Revisiting Automorphisms

One key idea of Galois theory is how we can relate field extensions to field automorphisms. Of course, we should have a working definition of what we mean by field automorphisms first.

Let $\mathbb F$ be any field and $\mathbb K \supseteq \mathbb F$ denote a field extension.

Theorem 1. Let $u, v \in \mathbb K$ be algebraic over $\mathbb F$ . Define the map $\Psi_{u,v} : \mathbb F(u) \to \mathbb F(v)$ by

$\displaystyle \Psi_{u,v}\left( \sum_{i=0}^n a_iu^i \right) = \sum_{i=0}^n a_iv^i.$

The following are equivalent.

$\Psi_{u,v}$ is a homomorphism.
$\Psi_{u,v}$ is an isomorphism.
$f_{\mathbb K}^u = f_{\mathbb K}^v$ .

In the latter, we say that $u$ and $v$ are conjugates.

Proof. If $\Psi_{u,v}$ is an isomorphism then it is immediately a homomorphism. Define the evaluation map $\Psi_v : \mathbb K[x] \to \mathbb K(v)$ by $\Psi(x) := v$ and extended to a homomorphism. Define $\Psi_u$ similarly, so that $\Psi_v = \Psi_{u, v} \circ \Psi_u$ . In particular,

$\mathbb F[x] \neq \ker(\Psi_v) \supseteq \langle f_{\mathbb K}^u \rangle \quad \Rightarrow \quad \ker(\Psi_v) = \langle f_{\mathbb K}^u \rangle,$

inducing an isomorphism $\tilde{\Psi}_v : \mathbb F[x]/ \langle f_{\mathbb K}^u \rangle \to \mathbb F(v)$ . By definition, $\tilde{\Psi}_u = \Psi_{u,v} \circ \tilde{\Psi}_v$ , so that $\Psi_{u,v} = \tilde{\Psi}_u \circ \tilde{\Psi}_v^{-1}$ is the required isomorphism if it is a homomorphism. Finally, $\Psi_{u,v}$ is an isomorphism if and only if

$\mathbb F[x]/\langle f_{\mathbb K}^u \rangle \cong \mathbb F[x]/\langle f_{\mathbb K}^v \rangle,$

which holds if and only if $f_{\mathbb K}^u = f_{\mathbb K}^v$ .

Recall that given the field extensions $\mathbb K_1 \supseteq \mathbb F$ and $\mathbb K_2 \supseteq \mathbb F$ , a field homomorphism $\phi : \mathbb K_1 \to \mathbb K_2$ is a $\mathbb F$ –homomorphism if $\phi|_{\mathbb F} = \mathrm{id}$ .

Corollary 1. Let $u$ be algebraic over a field $\mathbb F$ . Then for any conjugate $v$ of $u$ , there exists a unique $\mathbb F$ -homomorphism $\phi : \mathbb F(u) \to \bar{\mathbb F}$ such that $\phi(u) = v$ .

When homomorphisms are bijective, we call them isomorphisms. When isomorphisms map fields to themselves, we call them automorphisms. Denote the set of automorphisms on a field $\mathbb F$ by $\mathrm{Aut}(\mathbb F)$ .

Definition 1. For any automorphism $\sigma : \mathbb K \to \mathbb K$ , we say that $u \in \mathbb K$ is fixed under $\sigma$ if $\sigma(u) = u$ . Furthermore, for any subfield $\mathbb F \subseteq \mathbb K$ , $\sigma$ is called an $\mathbb F$ -automorphism if $\sigma|_{\mathbb F} = \mathrm{id}$ . Define

$\mathrm{Aut}_{\mathbb F}(\mathbb K) := \{ \sigma \in \mathrm{Aut}(\mathbb K) : \sigma \text{ is an }\mathbb F\text{-automorphism}\}.$

Example 1. For any $\sigma \in \mathrm{Aut}(\mathbb F)$ , $\sigma$ is a $\mathbb Q$ -automorphism.

Lemma 1. Given $\Sigma \subseteq \mathrm{Aut}(\mathbb F)$ , the set

$\mathbb F^{\Sigma} := \{ u \in \mathbb K : \sigma(u) = u \text{ for all } \sigma \in \Sigma\}$

forms a sub-field of $\mathbb K$ , called the fixed field of $\Sigma$ in $\mathbb F$ .

Given any isomorphism $\sigma : \mathbb F_1 \to \mathbb F_2$ , define the set

$\Sigma_{\mathbb F_1, \mathbb F_2}^{\sigma} := \{ \phi \in \mathrm{Hom}(\bar{\mathbb F}_1, \bar{\mathbb F}_2 ) : \phi|_{\mathbb F_1} = \sigma \}.$

Lemma 2. Suppose $\mathbb K \supseteq \mathbb F$ is a finite extension. For any isomorphism $\sigma : \mathbb F \to \mathbb L$ ,

$|\Sigma_{\mathbb K, \bar{\mathbb L}}^{\sigma}| = | \Sigma_{\mathbb K, \bar{\mathbb F}}^{ \mathrm{id} } |.$

Proof. Let $\phi : \bar{\mathbb L} \to \bar{\mathbb F}$ extend $\sigma^{-1} : \mathbb L \to \mathbb F$ . For any $\Psi \in \Sigma_{\mathbb K, \bar{\mathbb L}}^{\sigma}$ , $\Psi : \bar{\mathbb K} \to \bar{\mathbb L}$ is a homomorphism that extends $\sigma$ . Then, for any $x \in \mathbb F \subseteq \mathbb K$ , since $\sigma(x) \in \mathbb L$ ,

$(\phi \circ \Psi)(x) = \phi^{-1}(\Psi(x)) = \phi(\sigma(x)) = \sigma^{-1}(\sigma(x)) = x.$

Therefore, $(\phi \circ \Psi)|_{\mathbb F} = \mathrm{id}$ , i.e. $\phi \circ \Psi : \mathbb K \to \bar{\mathbb F}$ is a uniquely-defined $\mathbb F$ -homomorphism. Similarly, for any $\Psi_0 \in \Sigma_{\mathbb K, \bar{\mathbb F}}^{\mathrm id}$ , there exists a uniquely-defined extension of $\sigma$ defined by $\phi^{-1} \circ \Psi_0$ .

Definition 2. If $\mathbb K \supseteq \mathbb F$ is a finite extension, we define $\{ \mathbb K : \mathbb F\} := | \Sigma_{\mathbb K, \bar{\mathbb F}}^{ \mathrm{id} } |$ . By Lemma 2, for any isomorphism $\sigma : \mathbb F \to \mathbb L$ , $\{ \mathbb K : \mathbb F\} = |\Sigma_{\mathbb K, \bar{\mathbb L}}^{\sigma}|$ .

Theorem 2. For any $u$ algebraic over $\mathbb F$ , $\{\mathbb F(u) : \mathbb F\}$ equals the number of distinct roots of $f_{\mathbb F}^u$ .

Proof. By Definition 2, $\{\mathbb F(u) : \mathbb F\} = |\Sigma_{\mathbb F(u), \bar{\mathbb F}}^{\mathrm{id}}|$ . The right-hand side counts the number of $\mathbb F$ -homomorphisms from $\mathbb F(u)$ to $\bar{\mathbb F}$ . Define

$\mathcal R := \{v \in \bar{\mathbb F} : f_{\mathbb F}^u(v) = 0\},$

where $f_{\mathbb F}^u(x) = \sum_{j=0}^s b_j x^j$ .

We claim that $|\Sigma_{\mathbb F(u), \bar{\mathbb F}}^{\mathrm{id}}| = |\mathcal R|$ . Fix $\Psi \in \Sigma_{\mathbb F(u), \bar{\mathbb F}}^{\mathrm{id}}$ . Then

$\displaystyle f_{\mathbb F}^u(\Psi(u)) = \sum_{j=0}^s b_j \Psi(u)^j = \Psi\left( \sum_{j=0}^s b_j u^j \right) = \Psi(f_{\mathbb F}^u(u)) = \Psi(0) = 0,$

so that each $\Psi \in \Sigma_{\mathbb F(u), \bar{\mathbb F}}$ yields a unique root $\Psi(u) \in \mathcal R$ .

Conversely, given any root $v \in \mathcal R$ , define the evaluation map $\phi_v : \bar{\mathbb F}[x] \to \bar{\mathbb F}$ by

$\phi_v|_{\mathbb F} = \mathrm{id},\quad \phi_v(x) = v,$

as well as the inclusion map $\psi : \mathbb F[x] \to \bar{\mathbb F}[x]$ . Then $\phi_v \circ \psi : \mathbb F[x] \to \bar{\mathbb F}$ is a ring homomorphism such that $\langle f_{\mathbb F}^u \rangle \subseteq \ker(\phi \circ \psi)$ . Since $\langle f_{\mathbb F}^u \rangle$ is maximal in $\mathbb F[x]$ ,

$\begin{aligned} ((\phi_v \circ \psi)(1) = 1 \quad &\Rightarrow \quad \ker(\phi_v \circ \psi) \neq \mathbb F[x]) \\ \quad &\Rightarrow \quad \ker(\phi_v \circ \psi) = \langle f_{\mathbb F}^u\rangle. \end{aligned}$

By the first isomorphism theorem, $\phi_v$ induces the ring isomorphism

$\mathbb F(u) \cong \mathbb F[x]/\langle f_{\mathbb F}^u \rangle = \mathbb F[x]/{\ker(\phi_v \circ \psi)} \cong \bar{\mathbb F}.$

In particular, $\phi_v$ induces a ring isomorphism $\Phi_v : \mathbb F(u) \to \bar{\mathbb F}$ . Furthermore, $\Phi_v$ is an $\mathbb F$ -homomorphism. Therefore, each root $v \in \mathcal R$ yields a unique $\mathbb F$ -homomorphism $\Phi_v \in \Sigma_{\mathbb F(u), \bar{\mathbb F}}^{\mathrm{id}}$ .

Corollary 2. For the finite extensions $\mathbb F \subseteq \mathbb K \subseteq \mathbb L$ ,

$\{ \mathbb L : \mathbb F \} = \{ \mathbb L : \mathbb K \} \cdot \{ \mathbb K : \mathbb F \},$

i.e. the tower rule for the separable degree of finite extensions.

Proof. Denote $p:= \{ \mathbb L : \mathbb F \}$ , $q:= \{ \mathbb L : \mathbb K \}$ , and $r := \{ \mathbb K : \mathbb F \}$ . Hence,

for $1 \leq i \leq p$ , there exists an $\mathbb F$ -homomorphism $\sigma_i : \mathbb L \to \bar{\mathbb F}$ ,
for $1 \leq j \leq q$ , there exists an $\mathbb K$ -homomorphism $\rho_j : \mathbb L \to \bar{\mathbb K}$ ,
for $1 \leq k \leq r$ , there exists an $\mathbb F$ -homomorphism $\mu_k : \mathbb K \to \bar{\mathbb F}$ .

For each $k$ , extend $\mu_k$ to an $\mathbb F$ -homomorphism $\tilde{\mu}_k : \bar{\mathbb K} \to \bar{\mathbb F}$ . Then $\rho_j \circ \tilde{\mu}_k : \mathbb L \to \bar{\mathbb F}$ forms an $\mathbb F$ -homomorphism. Therefore, $q \cdot r \leq p$ .

Likewise, for each $i$ , consider the $\mathbb F$ -homomorphism $\sigma_i : \mathbb L \to \bar{\mathbb F}$ . Now $\sigma_i|_{\mathbb K} : \mathbb K \to \bar{\mathbb F}$ must be an $\mathbb F$ -homomorphism, so that $\sigma_i|_{\mathbb K} = \mu_k$ for some $k$ . In particular, $\sigma_i$ extends $\mu_k$ , so that $\sigma_i = \rho_j \circ \tilde{\mu}_k$ for some $j$ . Therefore, $p \leq q \cdot r$ .

Lemma 3. For any monic and irreducible polynomial $f \in \mathbb F[x]$ , there exist $u_1,\dots,u_r$ and some positive integer $\alpha$ such that, regarded as a polynomial in $\bar{\mathbb F}$ ,

$\displaystyle f(x) = \prod_{i=1}^r (x-u_i)^\alpha.$

Proof. Write $f(x) = \sum_{i=1}^r a_i x^i$ . Factorise $f$ into its roots $u_1,\dots, u_n$ when regarded as a polynomial in $\bar{\mathbb F}$ :

$\displaystyle f(x) = \prod_{i=1}^n (x-u_i)^{\alpha_i}.$

The crucial claim is that the $\alpha_i$ terms all equal each other. To establish that claim, We first regard $f \in \mathbb F(u_1,\dots, u_n)[x]$ . For any $k \neq \ell$ , the map $\Psi_{u_k,u_\ell} : \mathbb F(u_k) \to \mathbb F(u_\ell)$ as defined by Theorem 1 is an $\mathbb F$ -isomorphism. Fix any homomorphism

$\tilde{\Psi}_{u_k,u_\ell} : \mathbb F(u_1,\dots, u_n) \to \bar{\mathbb F}$

that extends $\Psi_{u_k,u_\ell}$ . Then evaluating $\tilde{\Psi}_{u_k,u_\ell}(f)$ and simplifying in two different ways,

$\begin{aligned} \prod_{j=1}^n (x-u_j)^{\alpha_j} = \tilde{\Psi}_{u_k,u_\ell}(f) &= \prod_{j=1}^n (x- \tilde{\Psi}_{u_k,u_\ell}(u_j) )^{\alpha_j} \\ &= (x-u_\ell)^{\alpha_k } \prod_{j \neq k}^n (x- \tilde{\Psi}_{u_k,u_\ell}(u_j) )^{\alpha_j}. \end{aligned}$

Since there are $\alpha_\ell$ copies of $(x-u_\ell)$ on the left-hand side, we have $\alpha_k \leq \alpha_\ell$ . Since this inequality holds for any $k, \ell$ , we also must have $\alpha_\ell \leq \alpha_k$ . Therefore, $\alpha_k = \alpha_\ell =: \alpha$ , so that

$\displaystyle f(x) = \prod_{i=1}^n (x-u_i)^{\alpha_i} = \prod_{i=1}^n (x-u_i)^{\alpha}.$

Corollary 3. For the finite extension $\mathbb F \subseteq \mathbb K$ , $\{ \mathbb K : \mathbb F \} \mid [\mathbb K : \mathbb F]$ .

Proof. Suppose $[\mathbb K : \mathbb F] = n$ . Define $\mathbb F_0 := \mathbb F$ , $\mathbb F_r = \mathbb F(u_1,\dots, u_r)$ , so that $\mathbb F_n = \mathbb K$ . Observe that

$\mathbb F \subseteq \mathbb F(u_1) \subseteq \cdots \subseteq \mathbb F(u_1,\dots, u_{n-1}) \subseteq \mathbb K.$

By the tower rule for separable degrees,

$\displaystyle \{\mathbb K : \mathbb F\} = \prod_{i=1}^n \{ \mathbb F_i : \mathbb F_{i-1} \}.$

It suffices to show that $m := \{ \mathbb F_i : \mathbb F_{i-1} \} \mid [ \mathbb F_i : \mathbb F_{i-1} ]$ . By Theorem 2, $\mathbb F_i = \mathbb F_{i-1}(u_i)$ implies that $m = \{ \mathbb F_{i-1}(u_i): \mathbb F_{i-1} \}$ is given by the number of distinct roots of $f_{\mathbb F_{i-1}}^{u_i}$ . By Lemma 3, factorise this polynomial into

$\displaystyle f_{\mathbb F_{i-1}}^{u_i}(x) = \prod_{j=1}^m (x - v_j)^{\alpha}.$

Therefore, $\alpha \cdot m = \deg(f_{\mathbb F_{i-1}}^{u_i}) = [ \mathbb F_i : \mathbb F_{i-1} ]$ , as required.

Definition 2 helps us define the notion of a separable extension—a central notion in describing the “Galois-ness” of a field.

Definition 3. A field extension $\mathbb K \supseteq \mathbb F$ is separable if $\{\mathbb K : \mathbb F\} = [\mathbb K : \mathbb F]$ . An element $u \in \mathbb K$ is separable over $\mathbb F$ if $\mathbb F(u) \supseteq \mathbb F$ is separable.

Theorem 3. Suppose $\mathbb F \subseteq \mathbb K \subseteq \mathbb L$ is finite. Then $\mathbb L \supseteq \mathbb F$ is separable if and only if $\mathbb K \supseteq \mathbb F$ and $\mathbb L \supseteq \mathbb K$ are separable.

Proof. Recall the tower rule for usual and separable degrees:

$\begin{aligned} [ \mathbb L : \mathbb F ] &= [ \mathbb L : \mathbb K ] \cdot [ \mathbb K : \mathbb F ], \\ \{ \mathbb L : \mathbb F \} &= \{ \mathbb L : \mathbb K \} \cdot \{ \mathbb K : \mathbb F\}. \end{aligned}$

Taking ratios, each ratio is an integer by Corollary 3:

$\begin{aligned} \frac{ [ \mathbb L : \mathbb F ] }{ \{ \mathbb L : \mathbb F \} }&= \frac{ [ \mathbb L : \mathbb K ] }{ \{ \mathbb L : \mathbb K \} } \cdot \frac{ [ \mathbb K : \mathbb F ] }{ \{ \mathbb K : \mathbb F \} }. \end{aligned}$

The direction $(\Leftarrow)$ is obvious by the usual tower rule. For the direction $(\Rightarrow)$ , the left-hand side would be $1$ , so that must both ratios on the right side are also equal to $1$ , as required.

Corollary 4. Suppose $\mathbb K \supseteq \mathbb F$ is finite. Then the following statements are equivalent:

Every element in $\mathbb K$ is separable over $\mathbb F$ .
There exist $u_1,\dots, u_r \in \mathbb K$ separable over $\mathbb F$ such that $\mathbb K = \mathbb F(u_1,\dots, u_r)$ .
$\mathbb K$ is separable over $\mathbb F$ .

Proof. Write $\mathbb K = \mathbb F(u_1,\dots, u_r)$ and apply Theorem 3 repeatedly.

Corollary 5. Suppose $\mathbb K \supseteq \mathbb F$ is algebraic. Then

$\mathbb K_{\mathrm s} := \{u \in \mathbb K : u\text{ is separable over }\mathbb F\}$

forms a subfield of $\mathbb K$ , called the separable closure of $\mathbb F$ in $\mathbb K$ .

Proof. Given $u,v \in \mathbb K_s$ , $\mathbb F(u,v) \supseteq \mathbb F$ is separable by Corollary 4.

Eventually, we would describe a special subset of separable extensions as Galois, but to do that we need to explore the notion of splitting fields and normal extensions.

—Joel Kindiak, 14 Apr 26, 1858H

KindiakMath

Revisiting Automorphisms

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Revisiting Automorphisms

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