Notes - Galois Theory HT25, Solvability by radicals

There exists a chain of intermediate subfields

\[F = F _ 0 \subset F _ 1 \subset \cdots \subset F _ n = K\]

such that for each $i = 1, \ldots, n$, there exist $\alpha _ i \in F _ i$ and positive integers $d _ i$ with

\[F _ i = F _ {i-1}(\alpha _ i)\]

and for each $1 \le i \le n$:

\[\alpha _ i^{d _ i} \in F _ {i-1}\]

Suppose:

• $F$ is a field of characteristic zero
• $\alpha$ is algebraic over $F$

Then:

• $\alpha$ lies in a radical extension of $F$ if and only if $\text{Gal} _ F(m _ {F, \alpha})$ is solvable.

Whenever $g \in F[t]$ is an irreducible polynomial such that $g$ has a root in $K$, $g$ splits completely in $K[t]$.

“One out, all out!”, Ian Stewart

$K/F$ is Galois $\implies$ $K/F$ is normal

Choose a separable $f \in F[t]$ such that $K$ is a splitting field of $f$. Pick $g \in F[t]$ irreducible where $g$ has a root in $K$.

Since any polynomial has a splitting field, find a splitting field of $fg \in K[t]$, say $L$, where $L/F$. Suppose that $\theta _ 1$ and $\theta _ 2$ are zeroes of $g$ in $L$, and consider the following diagram:

We aim to conclude that $[K(\theta _ i) : K] = 1$, so that we know $\theta _ i \in K$.

By two applications of the tower law, we have that for $i = 1$ or $i = 2$,

\[[K(\theta _ i) : F] = [K(\theta _ i) : K] [K:F] =\]

and

\[[K(\theta _ i): F] = [K(\theta _ i) : F(\theta _ i)][F(\theta _ i): F]\]

corresponding to the two paths through the diagram:

Since $g$ is irreducible,

\[g = m _ {F, \theta _ 1} = m _ {F, \theta _ 2}\]

and hence:

\[[F(\theta _ 1):F] = \deg g = [F(\theta _ 2) : F]\]

by the lemma that says the degree of an algebraic extension like this agrees with the degree of the minimal polynomial.

By the result which says

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

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

by taking $\varphi$ as the identity map, we can find an $F$-linear isomorphism

\[\varphi : F(\theta _ 1) \to F(\theta _ 2)\]

where $\theta _ 1 \mapsto \theta _ 2$.

Since $K(\theta _ i)$ is a splitting field for $f$ over $F(\theta _ i)$ and $f$ is separable over $F(\theta _ i)$ (by the result that says if a polynomial is separable, it separable over any field extension), by the result that says

If:

• $\varphi : F \to \tilde F$ 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 $\varphi$.

We can extend $\varphi$ to an isomorphism of fields

\[K(\theta _ 1) \to K(\theta _ 2)\]

Therefore

\[[K(\theta _ 1) : F(\theta _ 1)] = [K(\theta _ 2) : F(\theta _ 2)]\]

and hence

\[\begin{aligned} \text{}[K(\theta _ 1) : K] &= \frac{[K(\theta _ 1) : F(\theta _ 1)][F(\theta _ 1) : F]}{[K : F]} \\\\ &= \frac{[K(\theta _ 2) : F(\theta _ 2)][F(\theta _ 2) : F]}{[K : F]} \\\\ &= [K(\theta _ 2) : F] \end{aligned}\]

and so if $\theta _ 1$ is a root of $g$ which lies in $K$,

\[[K(\theta _ 2) : K] = [K(\theta _ 1) : K]\]

for any other root of $\theta _ 2$ of $g$ in $L$.

Hence all roots of $g$ lie in $K$, so $L = K$ and $g$ splits completely over $K$.

Suppose:

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

Then:

• $m _ {F, \alpha}$ is separable, and
• There is a surjective group homomorphism $\text{Gal}(K/F) \to \text{Gal} _ F(m _ {F, \alpha})$

We induct on $[K : F]$. Since $K/F$ is radical, there is some $\alpha \in K/F$ and $d \ge 2$ such that $\alpha^d \in F$.

Choose the pair $(\alpha, d)$ such that $d$ is as small as possible.

If $d$ is not prime, then $d = mn$ with $1 < m, n < d$.

Then $\alpha^m \notin F$, since $m < d$. But then $(\alpha^m)^n \in F$ provides a counterexample to the minimality of $d$.

Hence $d$ is prime.

Since $\alpha \notin F$, $\deg m _ {F, \alpha} \ge 2$.

Thus $m _ {F, \alpha}$ splits in $K$ by the result that says:

If $K/F$ is Galois, then $K/F$ is normal.

Then by the result that says:

If $K/F$ is a finite Galois extension and $\alpha \in K$, then $m _ {F, \alpha}$ is separable.

Hence by the result that says:

If $f \in F[t]$ is a separable irreducible polynomial and $f$ splits completely over $K/F$, $f$ has $\deg f$ distinct roots in $K$.

It follows that $\exists \beta \ne \alpha \in K$ such that $m _ {F, \alpha}(\beta) = 0$. Define

\[a \alpha^p \in F\]

Then $m _ {F, \alpha} \mid t^p - a$, and hence $\beta$ is a root of $t^p - a$ also.

Hence $\alpha^p = \beta^p = a$. Then by the result (the 3rd main theorem of Galois theory, [[Notes - Galois Theory HT25, Main theorems of Galois theory]]^U) that says:

If $K/F$ is a Galois extension with $G = \text{Gal}(K/F)$ and $L$ is an intermediate subfield with $H := \text{Gal}(K/F)$ and $L$ is an intermediate subfield with $H := \text{Gal}(K/L)$, then:

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 an isomorphism $G/H \cong \text{Gal}(L/F)$.

It follows that $G = \text{Gal}(K/F)$ has normal subgroups

\[H _ 1 = \text{Gal}(K/L) \trianglerighteq H _ 2 = \text{Gal}(K/M)\]

such that:

• $H _ 2$
• $H _ 1/H _ 2 \cong \text{Gal}(M/L)$
• $G/H _ 1 \cong \text{Gal}(L/F)$ are all solvable

Hence $G$ is also solvable.

Suppose:

• $K/F$ is a finite radical extension
• $\text{char }F = 0$

Then:

• There exists a finite Galois radical extension $M/F$ such that $M \supseteq K$.

@todo, come back to once I have done sheet 3

We induct on the length $r$ of the radical chain

\[F = F _ 0 \subseteq F _ 1 \subseteq F _ {r-1} \subseteq F _ r = K\]

Base case: When $r = 0$, this is the case when $K = F$. Since any field has a finite Galois radical extension (e.g. adjoining a $p$-th root of unity for a cyclotomic extension), the result is immediate.

Inductive step: Since $F _ {r-1} / F$ is radical, by the inductive hypothesis, there is some radical Galois finite extension $L$ of $F$ containing $F _ {r-1}$.

Let $\alpha$ be the element such that $F _ r = F _ {r-1}(\alpha)$ where $\alpha^d = \theta$ for some $\theta \in F _ {r-1}$, and let $G = \text{Gal}(L/F)$. By the result that says:

If $H$ is a finite group of automorphisms of a field $L$ and $X \subseteq L$ is a finite subset, then

\[f _ X := \prod _ {y \in X} (t-y) \in L[t]\]

is a separable polynomial, and if $X$ is $H$-stable, $f _ X$ has coefficients in $L^H$.

we can consider

\[f _ {G \cdot \theta} := \prod _ {\psi \in G \cdot \theta} (t - \psi) \in L[t]\]

and then $f _ {G \cdot \theta}$ has coefficients in $L^G$. By the result that says:

If $K/F$ is a finite extension and $G = \text{Gal}(K/F)$ and $K/F$ is Galois, then $F = K^G$.

we have that $L^G = F$. Thus

\[g(t) := f _ {G \cdot \theta} (t^d)\]

also lies in $F[t]$. Using the result that says and polynomial has a splitting field, choose a splitting field $M$ of $g$ containing $L$.

This construction means that $M$ is generated as an extension of $L$ by roots of $t^d - \psi$ for each $\psi \in G \cdot \theta$.

Hence $M/L$ is radical (@todo, why?).

Since $L/F$ is radical by induction, so is $M/F$.

Since $L/F$ is Galois, it is a splitting field of some $h \in F[t]$.

Since $hg \in F[t]$ and $M$ is generated by the roots of $hg$ together with $F$, it is also a splitting field of $hg$ over $F$.

Because $F$ has characteristic $0$ (by assumption), $hg$ is separable (since any polynomial over a field of zero characteristic).

So $M/F$ is a radical and Galois extension.

We want to show that there is then an embedding $F _ r \hookrightarrow M$ (recall an embedding is an injective homomorphism) that makes the following diagram

commute.

Since $\alpha^d = \theta$, $m _ {F _ {r-1}, \alpha} \mid t^d - \theta$ in $F _ {r-1}[t]$. But $t^d - \theta$ divides $g$ in $L[t]$, so there is some $q \in L[t]$ such that

\[g = m _ {F _ {r-1}, \alpha} q\]

Since $g$ splits completely in $M[t]$, we can find some $\beta \in M$ such that $m _ {F _ {r-1}, \alpha}(\beta) = 0$. By the result that says

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

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

there is an embedding

\[F _ r = F _ {r-1} (\alpha) \hookrightarrow F _ {r-1} (\beta) \subseteq M\]

that sends $\alpha$ to $\beta$, as required.

By the result that lets us enlarge radical extensions:

If:

• $K/F$ is a finite radical extension
• $\text{char }F = 0$ Then:
• There exists a finite Galois radical extension $M/F$ such that $M \supseteq K$.

we can enlarge $K$ and therefore assume that $K/F$ is Galois over $F$. Then $G := \text{Gal}(K/F)$ is a solvable group by the result that says

If $K/F$ is a radical Galois extension then $\text{Gal}(K/F)$ is solvable.

Since $\text{Gal} _ F(m _ {F, \alpha})$ is a homomorphic image of $G$ by the result that says

Suppose:

• $K/F$ is a finite Galois extension
• $\alpha \in K$ Then:
• $m _ {F, \alpha}$ is separable, and
• There is a surjective group homomorphism $\text{Gal}(K/F) \to \text{Gal} _ F(m _ {F, \alpha})$

it follows that it is also solvable (why? @todo).

Let $V(f)$ be the set of roots of $f$. Then the action of $G$ on $K$ restricts to an action on $V(f)$. Hence we can consider

\[\rho : G \to \text{Sym}(V(f)) \cong S _ n\]

This map is injective, since if $\rho(\sigma) = \text{id}$ for some $\rho \in G$, then $\sigma$ fixes $V(f)$ pointwise. But $V(f)$ generates $K$ as a field together with $F$, so $\sigma$ fixes all elements of $K$ and hence $\sigma = 1$. Therefore

\[G \cong \rho(G) \le S _ n\]

By Eisenstein’s Criterion at $p = 3$ together with Gauss’ lemma, $f$ is irreducible over $\mathbb Q$ and it is also separable over $\mathbb Q$. Therefore it has 5 distinct roots in a splitting field $K$.

Since $\mathbb C$ is algebraically closed, we will identify $K$ with a subfield of $\mathbb C$ and will identify

\[G = \text{Gal} _ {\mathbb Q}(f) = \text{Gal}(K/\mathbb Q)\]

with a subgroup of $S _ 5$.

Since:

• $f(-2) = -17 < 0$
• $f(0) = 3 > 0$
• $f(1) = -2 < 0$
• $f(2) = 23 > 0$

It follows that $f$ has a real root in each of $(-2, 0)$, $(0, 1)$ and $(1, 2)$.

Suppose that $f$ has five real roots. Then by the Mean Value Theorem,

\[f' = 5t^4 - 6\]

would have at least four real roots, which is not the case.

Hence $f$ has precisely three real roots. So complex conjugation $c : \mathbb C \to \mathbb C$ preserves $K$ and hence $c \vert _ K \in G$ and $c \vert _ K$ is a transposition: $c$ fixes the three real roots and swaps the two non-real roots of $f$.

As $f$ is irreducible over $\mathbb Q$, $[\mathbb Q(z) : \mathbb Q] = 5$ for any $z \in V(f)$. Hence $5$ divides $[K : \mathbb Q]$, which is equal to $ \vert G \vert $.

By Cauchy’s Theorem, $G$ then contains an element $\sigma$ of order $5$. Then $\sigma$ is necessarily a $5$-cycle.

Since $S _ 5$ is generated by a transposition and a $5$-cycle, it follows that $G = S _ 5$.

But $S _ 5$ is not solvable, so no roots of $f$ can lie in a radical extension of $\mathbb Q$.

Suppose:

• $K/F$ is a Galois extension
• $[K : F] = p$ where $p$ prime
• For some $\varepsilon \in F$ where $\varepsilon \ne 1$, we have $\varepsilon^p = 1$

Then there exists $u \in K$ such that $u^p \in F$ and $K = F(u)$.

Let $K/F$ be a splitting field of $m _ {F, \alpha}$. Proceed by induction on $[K : F]$.

Suppose

\[G := \text{Gal} _ F(m _ {F, \alpha}) = \text{Gal}(K/F)\]

is solvable. Then we can find a normal subgroup $H$ of $G$ such that $p := [G : H]$ is prime (why, @todo).

Let $L = K^H$ be the corresponding intermediate subfield.

Let $M$ be a splitting field of $(t^p - 1) m _ {F, \alpha}$ containing $K$. Since $\text{char } F = 0$, $M$ is Galois over $F$.

Let $\varepsilon \in M$ be a root of $t^p - 1$ such that $\varepsilon \ne 1$. Then $M = K(\varepsilon)$. By the result that says:

Let $K/F$ be a Galois extension and let $F \subseteq L \subseteq K$ be an intermediate field. Then $K/L$ is also Galois.

$K(\varepsilon)$ is Galois over $L(\varepsilon)$. But then using the result that

If $K/F$ is a finite Galois extension and $\alpha \in K$, then $m _ {F, \alpha}$ is separable and there is a surjective group homomorphism $\text{Gal}(K/F) \to \text{Gal} _ F(m _ {F, \alpha})$.

there is a surjective group homomorphism

\[\text{Gal}(K(\varepsilon) / L(\varepsilon)) \to \text{Gal} _ {L(\varepsilon)}(m _ {L(\varepsilon), \alpha})\]

Since $K$ is also Galois over $L$, by the result that says

Let $K/F$ be a Galois extension with $G = \text{Gal}(K/F)$. Let $L$ be an intermediate field. Then $L/F$ is Galois if and only if $L$ is $G$-stable.

it follows that the subfield $K$ of $K(\varepsilon)$ is $\text{Gal}(K(\varepsilon)/L)$-stable.

This gives a well-defined restriction map

\[\text{Gal}(K(\varepsilon)/L(\varepsilon)) \hookrightarrow \text{Gal}(K(\varepsilon) / L) \to \text{Gal}(K/L)\]

This restriction map is injective, since an $L(\varepsilon)$-linear automorphism of $K(\varepsilon)$ fixing $K$ must fix all of $K(\varepsilon)$.

Hence $\text{Gal}(K(\varepsilon)/L(\varepsilon))$ is isomorphic to a subgroup of $\text{Gal}(K/L)$. Hence

\[\text{Gal} _ {L(\varepsilon)}(m _ {L(\varepsilon), \alpha})\]

is isomorphic to a subquotient (@todo, what’s a subquotient) of the finite solvable group $\text{Gal}(K/F)$ and it is therefore solvable. Hence

\[\begin{aligned} |\text{Gal} _ {L(\varepsilon)}(m _ {L(\varepsilon), \alpha})| &\le |K(\varepsilon) : L(\varepsilon)| \\\\ &\le [K : L] \\\\ &= \frac 1 p [K : F] \end{aligned}\]

Hence by induction, there exists a radical extension $R$ of $L(\varepsilon)$ containing $\alpha$. This can be summarised by the diagram

Consider the extension $L(\varepsilon) / F$. Since $L = K^H$ and $H$ is Galois over $F$ by the theorem that says

Let $K/F$ be a Galois extension with $G = \text{Gal}(K/F)$ and let $L$ be an intermediate subfield with $H := \text{Gal}(K/L)$. Then:

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

Also $F(\varepsilon) / F$ is Galois by the result that says

If:

• $p$ is a prime
• $L / F$ is a field extension such that $L = F(\varepsilon)$ where $\varepsilon \in L$ and $\varepsilon^p = 1$ Then:
• $L/F$ is Galois
• $\text{Gal}(L/F)$ is abelian

Let $\sigma \in \text{Gal}(K(\varepsilon) / F)$. Then $\sigma$ preserves both $L$ and $F(\varepsilon)$, so it preserves $L(\varepsilon)$. Hence $L(\varepsilon)$ is Galois over $F$ by the result that says:

$L/F$ is Galois iff $L$ is $G$-stable.

Then the restriction map

\[\text{Gal}(L(\varepsilon) / F(\varepsilon)) \to \text{Gal}(L/F)\]

is injective. By the result that says

If $K$ is Galois over $F$, then $ \vert G \vert = [K : F]$.

We know $[L(\varepsilon) : F(\varepsilon)]$ divides $[L : F] = p$, so it is $1$ or $p$. Hence $L(\varepsilon) / F(\varepsilon)$ is radical by the result that says:

If:

• $K/F$ is a Galois extension
• $[K : F] = p$ where $p$ prime
• For some $\varepsilon \in F$ where $\varepsilon \ne 1$, we have $\varepsilon^p = 1$ then there exists $u \in K$ such that $u^p \in F$ and $K = F(u)$.

and $F(\varepsilon) / F$ is radical by definition. Thus $R/F$ is also radical.