05.7.ガロアの基本定理

ガロアの基本定理の補題

Statement

$E/F$:Field Extension
$M,K$:field
$E \supset M \supset K \supset F$
$[E:M] = [E:K]$
then
$M = K$

Proof

Assume $M \supsetneq K$
then
$\exists \alpha \in M - K$
$[K(\alpha):K] \ge 2$
$\therefore [M:K] = [M:K(\alpha)][K(\alpha):K] \ge 2$
$[E:K] = [E:M][M:K]$
$[E:K] \gt [E:M]$ leads to the contradiction
$\therefore M = K$

ガロアの基本定理

Statement

$E/F$:ガロア拡大
$G = \text{Gal}(E/F)$
$\mathbb{H} = \{H:\text{Group}, H \subset G\}$
$\mathbb{M} = \{M:\text{Field}, F \subset M \subset E\}$
$E^H = \{\alpha \in E | \forall h \in H, \alpha^h (= h(\alpha)) = \alpha\}$
$\text{Aut}(E/M) = \{g \in G| \forall m \in M, m^g(=g(m)) = m\}$

obviously
$(F \subset E^H \subset E)$
$(\text{Aut}(E/M) \subset G)$
then
(0)
$E^H$:Field
$\text{Aut}(E/M)$:Group
(1)
$E/M$ is Galois Expansion.
(2)
$\text{Aut}(E/E^H) = H$
$E^{\text{Aut}(E/M)} = M$
(3)
$\exists$ Bijection between $\mathbb{H}$ and $\mathbb{M}$
(4)
$F \subset M, M_1, M_2 \subset E$
$G \supset H, H_1, H_2 \supset \{id\}$
then
(4.1)
(i)$M_1 \supset M_2 \Leftrightarrow \text{Aut}(E/M_1) \subset \text{Aut}(E/M_2)$
(ii)$H_1 \subset H_2 \Leftrightarrow E^{H_1} \supset E^{H_2}$

(4.2)
let
$E^{H_1} = M_1 \quad(\text{Aut}(E/M_1) = H_1)$
$E^{H_2} = M_2 \quad(\text{Aut}(E/M_2) = H_2)$
then
(i)$E^{\langle H_1 \cup H_2 \rangle} = M_1 \cap M_2 \quad(\text{Aut}(E/(M_1 \cap M_2)) = \langle H_1 \cup H_2 \rangle)$
(ii)$E^{H_1 \cap H_2} = \langle M_1 \cup M_2 \rangle \quad(\text{Aut}(E/\langle M_1 \cup M_2 \rangle) = H_1 \cap H_2)$

(5)
$\sigma \in G, \text{Aut}(E/M^\sigma) = \sigma \text{Aut}(E/M) \sigma^{-1}$
(6)
(6.1)$M/F$:Galois Extension $\Leftrightarrow \text{Gal}(E/M) \triangleleft G$
then
(6.2)$\text{Gal}(M/F) \simeq G/\text{Gal}(E/M)$

Proof

(0)
$e_1,e_2 \in E^H$
$\forall h \in H$
$h(e_1 + e_2) = h(e_1) + h(e_2) = e_1 + e_2 \therefore e_1 + e_2 \in E^H$
$h(e_1 e_2) = h(e_1) h(e_2) = e_1 e_2 \therefore e_1 e_2 \in E^H$
$h(-e_1) + e_1 = h(-e_1) + h(e_1) = h(-e_1 + e_1) = h(0) = 0 \therefore h(-e_1) = -e_1 \therefore -e_1 \in E^H$
$h(e_1^{-1}) e_1 = h(e_1^{-1}) h(e_1) = h(e_1^{-1} e_1) = h(1) = 1 \therefore h(e_1^{-1}) = e_1^{-1} \therefore e^{-1} \in E^H$

$h_1,h_2 \in \text{Aut}(E/M)$
$h_1:E \rightarrow E, h_2:E \rightarrow E$
$h_1,h_2$:Bijective
$\therefore h_1 \circ h_2:E \rightarrow E$
$m \in M \Rightarrow h_1 \circ h_2(m) = h_1(m) = m$
$\therefore h_1 \circ h_2 \in \text{Aut}(E/M)$
(1)
$E/M:\text{Separable} \quad(\because\text{Prop.中間体と分離拡大})$
$E/M:\text{Normal} \quad(\because\text{Prop.正規拡大の中間体})$
$\therefore E/M$:Galois Extension
(2)
$E/E^H$:Galois Extension $\quad(\because\text{Prop.}アルティンの定理)$
$H = \text{Gal}(E/E^H) = \text{Aut}(E/E^H)$

$m \in M$
$\forall h \in \text{Aut}(E/M), h(m) = m$
$\therefore m \in E^{\text{Aut}(E/M)}$
$\therefore M \subset E^{\text{Aut}(E/M)}$

$\text{Aut}(E/M) \subset \text{Aut}(E)$
$[E:E^{\text{Aut}(E/M)}] = |{\text{Aut}(E/M)}| \quad(\because\text{Prop.アルティンの定理})$
$=|\text{Gal}(E/M)| \quad(\because E/M:\text{Galois Extension})$
$=[E:M]$
$\therefore E^{\text{Aut}(E/M)} = M \quad(\because\text{Prop.ガロアの基本定理の補題})$
(3)
Define
$\mathcal{G}:\mathbb{H} \rightarrow \mathbb{M}$
$\mathcal{G}:H \in \mathbb{H} \mapsto E^H \in \mathbb{M}$
$\mathcal{F}:\mathbb{M} \rightarrow \mathbb{H}$
$\mathcal{F}:M \in \mathbb{M} \mapsto \text{Aut}(E/M) \in \mathbb{H}$
then
$\mathcal{G} \circ \mathcal{F}:H \longrightarrow E^H \longrightarrow \text{Aut}(E/E^H) = H$
$\therefore \mathcal{G} \circ \mathcal{F} = id$
$\mathcal{F} \circ \mathcal{G}:M \longrightarrow \text{Aut}(E/M) \longrightarrow E^{\text{Aut}(E/M)} = M$
$\therefore \mathcal{F} \circ \mathcal{G} = id$
$\therefore \mathcal{F}^{-1} = \mathcal{G}$
$\therefore \mathcal{F}:\text{Bijective} \quad(\because\text{Prop.全単射の必要十分条件})$
$\therefore \mathcal{G}^{-1} = \mathcal{F}$
$\therefore \mathcal{G}:\text{Bijective} \quad(\because\text{Prop.全単射の必要十分条件})$
(4.1)
(i)
$(\Rightarrow)$
$\sigma_1 \in \text{Aut}(E/M_1)$
$\Rightarrow \forall m_2 \in M_2, \sigma_1(m_2) = m_2 \quad(\because m_2 \in M_2 \subset M_1)$
$\therefore \sigma_1 \in \text{Aut}(E/M2)$

$(\Leftarrow)$
$m_2 \in M_2$
$\Rightarrow \forall \sigma_1 \in \text{Aut}(E/M_1), \sigma_1(m_2) = m_2 \quad(\because \sigma_1 \in \text{Aut}(E/M_1) \subset \text{Aut}(E/M_2))$
$\therefore m_2 \in E^{\text{Aut}(E/M_1)} = M_1$

(ii)
$(\Rightarrow)$
$e_2 \in E^{H_2}$
$\Rightarrow \forall h_1 \in H_1, h_1(e_2) = e_2 \quad(\because h_1 \in H_1 \subset H_2)$
$\therefore e_2 \in E^{H_1}$

$(\Leftarrow)$
$h_1 \in H_1$
$\Rightarrow \forall e_2 \in E^{H_2},h_1(e_2) = e_2\quad(\because e_2 \in E^{H_2} \subset E^{H_1})$
$\therefore h_1 \in \text{Aut}(E/E^{H_2}) = H_2$
(4.2)
$F \subset (M_1 \cap M_2) \subset M_1 \subset \langle M_1 \cup M_2 \rangle \subset E$
$F \subset (M_1 \cap M_2) \subset M_2 \subset \langle M_1 \cup M_2 \rangle \subset E$
$G \supset \langle H_1 \cup H_2 \rangle \supset H_1 \supset (H_1 \cap H_2) \supset \{id\}$
$G \supset \langle H_1 \cup H_2 \rangle \supset H_2 \supset (H_1 \cap H_2) \supset \{id\}$
$\therefore$
$\text{Aut}(E/M_1 \cap M_2) \supset \text{Aut}(E/M_1) = H_1 \supset \text{Aut}(E/\langle M_1 \cup M_2 \rangle)$
$\text{Aut}(E/M_1 \cap M_2) \supset \text{Aut}(E/M_2) = H_2 \supset \text{Aut}(E/\langle M_1 \cup M_2 \rangle)$
$E^{\langle H_1 \cup H_2 \rangle} \subset E^{H_1} = M_1 \subset E^{(H_1 \cap H_2)}$
$E^{\langle H_1 \cup H_2 \rangle} \subset E^{H_2} = M_2 \subset E^{(H_1 \cap H_2)}$
$\therefore$
$\text{Aut}(E/M_1 \cap M_2) \supset \langle H_1 \cup H_2 \rangle$---(*1)
$H_1 \cap H_2 \supset \text{Aut}(E/\langle M_1 \cup M_2 \rangle)$---(*2)
$E^{\langle H_1 \cup H_2 \rangle} \subset M_1 \cap M_2$---(*3)
$\langle M_1 \cup M_2 \rangle \subset E^{(H_1 \cap H_2)}$---(*4)
$\therefore$
$E^{\text{Aut}(E/M_1 \cap M_2)} \subset E^{\langle H_1 \cup H_2 \rangle}$
$E^{H_1 \cap H_2} \subset E^{\text{Aut}(E/\langle M_1 \cup M_2 \rangle)}$
$\text{Aut}(E/E^{\langle H_1 \cup H_2 \rangle}) \supset \text{Aut}(E/(M_1 \cap M_2))$
$\text{Aut}(E/\langle M_1 \cup M_2 \rangle) \supset \text{Aut}(E/E^{(H_1 \cap H_2)})$
$\therefore$
$(M_1 \cap M_2) \supset E^{\langle H_1 \cup H_2 \rangle}$---(*5)
$E^{H_1 \cap H_2} \supset \langle M_1 \cup M_2 \rangle$---(*6)
$\langle H_1 \cup H_2 \rangle \supset \text{Aut}(E/(M_1 \cap M_2))$---(*7)
$\text{Aut}(E/\langle M_1 \cup M_2 \rangle) \supset (H_1 \cap H_2)$---(*8)

$\therefore$
$\text{Aut}(E/(M_1 \cap M_2)) = \langle H_1 \cup H_2 \rangle \quad(\because *1 *7)$
$\text{Aut}(E/\langle M_1 \cup M_2 \rangle) = (H_1 \cap H_2) \quad(\because *2 *8)$
$E^{\langle H_1 \cup H_2 \rangle} = (M_1 \cap M_2) \quad(\because *3 *5)$
$E^{H_1 \cap H_2} = \langle M_1 \cup M_2 \rangle \quad(\because *4 *6)$

(5)
$g \in \text{Aut}(E/M^\sigma) \Leftrightarrow g \in \text{Aut}(E) \text{ s.t. } \forall m \in M, g(\sigma(m)) = \sigma(m)$
$\Leftrightarrow g \in \text{Aut}(E) \text{ s.t. } \forall m \in M, \sigma^{-1}(g(\sigma(m))) = m$
$\Leftrightarrow \sigma^{-1}g\sigma \in \text{Aut}(E/M)$
$\Leftrightarrow g \in \sigma\text{Aut}(E/M)\sigma^{-1}$
$\therefore \text{Aut}(E/M^\sigma) = \sigma\text{Aut}(E/M)\sigma^{-1}$

(6.1)
Prove
$M/F$:Galois Extension $\Leftrightarrow \forall \sigma \in \text{Gal}(E/F), \sigma(M) = M$

$\Rightarrow$
Suppose $\exists \sigma \in \text{Gal}(E/F), \sigma(M) \neq M$
then
$\exists m \in M, \sigma(m) \notin M$
let $p(x) \in F[x]$:Minimal Polynomial of $m$
then $p(m) = 0$ and $p(\sigma(m)) = \sigma(p)(\sigma(m)) = \sigma(p(m)) = \sigma(0) = 0$
$\therefore M/F$ is not Normal Extension $\Rightarrow$ contradiction
$\Leftarrow$
$E/F$:Separable $\Rightarrow M/F$:Separable ($\because\text{Prop.中間体と分離拡大}$)
$\forall m \in M$
let $p(x) \in F[x]$:Minimal Polynomial of $m$
$p(x) = (x - m_1)(x - m_2) \cdots (x - m_n) \quad(m = m_1)$
then $m_1, \cdots m_n \in E \quad (\because E/F:\text{Normal Extension})$
$\forall m_i, \exists \sigma_i \in \text{Gal}(E/F), \sigma_i(m) = m_i$
$\sigma_i(m) \in \sigma_i(M) = M \therefore m_i \in M$
$\therefore M/F$:Normal Extension
$\therefore M/F$:Galois Extension

$\therefore$
$M/F$:Galois Extension $\Leftrightarrow \forall \sigma \in \text{Gal}(E/F), \sigma(M) = M$
$\Leftrightarrow \forall \sigma \in \text{Gal}(E/F), \text{Aut}(E/\sigma(M)) = \text{Aut}(E/M)$
$\Leftrightarrow \forall \sigma \in \text{Gal}(E/F), \sigma\text{Aut}(E/M)\sigma^{-1} = \text{Aut}(E/M) \quad(\because(5))$
$\Leftrightarrow \text{Aut}(E/M) \triangleleft \text{Gal}(E/F)$

(6.2)
let
$\phi:\text{Gal}(E/F) \rightarrow \text{Gal}(M/F)$
$\phi:\sigma \mapsto \sigma|_M$
then
$\phi$:surjection
$\text{Ker}\phi = \text{Gal}(E/M)$
$\therefore G/\text{Aut}(E/M) = \text{Gal}(E/F)/ \text{Gal}(E/M) = \text{Gal}(E/F)/ \text{Ker}\phi \simeq \text{Im}\phi = \text{Gal}(M/F)$
QED.