Notes - Galois Theory HT25, Main theorems of Galois theory

Bijective correspondence of subgroups and subfields: The function $L \mapsto \text{Gal}(K/L)$ is a bijection with inverse $H \mapsto K^H$, and establishes a correspondence between the intermediate fields $F \subseteq L \subseteq K$ and the subgroups of $H \le \text{Gal}(K/F)$.

The intermediate subfields which are Galois correspond to the normal subgroups of $G$: Intermediate subfields $F \subseteq L \subseteq K$ that are Galois over $F$ correspond precisely with the normal subgroups $H$ of $G$ under the above mapping, and for any Galois extension $L/F$ we have:

\[\text{Gal}(L/F) \cong \frac{\text{Gal}(K/F)}{\text{Gal}(K/L)}\]

The correspondences are inclusion reversing: If $F \subseteq L _ 1 \subseteq L _ 2 \subseteq K$, then $\text{Gal}(K/L _ 2) \le \text{Gal}(K/L _ 1)$, and if $H _ 1 \le H _ 2 \le G$, then $K^{H _ 2} \le K^{H _ 1}$.

The indexes of the intermediate fields correspond to the indexes of the quotient groups: If $F \subseteq L \subseteq K$ is an intermediate field and $H = \text{Gal}(K/L)$, then

\[[L : F] = |G/H|\]

and

\[[K : L] = |H|\]

Since $K/F$ is Galois, by definition $K$ is a splitting field for some separable polynomial $f \in F[t]$. Since the roots of $f$ in $K$ still generate $K$ as a field and $f$ is still separable when viewed as an element of $L[t]$ (by a problem sheet result, @todo), it follows that $K/L$ is also Galois.

\[L = K^{\text{Gal}(K/L)}\]

This comes from two previous results:

If:

• $K/F$ is a Galois extension
• $F \subseteq L \subseteq K$ Then: $K/L$ is also Galois.

and

If:

• $K/F$ is a finite extension
• $G = \text{Gal}(K/F)$ Then the following are equivalent:
• $K/F$ is Galois.
• $ \vert G \vert = [K : F]$
• $F = K^G$

In the above, $K/L$ is also Galois. But then $L = K^{\text{Gal}(K/L)}$.

1: Consider the polynomial $f _ {H \cdot z}$ defined by

\[f_X := \prod_{y \in H \cdot z} (t - y)\]

Since $f _ {H \cdot z}(z) = 0$, it follows $z$ is algebraic over $K^H$.

2: Since $f _ {H \cdot z}(z) = 0$, $m _ {K^H, z}$ must divides $f _ {H \cdot z}$.

But also, $K \le \text{Gal}(K/K^H)$, so $H$ acts on the roots of $m _ {K^H, z}$ in $K$. Since $z$ is a root of $m _ {K^H, z}$ in $K$, it follows that the entire $H$-orbit $H\cdot z$ is contained in the set of roots of $m _ {K^H, z}$. Hence $f _ {H \cdot z} \mid m _ {K^H, z}$.

Since both $f _ {H \cdot z}$ and $m _ {K^H, z}$ are monic, there must be equality.

3. This follows from 2.

\[\text{Gal}(K/K^H) = H\]

Let $J := \text{Gal}(K/K^H)$. This is a finite group since $K/K^H$ is a finite extension (really? @todo), and $H \le J$, since any $K^H$-linear automorphism of $K$ will also be a $F$-linear automorphism of $K$.

Choose some $k \in K$ such that $\text{Stab} _ J(z) = \{1\}$, this is always possible by previous results (lower bounds on the size of Galois groups). Then by the orbit-stabiliser theorem,

\[|H| = |H\cdot z| = \deg m_{K^H, z}\]

and

\[|J| = |J \cdot z| = \deg m_{K^J, z}\]

Hence it is enough to show that $K^J = K^H$ (why?). Since $K/F$ is a Galois extension, the extension $K/K^H$ is also Galois with Galois group $\text{Gal}(K/K^H) = J$. By the result that says

If:

• $K/F$ is a finite extension
• $G = \text{Gal}(K/F)$ Then the following are equivalent:
• $K/F$ is Galois.
• $ \vert G \vert = [K : F]$
• $F = K^G$

It then follows that $K^J = K^H$.

The restriction map

\[r : \text{Gal}(K/F) \to \text{Gal}(L/F)\]

is a well-defined surjective group homomorphism.

Let $\sigma \in \text{Gal}(K/F)$.

Then $r(\sigma) = \sigma \vert _ L : L \to K$ is a well-defined element of $\text{Gal}(L/F)$ since $L$ is assumed to be $\text{Gal}(K/F)$-stable.

Therefore $r$ is well-defined and is a group homomorphism by its construction. It is also surjective (@todo).

Forward direction: Suppose $L/F$ is Galois. Then it is a splitting field of a separable polynomial $g \in F[t]$, so the roots $V(g)$ of $g$ in $L$ generate $L$. But $V(g)$ is $G$-stable by a previous result, and so $L$ is also $G$-stable.

Backward direction: Suppose that $L$ is $G$-stable. Then by the result that says

If:

• $K/F$ is a finite extension
• $G = \text{Gal}(K/F)$ Then the following are equivalent:
• $K/F$ is Galois.
• $ \vert G \vert = [K : F]$
• $F = K^G$

it is enough to show that $L^{\text{Gal}(L/F)} = F$. But $r(G) \le \text{Gal}(L/F)$ since the restriction map $r$ is a surjective group homomorphism by the result that says:

If

• $K/F$ is a Galois extension
• $L$ is an intermediate field
• $L$ is $\text{Gal}(K/F)$-stable Then the restriction map

\[r : \text{Gal}(K/F) \to \text{Gal}(L/F)\]

is a well-defined surjective group homomorphism.

Hence $L^{\text{Gal}(L/F)} \subseteq L^{r(G)} = L \cap K^G$. Since $K^G = F$ by the first result above, it follows that $L^{\text{Gal}(L/F)} = F$.

\[K^{\varphi H \varphi^{-1} } = \varphi(K^H)\]

Let $x \in K$ and $\psi \in H$. Then $(\varphi \psi \varphi^{-1}) \cdot x = x$ iff $\psi(\varphi^{-1} x) = \varphi^{-1}x$. So $x \in K^{\varphi H \varphi^{-1}} \iff \varphi^{-1}(x) \in K^H \iff x \in \varphi(K^H)$.

1. $H$ is normal in $G$ iff $L$ is Galois over $F$.
2. If $H$ is normal in $G$, the restriction map $\text{Gal}(K/F) \to \text{Gal}(L/F)$ induces a group isomorphism $G/H \cong \text{Gal}(L/F)$.

(1): By the result that says

Suppose:

• $K/F$ is a Galois extension
• $L$ in $F \subseteq L \subseteq K$ is an intermediate field Then:

\[L = K^{\text{Gal}(K/L)}\]

we have $L = K^H$. Then $H$ is normal in $G$ iff $\varphi H \varphi^{-1} = H$ for all $\varphi \in G$. By the above theorem and the result that says:

If:

• $K/F$ is a Galois extension
• $H \le \text{Gal}(K/F)$ Then:

\[\text{Gal}(K/K^H) = H\]

This implies $K^H = K^{\varphi H \varphi^{-1}}$ for all $\varphi \in G$. But since $K^{\varphi H \varphi^{-1}} = \varphi(K^H)$ by a previous result, this is equivalent to $K^H$ being