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 $K^{H _ 2} \subseteq K^{H _ 1}$.

---

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

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

$K/L$ is also Galois.

• Since $K/F$ is Galois, by definition $K$ is a splitting field for some separable polynomial $f \in F[t]$.
• As the roots of $f$ in $K$ still generate $K$ as a field and $f$ is still separable when viewed as an element of $L[t]$ (this follows from previous results), $K/L$ is also Galois.

@important~ (main theorem)

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

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

\[f _ {H \cdot z}(t) := \prod _ {y \in H \cdot z} (t - y)\]

Since $f _ {H \cdot z}(z) = 0$ and $f _ {H \cdot z}$ has coefficients in $K^H$ (by previous results), 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 and $H \le J$, since almost tautologically

\[\begin{aligned} \varphi \in H &\implies \varphi \text{ is a }K^H\text{-linear aut. of }K \\\\ &\implies \varphi \in \text{Gal}(K/K^H) = J \end{aligned}\]

We aim to show that $ \vert H \vert = \vert J \vert $, since this together with $H \le J$ implies $H = J$.

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

\[|H| = |H\cdot z| \cdot |\lbrace \text{id}\rbrace| = \deg m _ {K^H, z}\]

and

\[|J| = |J \cdot z| \cdot |\lbrace \text{id}\rbrace| = \deg m _ {K^J, z}\]

Hence it is enough to show that $K^J = K^H$, since then these polynomials will be the same and $ \vert H \vert = \vert J \vert $. 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 $K^J = K^H$.

@important~ (main theorem)

The restriction map

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

is a well-defined surjective group homomorphism.

• Well-defined: If $\sigma \in \text{Gal}(K/F)$, then $r(\sigma) = \sigma \vert _ L : L \to K$ is an element of $\text{Gal}(L/F)$ since $L$ is $\text{Gal}(K/F)$-stable.
• Group homomorphism: This is by construction.
• Surjective: Any $F$-stable automorphism of $L$ extends to an automorphism of $K$ by the theorem:

Suppose:

• $\varphi : F \to \tilde F$ is an isomorphism of fields
• $f \in F[t]$ is a separable polynomial
• $K$ is a splitting field for $f$
• $\tilde f = \varphi(f)$
• $\tilde K$ is a splitting field for $\tilde f$

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

• $L/F$ is Galois iff $L$ is $G$-stable.
• $L/F$ is Galois iff $\text{Gal}(K/L) \trianglelefteq G$

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, 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 $L^{\text{Gal}(L/F)} = F$. Recall also the result that:

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 $r(G) \le \text{Gal}(L/F)$. But then

\[\begin{aligned} L^{\text{Gal}(L/F)} &\subseteq L^{r(G)} &&(\text{since } r(G) \le \text{Gal}(L/F)) \\\\ &= \lbrace x \in L \mid \tau(x) = x \text{ for all } \tau \in r(G) \rbrace \\\\ &= \lbrace x \in L \mid \sigma(x) = x \text{ for all } \sigma \in G \rbrace && (\tau(x) = x \iff \sigma(x) = x \text { when }x \in L) \\\\ &= L \cap \lbrace x \in K \mid \sigma(x) = x \text{ for all } \sigma \in G \rbrace \\\\ &= L \cap K^G \\\\ &= L \cap F &&(G\text{ is Galois}) \\\\ &= F \end{aligned}\]

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

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 $x \in K^{\varphi H \varphi^{-1} } \iff \varphi^{-1}(x) \in K^H \iff x \in \varphi(K^H)$.

$H$ is normal in $G$ iff $L$ is Galois over $F$.

• The restriction map $\text{Gal}(K/F) \to \text{Gal}(L/F)$ induces a group isomorphism $G/H \cong \text{Gal}(L/F)$.

(1):

\[\begin{align} H \trianglelefteq G &\iff \varphi H \varphi^{-1} = H \text{ for all } \varphi \in G \\\\ &\iff K^{\varphi H \varphi^{-1} } = K^H \text{ for all } \varphi \in G && \\\\ &\iff \varphi(K^H) = K^H \text{ for all }\varphi \in G && (1\star)\\\\ &\iff K^H \text{ is }G\text{-stable} \\\\ &\iff K^H / F \text{ is Galois} &&(2\star) \\\\ &\iff L/F \text{ is Galois} &&(3\star) \end{align}\]

where:

• $(1\star)$ is justified by the general result that $K^{\varphi H \varphi^{-1} } = \varphi(K^H)$.
• $(2\star)$ is justified by the result that says if $K/F$ is a Galois extension with $G = \text{Gal}(K/F)$ and $L$ is an intermediate field, then $L/F$ is Galois iff $L$ is $G$-stable
• $(3\star)$ is justified by one of the main theorems of Galois theory; namely that if $K/F$ is a Galois extension and $F \subseteq L \subseteq K$ is an intermediate field, then $L = K^{\text{Gal}(K/L)}$.

(2): Suppose that $H$ is normal in $G$. Then $L$ is Galois over $F$ by (1), so $L$ is $G$-stable.

Hence $r : G \to \text{Gal}(L/F)$ is a well-defined surjective group homomorphism. The kernel of this map is

\[\begin{aligned} \ker r &= \\{\varphi \in G : \varphi| _ L = 1 _ L\\} \\\\ &= \text{Gal}(K/L) \\\\ &= H \end{aligned}\]

Then by the first isomorphism theorem,

\[G/(\ker r) = G/H \cong \text{Gal}(L/F)\]

as required.

@important~ (main theorem)

Since $G := \text{Gal}(K/F)$ is cyclic, it has exactly one subgroup for each divisor of $60$. Since $60 = 2^2 \times 3 \times 5$, there are 12 intermediate fields.

@exam~ @example~

Let:

• $F = \mathbb F _ p(t _ 1, t _ 2)$, i.e. the field of fractions of $\mathbb F _ p[t _ 1, t _ 2]$
• $K = F(\alpha, \beta)$, where:
• $\alpha$ is a root of $x^p - t _ 1$
• $\beta$ is a root of $x^p - t _ 2$

Then $K/F$ is a field extension. Now consider intermediate extensions $L = F(\alpha + \lambda \beta)$ for $\lambda \in F$.

Note that as $F$ is infinite, we may consider an infinite number of fields $F(\alpha + \lambda \beta)$, so it suffices to show that

\[F(\alpha + \lambda \beta) \ne F(\alpha + \mu \beta)\]

for distinct $\lambda, \mu \in F$

Suppose for a contradiction that $F(\alpha + \lambda \beta) = F(\alpha + \mu \beta)$. Then $\alpha + \mu \beta \in F(\alpha + \lambda \beta)$, so

\[\begin{aligned} (u - \lambda) \beta &= (\alpha - \alpha) + (\mu - \lambda)\beta \\\\ &= (\alpha + \mu \beta) - (\alpha + \lambda \beta) \\\\ &\in F(\alpha + \lambda \beta) \end{aligned}\]

This implies $\beta \in F(\alpha + \lambda \beta)$, as $u - \lambda$ is an element of $F$.

This implies also that $\alpha = \alpha + \lambda \beta - \lambda \beta \in F(\alpha + \lambda \beta)$.

Thus $F(\alpha + \lambda \beta) = F(\alpha, \beta)$. But this is impossible as

\[(\alpha + \lambda \beta)^p = \alpha^p + \lambda^p \beta^p = t_1 + \lambda^p t_2 \in F\]

and therefore $m _ {F, a + \lambda \beta}(x) = x^p - (t _ 1 + \lambda^p t _ 2)$, so:

\[[F(\alpha + \lambda \beta) : F] = p\]

but

\[[F(\alpha + \lambda \beta) : F] = [F(\alpha, \beta) : F] = p^2\]

and $p \ne p^2$.

So each $F(\alpha + \lambda \beta)$ is distinct, and so there are infinitely many subfields.

@exam~