Galois Theory HT25, Bounds on the size of the Galois group

$G \cdot x$ is finite since $X$ is finite (there’s only finitely many places that can be mapped onto). Let $G \cdot x = \{g _ 1 \cdot x, \ldots, g _ n \cdot x\}$. Then:

\[\pi_x^{-1}(g_i \cdot x) = g_i \text{Stab}_G(x)\]

which is a finite set for each $i$, by assumption.

But then:

\[G = \pi_x^{-1} (g_1 \cdot x) \cup \cdots \cup \pi_x^{-1}(g_n \cdot x)\]

This is a finite union of finite sets and is therefore finite.

It is also finite.

Let $\{\alpha _ 1, \ldots, \alpha _ n\}$ be a basis for $K$ as an $F$-vector space. Each $\alpha _ i$ is algebraic over $F$ by the result that says

Suppose:

• $K/F$ is a finite field extension
• $\alpha \in K$

Then $\alpha$ is algebraic over $F$.

Let $m _ i := m _ {F, \alpha _ i}$ be the minimal polynomial of each $\alpha _ i$ over $F$, and consider the corresponding finite sets of roots $V(m _ i) := \{\beta \in K \mid m _ i(\beta) = 0\}$. Then consider the induced “diagonal action” of $G$ on the set $X = V(m _ 1) \times \cdots V(m _ n)$.

We’re looking to apply the result that says:

Suppose:

• $G$ is a (possibly infinite) group
• $X$ is a finite set
• There exists $x \in X$ such that $\text{Stab} _ G(x)$ is a finite set

Then $G$ is finite.

Note that $\text{Stab} _ G((\alpha _ 1, \ldots, \alpha _ n)) = \{1\}$, since if it fixes each $\alpha _ i$ it must fix every element of $K$ as $\{\alpha _ 1, \ldots, \alpha _ n\}$ is an $F$-vector space basis for $K$. But if it fixes every element, it is the identity map. So we have found an element of $x \in X$ such that $\text{Stab} _ G(x)$ is a finite set, and hence $G$ must also be finite.

It follows that $\text{Gal}(K/F)$ is finite.

\[K_1 \cup \cdots \cup K_m < K\]

Suppose for a contradiction that $K = K _ 1 \cup \cdots \cup K _ m$. We can assume that $m \ge 2$ (otherwise it is trivial) and that $m$ is minimal with this property (otherwise combine two subsets together).

Choose $y \in K \setminus K _ 1$. We aim to show that $K _ 1 \subseteq K _ 2 \ldots \cup K _ m$, since that implies $K = K _ 2 \cup \ldots \cup K _ m$, which contradicts the minimality of $m$.

Let $x \in K _ 1$. Since $F$ is infinite, there exists a finite subset $S \subset F$ where $ \vert S \vert = m+1$.

Then for each $a \in S$, there is some $i(\alpha) \in \{1, \ldots, m\}$ such that $x + \alpha y \in K _ {i(\alpha)}$ by the assumption that $K _ 1, \ldots, K _ m$ cover $K$.

$i : S \to \{1, \ldots, m\}$ cannot be injective since $ \vert S \vert = m+1$.

Hence $\exists \alpha, \beta \in S$ such that $\alpha \ne \beta$ and $x + \alpha y$ and $x + \beta y$ both lie in $K _ {i(\alpha)}$.

But since

\[y = \frac{(x + \alpha y) - (x + \beta y)}{\alpha - \beta}\]

It must be the case that $y \in K _ {i(\alpha)}$. But as $y \notin K _ 1$ by assumption, $i(\alpha) > 1$.

But since

\[x = (x + \alpha y) - \alpha y \in K_{i(\alpha)}\]

and $K _ {i(\alpha)} \subseteq K _ 2 \cup \ldots \cup K _ m$ for each $x \in K _ 1$, it follows $K _ 1 \subseteq K _ 1 \cup \cdots \cup K _ m$, contradicting the minimality of $m$.

\[|G| \le [F(z) : F]\]

Define

\[V(m_{F,z}) := \\{k \in K \mid m_{F,z}(k) = 0\\}\]

Then

\[\begin{aligned} |G| &= |G\cdot z | \times |\text{Stab} _ G(z)| &&\text{(orbit-stab)} \\\\ &= |G\cdot z| \times 1 &&(\text{assump. in question}) \\\\ &\le |V(m _ {F,z})| &&(\star 1) \\\\ &\le \deg(m _ {F,z}) &&(\text{fundamental theorem of algebra}) \\\\ &= [F(z) : F] &&(\star 2) \end{aligned}\]

where:

• $(\star 1)$ is justified like so: $G \cdot z$ is the orbit of $z$ under $G$. Since $z \in V(m _ {F, z})$ and the action on $K$ restricts to one on $V(m _ {F, z})$, any element of $G$ must send $z$ back into $V(m _ {F, z})$. So $G \cdot z$ can be no bigger than the size of $V(m _ {F, z})$.
• $(\star 2)$ is justified by the general result that the degree of the field extension is equal to the degree of the minimal polynomial.

\[|G| \le [K : F]\]

$G$ is finite by the result that says finite field extensions lead to finite Galois groups (reminder: this is a result that comes from considering how $G$ acts on $V(m _ {\alpha _ 1}) \times \cdots \times V(m _ {\alpha _ n})$). We consider two cases, one where $F$ is infinite, and one where $F$ is finite.

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Case 1: $F$ infinite.

Write $G = \{1, g _ 1, \ldots, g _ n\}$ where $g _ i$ distinct.

Since each $g _ i$ is non-trivial, each $K^{\langle g _ i \rangle}$ is a proper subfield of $K$ containing $F$. Since $F$ is infinite, there exists some $z$ in the set

\[K \setminus (K^{\langle g_1\rangle} \cup \cdots \cup K^{\langle g_n \rangle})\]

by the result that says

If:

• $K/F$ is a finite field extension
• $F$ is infinite
• $K _ 1, \ldots, K _ m$ are finitely many proper subfields of $K$ containing $F$.

then:

\[K _ 1 \cup \cdots \cup K _ m < K\]

But this means that $z$ is not fixed by any $g _ i$, so the only element of $G$ fixing $z$ is $1$.

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Case 2: $F$ finite.

Note that $K^\times$ is a finite abelian group. Hence by the structure theorem,

\[K^\times \cong \left( \frac{\mathbb Z}{p_i \mathbb Z} \right)^{k_1} \times \cdots \times \left( \frac{\mathbb Z}{p_n \mathbb Z} \right)^{k_n}\]

In fact, $k _ i = 1$ for all $i$. If $k _ i > 1$, then it would contain a subgroup of the form $\left(\frac{\mathbb Z}{p _ i \mathbb Z}\right)^2$. All elements of the subgroup must satisfy $x^{p _ i} = 1$. Therefore all elements are roots of $x^{p _ i} - 1$, but $x^{p _ i} - 1$ has at most $p _ i$ roots, not $p _ i^2$. Hence each $k _ i = 1$.

This implies that

\[K^\times \cong \frac{\mathbb Z}{p_i \mathbb Z} \times \cdots \times \frac{\mathbb Z}{p_n \mathbb Z}\]

and hence

\[K^\times \cong \frac{\mathbb Z}{p_1 \cdots p_n \mathbb Z}\]

and so $K = \langle z \rangle$ for some $z \in K$. Hence if $\varphi \in \text{Stab} _ G(z)$, then $\varphi(z) = z$ and $\varphi$ fixes all of $K$, so $\varphi = \text{id}$. Hence we have found the required $z$.

(this is pretty much just a proof that all extensions of finite fields are simple, and then using the fact that the generator has the required property).

There is a unique extension of $\varphi$ to $\varphi^\ast$ where

\[\varphi^\ast : F(\alpha) \to \tilde F(\tilde \alpha)\]

such that $\varphi^\ast (\alpha) = \tilde \alpha$.

In other words, if you’ve got an isomorphism between two fields, and two elements of an extension with the important “minimal polynomial preserving” property, then it extends to a field extension of the field adjoined with these two elements. This could be summarised by the diagram:

Existence. By assumption, $\varphi(m _ {F, \alpha})(\tilde \alpha) = 0$. Then $\varphi(m _ {F, \alpha}) \mid m _ {\tilde F, \tilde \alpha}$, but as $m _ {\tilde F, \tilde \alpha}$ is irreducible, it follows that $\varphi(m _ {F, \alpha}) = m _ {\tilde F, \tilde \alpha}$.

We have the isomorphisms

\[\begin{aligned} &\theta : \frac{F[t]}{\langle m_{F, \alpha} \rangle} \stackrel\cong\longrightarrow F(\alpha) \\\\ &\tilde \theta : \frac{\tilde F[t]}{\langle m_{\tilde F, \tilde \alpha}\rangle} \stackrel\cong\longrightarrow\tilde F(\tilde \alpha) \end{aligned}\]

And $\varphi : F \to \tilde F$ extends to an isomorphism

\[\varphi' : F[t] \stackrel\cong\longrightarrow\tilde F[t]\]

by fixing $t$. As $\varphi’$ sends $\langle m _ {F, \alpha}\rangle$ to $\langle m _ {\tilde F, \tilde \alpha} \rangle$, by the universal property of quotients this map descends to give

\[\overline \varphi : \frac{F[t]}{\langle m_{F, \alpha}\rangle} \longrightarrow \frac{\tilde F[t]}{\langle m_{\tilde F, \tilde \alpha}\rangle}\]

Chaining isomorphism together, we have

\[\begin{aligned} &\varphi^\ast : F(\alpha) \to \tilde F(\tilde \alpha) \\\\ &\varphi^\ast := \tilde \theta \circ \overline \varphi \circ \theta^{-1} \end{aligned}\]

and this gives the existence of the required isomorphism.

This can be summarised with the following commutative diagram, where $\tilde \varphi$ should be replaced by $\varphi^\ast$.

Uniqueness. Suppose that $\psi : F(\alpha) \to \tilde F(\tilde \alpha)$ is another isomorphism with the property that $\psi(\alpha) = \tilde \alpha$ (this is what we are assuming in the construction).

Then $\psi$ agrees with $\varphi^\ast$ on $F$ and $\alpha$.

Since $F$ and $\alpha$ generate $F[\alpha]$, and $F[\alpha] = F(\alpha)$ ($\alpha$ is algebraic) it follows that $\psi = \varphi^\ast$, so we have uniqueness as required.

There are at least $[K : F]$ distinct isomorphisms extending $\varphi$ to an isomorphism $\phi’ : K \to \tilde K$.

Overall idea: Inductively build on a previous result that establishes extensions of isomorphisms to $F(\alpha) \to F(\tilde \alpha)$ under the assumption that $\varphi(m _ {F, \alpha})(\tilde \alpha) = 0$. This is done by applying the result to the roots of $f$.

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We proceed by induction on $[K : F]$.

Base case: $[K:F] = 1$. Then $K \cong F$, so the chain of isomorphisms $K \cong F \cong \tilde F \cong \tilde K$ gives at least one isomorphism $K \to \tilde K$.

Inductive step: Let $g$ be a monic irreducible factor of $f$ in $F[t]$ with $\deg g \ge 2$. Then $g$ is separable in $F[t]$. (it’s not necessary to assume that the field is of characteristic zero, since one of the assumptions above is that $f$ is separable).

Then $\tilde g := \varphi(g)$ is also separable (since if it weren’t this would also imply $g$ wasn’t).

Since $\tilde K$ is a splitting field of $\tilde f$ and $\tilde g \mid \tilde f$, $\tilde g$ has exactly $n := \deg \tilde g$ roots, say $\beta _ 1, \ldots, \beta _ n \in \tilde K$ by the result that all roots of separable irreducible polynomials are simple.

Choose a root $\alpha \in K$ of $g$. Since $g(\alpha) = 0$, by the “division property” of minimal polynomials, $m _ {F, \alpha} \mid g$. But then by assumption $g$ is monic and irreducible over $F$, so it must actually be the case $m _ {F, \alpha} = g$.

Pick some $i = 1, \ldots, n$. Then $\phi(m _ {F, \alpha})(\beta _ i) = \phi(g)(\beta _ i) = \tilde g(\beta _ i) = 0$, so the conditions of the lemma that

If

• $\varphi : F \to \tilde F$ is an isomorphism between two fields
• $K / F$ and $\tilde F / \tilde K$ are finite extensions
• $\alpha \in K$, $\tilde \alpha \in \tilde K$
• $\varphi(m _ {F, \alpha})(\tilde \alpha) = 0$

Then there is a unique extension of $\varphi$ to

\[\varphi^\ast : F(\alpha) \to \tilde F(\tilde \alpha)\]

such that $\varphi^\ast (\alpha) = \tilde \alpha$.

are satisfied (where $\alpha$ is also $\alpha$ in the above, and $\beta _ i$ plays the role of $\tilde \alpha$) . Hence we obtain a unique isomorphism $\phi _ i : F(\alpha) \to \tilde F(\beta _ i)$ which extends $\phi : F \to \tilde F$ and sends $\alpha$ to $\beta _ i$.

Let $m := [K : F(\alpha)]$. As $m < [K : F]$, we can apply the inducitve hypothesis to $K/F(\alpha)$ and $\tilde K / \tilde F(\beta _ i)$ together with the isomorphism to find at least $m$ different extensions

\[\phi_i^{(j)} : K \to \tilde K \quad\text{ for }j = 1, \ldots, m\]

It remains to show that these are all distinct. If $\phi _ {i}^{(j)} = \phi _ {i’}^{(j’)}$ for some $1 \le i, i’ \le n$ and $1 \le j, j’ \le m$, we have

\[\begin{aligned} \beta_i &= \phi_i^{(j)}(\alpha) \\\\ &= \phi_{i'}^{(j')}(\alpha) \\\\ &= \beta_{i'} \end{aligned}\]

So $i = i’$ and by induction $j = j’$ (since they are distinct in the base case and the inductive hypothesis guarantees the distinctness of the isomorphisms $K/F(\alpha) \to \tilde K / \tilde F(\beta _ i)$ for each fixed $i$).

Therefore there are at least $mn$ different extensions ($n$ for the initial choice of $\beta _ i$, and $m$ constructed inductively).

Since $[F(\alpha):F] = \deg m _ {F, \alpha} = \deg g = \deg \tilde g = n$, it follows that $[K : F] = [K : F(\alpha)] [F(\alpha) : F] = mn$ and we are done.

Suppose:

• $K/F$ is a Galois extension

Then:

\[|\text{Gal}(K/F)| \ge [K : F]\]

Proof: Since $K/F$ is a Galois extension, by definition $K$ is the splitting field for some polynomial $f \in F[t]$. In the above result, take $\tilde F := F$ and $\varphi : F \to \tilde F$ as the identity map. Then there are at least $[K : F]$ distinct automorphisms of $K$ extending $1 : F \to F$.

Automorphisms of this type are exactly the $F$-linear automorphisms of $K$, and hence

\[|\text{Gal}(K/F)| \ge [K : F]\]

as required.

\[|\text{Gal}(K/F)| = [K : F]\]

Suppose:

• $K/F$ is a Galois extension

Then:

\[\vert \text{Gal}(K/F) \vert \ge [K : F]\]

and

Suppose:

• $K/F$ is a finite field extension

Then:

\[\vert \text{Gal}(K/F) \vert \le [K : F]\]

• $\text{Gal}(K/K^G) \subseteq G$: $K^G$ is a subfield containing $F$, so every $K^G$-linear automorphism $\sigma$ of $K$ is also $F$-linear, hence $\sigma \in G$.
• $\text{Gal}(K/K^G) \supseteq G$: If $\sigma : K \to K$ is $F$-linear, then $\sigma \in G$ and so $\sigma$ fixes $K^G$ pointwise. Therefore $\sigma$ is also $K^G$-linear.