Galois Extensions

We have almost reached our most important definition in abstract algebra. But to get there, we still need to discuss some polynomial-based observations.

Let $\mathbb F$ denote a field with algebraic closure $\bar{\mathbb F}$ .

Definition 1. Given $f \in \mathbb F[x] \subseteq \bar{\mathbb F}[x]$ , define the set of zeroes of $f$ by

$\mathcal Z_f := \{ \alpha \in \bar{\mathbb F} : f(\alpha) = 0 \}.$

Note that $\mathcal Z_f = \{\alpha_1,\dots,\alpha_n\}$ is finite. We call

$\mathbb F(\mathcal Z_f) \equiv \mathbb F(\alpha_1,\dots,\alpha_n)$

the splitting field of $f$ over $\mathbb F$ .

Of course, this definition generalises to families of functions rather naturally.

Definition 2. Given $\mathcal F \subseteq \mathbb F[x] \subseteq \bar{\mathbb F}[x]$ , define the set of zeroes of $\mathcal F$ by

$\displaystyle \mathcal Z_{\mathcal F} := \bigcup_{f \in \mathcal F} \mathcal Z_f.$

Recall that $\mathbb F(\mathcal Z_{\mathcal F})$ denotes the smallest field that contains $\mathbb F$ and $\mathcal Z_{\mathcal F}$ . We call $\mathbb F(\mathcal Z_{\mathcal F})$ the splitting field of $\mathcal F$ over $\mathbb F$ . Observe that $\mathbb F(\mathcal Z_f) = \mathbb F(\mathcal Z_{\{f\}})$ so that Definition 1 can be thought of as subsumed under Definition 2.

We say that a field $\mathbb K \supseteq \mathbb F$ is a splitting field over $\mathbb F$ if there exists some $\mathcal F \subseteq \mathbb F[x]$ such that $\mathbb K = \mathbb F(\mathcal Z_{\mathcal F})$ .

Example 1. The splitting field of $x^4 - 1$ over $\mathbb Q$ is $\mathbb Q(i)$ , since

$x^4 - 1 = (x-1)(x+1)(x^2+1)$

has the roots $1,-1,i,-i$ . Writing as a product of linear factors,

$x^4 - 1 = (x-1)(x+1)(x-i)(x+i).$

Hence, we can conceive the splitting field of $\mathcal F$ as the “smallest” field extension that includes all roots of the functions in $\mathcal F$ .

Now let $\mathbb K \supseteq \mathbb F$ denote any finite extension.

Definition 2. We say that the extension $\mathbb K \supseteq \mathbb F$ is normal if for any $u \in \mathbb K$ ,

$\mathbb F(\mathcal Z_{f_{\mathbb F}^u}) \subseteq \mathbb K.$

Lemma 1. The following are equivalent.

$\mathbb K \supseteq \mathbb F$ is normal.
There exists some nonzero $f \in \mathbb F[x]$ such that $\mathbb K = \mathbb F(\mathcal Z_f)$ .
For any $\mathbb F$ -homomorphism $\sigma : \mathbb K \to \bar{\mathbb F}$ , $\sigma(\mathbb K) = \mathbb K$ .

Proof. We will prove in a cycle.

Suppose $\mathbb K \supseteq \mathbb F$ is normal. Since the extension is finite, write $\mathbb K = \mathbb F(\alpha_1,\dots,\alpha_n)$ for some $\alpha_i \in \mathbb K$ . By Definition 2,

$\mathbb F(\mathcal Z_{f_{\mathbb F}^{\alpha_i}}) \subseteq \mathbb K$

for each $i$ . Define the polynomial $f := \prod_{i=1}^n f_{\mathbb F}^{\alpha_i} \in \mathbb F[x] \backslash \{0\}$ , whose zeroes are characterised by

$\displaystyle \mathcal Z_f = \bigcup_{i=1}^n \mathcal Z_{f_{\mathbb F}^{\alpha_i}} \subseteq \bigcup_{i=1}^n \mathbb F(\mathcal Z_{f_{\mathbb F}^{\alpha_i}}) \subseteq \bigcup_{i=1}^n \mathbb K = \mathbb K.$

On the other hand, since each $\alpha_i \in \mathcal Z_f$ , we have

$\mathbb K = \mathbb F(\alpha_1,\dots,\alpha_n) \subseteq \mathbb F (\mathcal Z_f) \subseteq \mathbb K.$

Therefore, $\mathbb K = \mathbb F(\mathcal Z_f)$ , as required. That is, $\mathbb K$ is a splitting field over $\mathbb F$ .

Now suppose there exists some $f \in \mathbb F[x] \backslash \{0\}$ such that $\mathbb K = \mathbb F(\mathcal Z_f)$ . Then for any $\alpha \in \mathcal Z_f$ ,

$f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0,$

$\displaystyle \sigma(\mathbb K) = \sigma(\mathbb F(\mathcal Z_f)) = \mathbb F(\sigma(\mathcal Z_f)) = \mathbb F(\mathcal Z_f) = \mathbb K.$

Finally, suppose for any $\mathbb F$ -homomorphism $\sigma : \mathbb K \to \bar{\mathbb F}$ , $\sigma(\mathbb K) = \mathbb K$ . Fix $u \in \mathbb K$ . We need to show that $\mathbb F(\mathcal Z_{f_{\mathbb F}^u}) \subseteq \mathbb K$ . It suffices to check that for any conjugate $v \in \bar{\mathbb F}$ of $u$ , $v \in \mathbb K$ . Since $u,v$ are conjugates, there exists an $\mathbb F$ -homomorphism

$\sigma : \mathbb F(u) \to \mathbb F(v), \quad \sigma(u) = v.$

Thus, there exists an extension $\tilde{\sigma} : \mathbb K \to \bar{\mathbb F}$ of $\sigma$ which is itself an $\mathbb F$ -homomorphism. By hypothesis,

$v = \sigma(u) \in \sigma(\mathbb K) = \mathbb K,$

as required.

Lemma 2. For any finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ , $[\mathbb K : \mathbb K^G] = |G|$ .

Proof. Delayed.

Theorem 1. There exists a finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ such that $\mathbb K^G = \mathbb F$ if and only if $\mathbb K \supseteq \mathbb F$ is normal and separable.

Remark 1. Recall that $\mathbb K^G$ denotes the fixed field

$\mathbb K^G := \{u \in \mathbb K : \sigma(u) = u, \sigma \in G\} \subseteq \mathbb K.$

Proof. As a finite extension, write $\mathbb K = \mathbb F(\alpha_1,\dots,\alpha_n)$ for distinct $\alpha_i \in \mathbb K$ .

$(\Rightarrow)$ Suppose $\mathbb K^G = \mathbb F$ for some finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ . Define the finite subset $\mathcal Z \subseteq \mathbb K$ and the polynomial $f \in \mathbb K[x]$ by

$\displaystyle \mathcal Z := \bigcup_{i=1}^n \bigcup_{\sigma \in G} \{ \sigma(\alpha_i) \},\quad f(x) = \prod_{ \alpha \in \mathcal Z} (x-\alpha).$

Clearly, $\mathbb K = \mathbb F(\mathcal Z_f)$ , where $\mathcal Z_f = \mathcal Z$ . We claim that $f \in \mathbb F[x]$ , so that $\mathbb K \supseteq \mathbb F$ is separable, and normal by Lemma 1. To that end, we claim that for any $\phi \in G$ , $\phi(\mathcal Z) = \mathcal Z$ . To see this result, observe that $\phi \circ G = G$ . Therefore,

$\begin{aligned} \phi(\mathcal Z) &= \bigcup_{i=1}^n \bigcup_{\sigma \in G} \{ (\phi \circ \sigma)(\alpha_i) \} \\ &= \bigcup_{i=1}^n \bigcup_{\psi \in \phi \circ G} \{ \psi (\alpha_i) \} \\ &= \bigcup_{i=1}^n \bigcup_{\psi \in G} \{ \psi (\alpha_i) \} = \mathcal Z. \end{aligned}$

In particular,

$\begin{aligned} \phi(f) &= \phi \left( \prod_{\alpha \in \mathcal Z} (\cdot - \alpha) \right) \\ &= \prod_{\alpha \in \mathcal Z} (\cdot - \phi(\alpha)) \\ &= \prod_{\beta \in \phi(\mathcal Z)} (\cdot - \beta) \\ &= \prod_{\beta \in \mathcal Z} (\cdot - \beta) = f. \end{aligned}$

Writing $f(x) = \sum_{i=0}^n c_i x^i$ , we have $\phi(c_i) = c_i$ . Since $\phi \in G$ is chosen arbitrarily, $c_i \in \mathbb K^G = \mathbb F$ , so that $f \in \mathbb F[x]$ , as required.

$(\Leftarrow)$ Suppose $\mathbb K \supseteq \mathbb F$ is normal and separable. Suppose without loss of generality that $f_{\mathbb F}^{\alpha_1},\dots, f_{\mathbb F}^{\alpha_n}$ are distinct products of degree-one terms. Define $f := \prod_{i=1}^n f_{\mathbb F}^{\alpha_i} \in \mathbb F[x]$ , clearly a product of degree-one terms. Since $\mathbb K \supseteq \mathbb F$ is normal, $\mathbb F( \mathcal Z_{f_{\mathbb F}^{\alpha_i}}) \subseteq \mathbb K$ for each $i$ . Since each $\alpha_i \in \mathcal Z$ ,

$\displaystyle \mathbb K \subseteq \mathbb F(\mathcal Z) \subseteq \bigcup_{i=1}^n \mathbb F( \mathcal Z_{f_{\mathbb F}^{\alpha_i}}) \subseteq \mathbb K.$

Since, $\mathcal Z = \mathcal Z_f$ , we have $\mathbb K = \mathbb F(\mathcal Z) = \mathbb F(\mathcal Z_f)$ . By Lemma 1, for any $\mathbb F$ -homomorphism $\sigma : \mathbb K \to \bar{\mathbb F}$ , $\sigma(\mathbb K) = \mathbb K$ , so that $\mathrm{Aut}_{\mathbb F}(\mathbb K) = \mathrm{Aut}(\mathbb K)$ . Since $\mathbb K \supseteq \mathbb F$ is separable,

$|\mathrm{Aut}_{\mathbb F}(\mathbb K)| = \{ \mathbb K : \mathbb F \} = [ \mathbb K : \mathbb F].$

By definition, $\mathbb K \supseteq \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)}$ is Galois. By the tower rule and Lemma 2,

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

Therefore, $[ \mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} : \mathbb F ] = 1$ implies that $\mathbb K^{\mathrm{Aut}_{\mathbb F}(\mathbb K)} = \mathbb F$ . Define $G := \mathrm{Aut}_{\mathbb F}(\mathbb K)$ , as required.

Definition 3. For any finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ , we say that the extension $\mathbb K \supseteq \mathbb K^G$ is Galois with Galois group $G =: \mathrm{Gal}(\mathbb K / \mathbb K^G)$ .

We say that an extension $\mathbb K \supseteq \mathbb F$ is Galois if there exists a finite subgroup $G \leq \mathrm{Aut}(\mathbb K)$ such that $\mathbb K^G = \mathbb F$ . In this case, $\mathbb K^{\mathrm{Gal}(\mathbb K/\mathbb F)} = \mathbb F$ . By Theorem 1, if $\mathbb K \supseteq \mathbb F$ is finite, then the extension $\mathbb K \supseteq \mathbb F$ is Galois if and only if the extension $\mathbb K \supseteq \mathbb F$ is normal and separable.

However, we haven’t proven Lemma 2 in order to arrive at this equivalence. Even worse; while Definition 3 is well defined, the Galois extension comment, strictly speaking, is not. How do we know that given two finite subgroups $G,H \leq \mathrm{Aut}(\mathbb K)$ , $\mathbb K^G = \mathbb K^H$ implies that $G = H$ ? Only after establishing this result can we properly define the Galois group $\mathrm{Gal}(\mathbb K/ \mathbb F)$ .

This is the content of the famous fundamental theorem of Galois theory, tying together the various ideas that we have explored regarding splitting fields. Once again, suppose $\mathbb K \supseteq \mathbb F$ is finite. 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 2. 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.

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

$\mathbb L_1 \subseteq \mathbb L_2 \quad \iff \quad \Phi(\mathbb L_1) \supseteq \Phi(\mathbb L_2).$

In this case, $[\mathbb L_2 : \mathbb L_1] = [\Phi(\mathbb L_1) : \Phi(\mathbb L_2)]$ .

Proof. Under Theorem 2,

$\begin{aligned} \mathbb L_1 \subseteq \mathbb L_2 \quad &\Rightarrow \quad \mathrm{Aut}_{\mathbb L_1}(\mathbb K) \supseteq \mathrm{Aut}_{\mathbb L_2}(\mathbb K) \\ &\Rightarrow \quad \Phi(\mathbb L_1) \supseteq \Phi(\mathbb L_2) \\ &\Rightarrow \quad \mathbb K^{\Phi(\mathbb L_1)} \subseteq \mathbb K^{\Phi(\mathbb L_2)} \\ &\Rightarrow \quad \Psi(\Phi(\mathbb L_1)) \subseteq \Psi(\Phi(\mathbb L_2)) \\ &\Rightarrow \quad (\Psi \circ \Phi)(\mathbb L_1) \subseteq (\Psi \circ \Phi)(\mathbb L_2) \\ &\Rightarrow \quad \mathbb L_1 \subseteq \mathbb L_2. \end{aligned}$

Furthermore,

Before proving this theorem, we should take some time to appreciate what this theorem is telling us.

Example 2. For any prime number $p$ , the extension $\mathbb Q(\sqrt p) \supseteq \mathbb Q$ is Galois with Galois group

$\mathrm{Gal}(\mathbb Q(\sqrt p)/\mathbb Q) \cong \mathbb Z_2.$

Proof. Since $\mathbb Q(\sqrt p)$ is the splitting field of $x^2-p$ over $\mathbb Q$ , the extension $\mathbb Q(\sqrt p) \supseteq \mathbb Q$ is normal by Lemma 1 and separable by the factorisation

$x^2-p = (x-\sqrt p)(x+\sqrt p).$

By Theorem 1, $\mathbb Q(\sqrt p) \supseteq \mathbb Q$ is Galois. For $i \in \{0, 1\}$ , define the $\mathbb Q$ -automorphisms $\sigma_i : \sqrt 2 \mapsto (-1)^i \sqrt 2$ . It is clear that

$\{ \sigma_0 , \sigma_1 \} \cong \mathbb Z_2.$

To show that $\mathrm{Aut}_{\mathbb Q}(\mathbb Q(\sqrt p)) = \{ \sigma_0, \sigma_1 \}$ , it suffices to prove the direction $(\subseteq)$ , since the direction $(\supseteq)$ is obvious. To that end, fix any $\mathbb Q$ -automorphism

$\sigma \in \mathrm{Aut}_{\mathbb Q}(\mathbb Q(\sqrt p)).$

Thus, for any $a , b \in \mathbb Q$ ,

$\begin{aligned} \sigma(a + b\sqrt 2) &= \sigma(a) + \sigma(b\sqrt 2) \\ &= \sigma(a) + \sigma(b) \sigma(\sqrt 2) \\ &= a + b \sigma(\sqrt 2). \end{aligned}$

Hence, $\sigma$ is determined by $\sigma(\sqrt p)$ . Suppose $\sigma(\sqrt p) = c + d\sqrt p$ for some $c, d \in \mathbb Q$ . Then

$\begin{aligned} \sigma(\sqrt p)^2 &= \sigma((\sqrt p)^2) \\ (c+d\sqrt p)^2 &= \sigma(2) \\ (c^2 + pd^2) + 2cd\sqrt p &= p. \end{aligned}$

If $d = 0$ , then $\sqrt p = \sigma^{-1}(c) = c \in \mathbb Q$ , a contradiction. Therefore, $2cd = 0$ implies that $c = 0$ . Therefore, $pd^2 = p$ implies that $d = \pm 1 = (-1)^i$ for $i \in \{0,1\}$ . Therefore,

$\sigma(\sqrt 2) = (-1)^i \sqrt p = \sigma_i(\sqrt p),$

and hence $\sigma \in \{\sigma_0, \sigma_1\}$ .

Example 3. For distinct primes $p, q$ , the extension $\mathbb Q(\sqrt p, \sqrt q) \supseteq \mathbb Q$ is Galois with Galois group

$\mathrm{Gal}(\mathbb Q(\sqrt p, \sqrt q)/\mathbb Q) \cong \mathbb Z_2 \times \mathbb Z_2.$

Proof. Since $\mathbb Q(\sqrt p, \sqrt q)$ is the splitting field of $(x^2-p)(x^2-q)$ over $\mathbb Q$ , the extension $\mathbb Q(\sqrt p, \sqrt q) \supseteq \mathbb Q$ is normal by Lemma 1 and separable by the factorisation

$(x^2-p)(x^2-q) = (x-\sqrt p)(x+\sqrt p)(x-\sqrt q)(x+\sqrt q).$

By Theorem 1, $\mathbb Q(\sqrt p, \sqrt q) \supseteq \mathbb Q$ is Galois. For $i,j \in \{0,1\}$ , define the $\mathbb Q$ -automorphisms

$\begin{aligned} \sigma_{ij} : (\sqrt p, \sqrt q) &\mapsto ((-1)^i \sqrt p, (-1)^j \sqrt q). \end{aligned}$

We leave it as an exercise to check in similar fashion to Example 2 that

$\mathrm{Aut}_{\mathbb Q}(\mathbb Q(\sqrt p, \sqrt q)) = \{\sigma_{ij} : i,j \in \{0, 1\}\},$

and that the map $\sigma_{ij} \mapsto ([i], [j])$ yields the desired group isomorphism.

Therefore, with finite Galois-ness, we can characterise field extensions in terms of familiar finite groups: subfields of the field extension are in direct correspondence with subgroups of the Galois group! This connection is the key insight of Galois theory that would help us encode the roots of higher-degree polynomials into familiar groups.

The grand idea to the non-existence of a quintic formula looks like this—if a quintic formula exists, then the corresponding Galois group needs to satisfy some properties. However, it fails to do so. By contrapositive or contradiction, a quintic formula cannot exist.

We will prove the rather elegant Theorem 2 in the next post. Ironically, we would need Lemma 2 first. But we will explore both ideas in the next time.

—Joel Kindiak, 23 Apr 26, 1926H

KindiakMath

Galois Extensions

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Galois Extensions

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