Notes - 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, @todo. This is Q6 of Sheet 1.