K) \subseteq \mathcal G \cap \mathrm{Aut}(\mathbb K) \subseteq \mathcal G.

Therefore,

|G| \leq |\mathcal G| = \{ \mathbb K : \mathbb K^G \} \leq [\mathbb K : \mathbb K^G].

On the other hand, we claim that |G| = [\mathbb K : \mathbb K^G]. Suppose for a contradiction that |G| < [\mathbb K : \mathbb K^G]. Write G = \{\sigma_1,\dots, \sigma_n\}, which is linearly independent over \mathbb F by Lemma 1. Therefore, \sum_{\sigma \in G} \sigma \neq 0, which means that there exists u \in \mathbb K such that \sum_{\sigma \in G} \sigma(u) \neq 0.

Furthermore, since [\mathbb K : \mathbb K^G] \geq n + 1, there exists a subset \{v_1,\dots, v_{n+1} \} \subseteq \mathbb K that is linearly independent over \mathbb K^G. For each j=1,\dots,n+1, define

\mathbf w_j := \begin{bmatrix} \sigma_1^{-1}(v_j) \\ \vdots \\ \sigma_n^{-1}(v_j) \end{bmatrix} \in \mathbb K^n.

The set \{\mathbf w_1,\dots, \mathbf w_n, \mathbf w_{n+1}\} \subseteq \mathbb K^n must be linearly dependent over \mathbb K, since it contains n+ 1 > n elements. Thus, there exist scalars c_j \in \mathbb K with c_1 \neq 0 without loss of generality such that

\displaystyle \sum_{j=1}^{n+1} c_j\begin{bmatrix} \sigma_1^{-1}(v_j) \\ \vdots \\ \sigma_n^{-1}(v_j) \end{bmatrix} = \mathbf 0.

By multiplying by u/c_1 \in \mathbb K, we can assume c_1 = u \neq 0 without loss of generality. For each i,

\displaystyle 0 = \sum_{j=1}^{n+1} c_j \sigma_i^{-1}(v_j) = \sigma_i^{-1} \left( \sum_{j=1}^{n+1} \sigma_i (c_j) v_j \right),

so that \sum_{j=1}^{n+1} \sigma_i(c_j) v_j = 0. In particular,

\displaystyle \begin{aligned} \sum_{j=1}^{n+1} \left( \sum_{\sigma \in G} \sigma(c_j)\right) v_j &= \sum_{j=1}^{n+1} \left( \sum_{i=1}^n \sigma_i(c_j)\right) v_j \\ &= \sum_{i=1}^n \left( \sum_{j=1}^{n+1} \sigma_i(c_j) v_j \right) = 0. \end{aligned}

Now observe that for any t, \{ \sigma_t \circ \sigma_1,\dots, \sigma_t \circ \sigma_n \} = G. Therefore,

\begin{aligned} \sigma_t \left( \sum_{\sigma \in G} \sigma(c_j) \right) &= \sigma_t \left( \sum_{i=1}^n \sigma_i(c_j) \right) \\ &= \sum_{i=1}^n (\sigma_t \circ \sigma_i)(c_j) \\ &= \sum_{\sigma \in G} \sigma(c_j) = \sum_{i = 1}^n \sigma_i(c_j). \end{aligned}

Therefore, \sum_{\sigma \in G} \sigma(c_j) \in \mathbb K^G for each j. By linear independence, \sum_{\sigma \in G} \sigma(c_j) = 0 for each j. In particular, for j = 1,

\displaystyle 0 \neq \sum_{\sigma \in G} \sigma(u)= \sum_{\sigma \in G} \sigma(c_1) = 0,

a contradiction.

Therefore, |G| = [\mathbb K : \mathbb K^G], which implies that

\mathcal G = G \subseteq \mathrm{Aut}_{\mathbb K^G}(\mathbb K) \subseteq \mathcal G.

Therefore, \displaystyle \mathrm{Aut}_{\mathbb K^G}(\mathbb K) = G, as required.

Remark 1. Upon proving Lemma 2, we can conclude that a field extension is Galois if and only if it is normal and separable.

Lemma 3. We have |\mathrm{Aut}_{\mathbb F}(\mathbb K)| \leq [\mathbb K : \mathbb F] with equality if and only if \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} = \mathbb F.

Proof. By the usual tower rule,

\begin{aligned} [\mathbb K : \mathbb F] &= [\mathbb K : \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)}] \cdot [\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F] \\ &= |\mathrm{Aut}_{\mathbb F}(\mathbb K)| \cdot [\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F] \\ &\geq |\mathrm{Aut}_{\mathbb F}(\mathbb K)|. \end{aligned}

In particular,

\begin{aligned} |\mathrm{Aut}_{\mathbb F}(\mathbb K)| = [\mathbb K : \mathbb F] \quad &\iff \quad [\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F] = 1 \\ &\iff \quad \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} = \mathbb F. \end{aligned}

We are now ready to prove Theorem 1.

Theorem 1 (Restated). Define the maps \Phi : \mathbb F_{\text{ext}} \to G_{\text{sub}} and \Psi : G_{\text{sub}} \to \mathbb F_{\text{ext}} by

\Phi(\mathbb L) := \mathrm{Aut}_{\mathbb L}(\mathbb K),\quad \Psi(H) := \mathbb K^H.

Then \Phi, \Psi are inverses of each other, so that

\mathbb K^{\mathrm{Aut}_{\mathbb L}(\mathbb K)} = \mathbb L, \quad \mathrm{Aut}_{\mathbb K^H}(\mathbb K) = H.

Elements of G_{\mathrm{sub}} are called intermediate fields.

Proof of Theorem 1. By Lemma 2, for any H \in G_{\mathrm{sub}},

\begin{aligned} (\Phi \circ \Psi)(H) &= \Phi(\Psi(H)) \\ &= \Phi(\mathbb K^H) \\ &= \mathrm{Aut}_{\mathbb K^H}(\mathbb K) \\ &= H. \end{aligned}

For the second claim, it suffices to check that |\mathrm{Aut}_{\mathbb L}(\mathbb K)| = [\mathbb K : \mathbb L] by Lemma 3, so that

\begin{aligned} (\Psi \circ \Phi)(\mathbb L) &= \Psi(\Phi(\mathbb L)) \\ &= \Psi(\mathrm{Aut}_{\mathbb L}(\mathbb K)) \\ &= \mathbb K^{\mathrm{Aut}_{\mathbb L}(\mathbb K)} \\ &= \mathbb L. \end{aligned}

To that end, define the collection

\mathcal S := \{\sigma|_{\mathbb L} : \sigma \in \mathrm{Aut}_{\mathbb F}(\mathbb K)\} \subseteq \mathrm{Aut}_{\mathbb F}(\mathbb L).

For any \phi \in \mathrm{Aut}_{\mathbb F}(\mathbb K),

\begin{aligned} \phi|_{\mathbb L} = \sigma|_{\mathbb L}\quad &\iff \quad (\phi^{-1} \circ \sigma)|_{\mathbb L} = \mathrm{id}_{\mathbb L}\\ &\iff \quad \phi^{-1} \circ \sigma \in \mathrm{Aut}_{\mathbb L}(\mathbb K) \\ &\iff \quad \phi \mathrm{Aut}_{\mathbb L}(\mathbb K) = \sigma \mathrm{Aut}_{\mathbb L}(\mathbb K) .\end{aligned}

Clearly, |\mathcal S| \leq |\mathrm{Aut}_{\mathbb F}(\mathbb L)| = \{\mathbb L : \mathbb F\}, so that

\displaystyle \{\mathbb L : \mathbb F\} \geq |\mathcal S| = |\mathrm{Aut}_{\mathbb F}(\mathbb K) / \mathrm{Aut}_{\mathbb L}(\mathbb K)| = \frac{|\mathrm{Aut}_{\mathbb F}(\mathbb K)|}{|\mathrm{Aut}_{\mathbb L}(\mathbb K)| }.

By the tower rule, and Lemma 3 again,

\begin{aligned} [ \mathbb K : \mathbb F ] &= [\mathbb K : \mathbb L] \cdot [\mathbb L : \mathbb F] \\ &\geq [\mathbb K : \mathbb L] \cdot \{\mathbb L : \mathbb F\} \\ &\geq |\mathrm{Aut}_{\mathbb L}(\mathbb K)| \cdot \{\mathbb L : \mathbb F\} \\ &\geq |\mathrm{Aut}_{\mathbb L}(\mathbb K)| \cdot \frac{|\mathrm{Aut}_{\mathbb F}(\mathbb K)|}{|\mathrm{Aut}_{\mathbb L}(\mathbb K)| } \\ &= |\mathrm{Aut}_{\mathbb F}(\mathbb K)| \\ &= [\mathbb K: \mathbb F].\end{aligned}

Therefore, we must have |\mathrm{Aut}_{\mathbb L}(\mathbb K)| = [\mathbb K : \mathbb L], as required.

Definition 1. We say that a finite extension \mathbb K \supseteq \mathbb F is Galois if there exists a (unique) subgroup G \leq \mathrm{Aut}(\mathbb K) such that \mathbb K^G = \mathbb L. In this case, we call \mathrm{Gal}(\mathbb K / \mathbb F) := G the Galois group of \mathbb K over \mathbb F. In particular,

\mathrm{Gal}(\mathbb K / \mathbb F) = \mathrm{Aut}_{\mathbb F}(\mathbb K),

where the latter is defined even if \mathbb K \supseteq \mathbb F is not Galois.

Suppose \mathbb K \supseteq \mathbb F is Galois.

Corollary 1. Given intermediate fields \mathbb L_1 \subseteq \mathbb L_2,

\displaystyle [\mathbb L_2 : \mathbb L_1] = \frac{|\mathrm{Aut}_{\mathbb L_2}(\mathbb K)|}{|\mathrm{Aut}_{\mathbb L_1}(\mathbb K)|}.

Proof. By Theorem 1, |\mathrm{Aut}_{\mathbb L_1}(\mathbb K)| = [\mathbb K : \mathbb L_1]. Therefore,

\displaystyle \frac{|\mathrm{Aut}_{\mathbb L_2}(\mathbb K)|}{|\mathrm{Aut}_{\mathbb L_1}(\mathbb K)|} = \frac{[\mathbb K : \mathbb L_2]}{[\mathbb K : \mathbb L_1]} = [\mathbb L_2 : \mathbb L_1].

Corollary 2. Suppose \mathbb K \supseteq \mathbb F is Galois. For any intermediate field \mathbb L, \mathbb K \supseteq \mathbb L is Galois by Theorem 1. Then \mathbb L \supseteq \mathbb F is Galois if and only if \mathrm{Gal}(\mathbb L/\mathbb F) \triangleleft \mathrm{Gal}(\mathbb K / \mathbb F).

Proof. Recall that \mathcal S \subseteq \mathrm{Aut}_{\mathbb F}(\mathbb L) and |\mathcal S| \geq [\mathbb L : \mathbb F]. It is not hard to check that

\begin{aligned} \mathrm{Gal}(\mathbb L/\mathbb F) \triangleleft \mathrm{Gal}(\mathbb K / \mathbb F) \quad &\iff \quad \mathcal S \subseteq \mathrm{Aut}_{\mathbb L}(\mathbb K). \end{aligned}

Furthermore,

\mathcal S \subseteq \mathrm{Aut}_{\mathbb L}(\mathbb K) \quad \iff \quad [\mathbb L:\mathbb F] = |\mathrm{Aut}_{\mathbb F}(\mathbb L)|,

so that \mathbb L \supseteq \mathbb F is Galois.

Corollary 3. Under the assumption of Corollary 2,

\mathrm{Gal}(\mathbb K/\mathbb F)/\mathrm{Gal}(\mathbb K/\mathbb L) \cong \mathrm{Gal}(\mathbb L/\mathbb F).

That is, we obtain some kind of third isomorphism theorem for Galois groups.

Proof. Check that the map \sigma \mapsto \sigma|_{\mathbb L} is a surjective group homomorphism with kernel \mathrm{Gal}(\mathbb K/\mathbb L), then invoke the first isomorphism theorem for groups.

The fundamental theorem of Galois theory is lovely, because it tells us that the intermediate fields in the finite extension \mathbb K \supseteq \mathbb F are in direct correspondence with the subgroups of automorphisms \mathrm{Aut}(\mathbb K). Therefore, we can translate our rather complicated ideas in field theory into relatively simple ones in group theory. Paradoxically, this connection was the historical motivation to even study group theory to begin with.

Next time, we use these mathematical mountaineering tools to begin our ascent in proving that there is no quintic formula, beginning with the group-theoretic connections with Galois extensions.

—Joel KIndiak, 24 Apr 26, 1435H

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