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 $1 \le i \le n$:

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

Overall idea: Use induction on the degree of the field extension. Find an intermediate Kummer and cyclotomic extension to start off the subnormal series, and then apply the inductive hypothesis to the the intermediate field to finish it off.

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We induct on $[K : F]$. If $[K : F] = 1$, then the result is immediate since $\text{Gal}(K/F)$ is just the trivial group.

Now suppose $[K : F] > 1$. Since $K/F$ is radical, there is some $\alpha \in K$ and $p \ge 2$ such that $\alpha^p \in F$. Choose a pair $(\alpha, p)$ such that $p$ is as small as possible. Suppose for a contradiction that $p$ were not prime, then $p = mn$ with $1 < m, n < p$. Then $\alpha^m \notin F$, as $m < p$. But then $(\alpha^m)^n \in F$ provides a counterexample to the minimality of $(\alpha, p)$, and so $p$ must be prime.

Since $\alpha \notin F$, $\deg m _ {F, \alpha} \ge 2$. Note $m _ {F, \alpha}$ splits in $K$ as $K/F$ is Galois and so in particular normal. Likewise, as $K/F$ is Galois, $m _ {F, \alpha}$ must also be 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 there exists some $\beta \ne \alpha \in K$ such that $m _ {F, \alpha}(\beta) = 0$. Define $a := \alpha^p \in F$. Since $m _ {F, \alpha} \mid t^p - a$, $\beta$ is a root of $t^p - a$ and so $\alpha^p = \beta^p = a$.

Let $\varepsilon := \alpha / \beta \in K$ and let $L := F(\varepsilon)$ and let $M := L(\alpha) = F(\varepsilon, \alpha)$. Since the polynomial $t^p - a \in F[t]$ splits completely in $M$ and has no repeated roots, it is separable and therefore $M$ is a Galois extension of $F$.

Since $\alpha \in M \setminus F$, we have $[M : F] > 1$, so $[K : M] < [K : F]$. Since $K/F$ is a radical Galois extension, so is $K/M$. Hence by induction, $\text{Gal}(K/M)$ is solvable.

Now we aim to prove the following is a subnormal series with abelian quotients:

\[G = \text{Gal}(K/F) \stackrel{(1)}\trianglerighteq \text{Gal}(K/L) \stackrel{(2)}{\trianglerighteq} \text{Gal}(K/M) \stackrel{(3)}{\trianglerighteq \cdots \trianglerighteq} \\{e\\}\]

• For 1:
  • It is a normal subgroup: By the Galois correspondence, $\text{Gal}(K/L)$ is a normal subgroup of $\text{Gal}(K/F)$ if $L/F$ is a Galois extension. $L/F$ is cyclotomic, so it is Galois.
  • Has an abelian quotient: By the Galois correspondence, the restriction map induces an isomorphism $\text{Gal}(K/F) / \text{Gal}(K/L) \cong \text{Gal}(L/F)$, which is abelian as $L/F$ is cyclotomic
• For 2:
  • It is a normal subgroup: Likewise, this follows from $M/L$ being Kummer, so it is Galois.
  • Has an abelian quotient: Likewise, this follows from $\text{Gal}(M/L)$ being abelian as $M/L$ is Kummer.
• For 3:
  • This follows from the inductive hypothesis.

Hence $G$ is solvable.

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Some results used, written more explicitly:

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

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

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

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

Suppose:

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

Then:

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

Overall idea: We induct on the length $r$ of the radical chain

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

In the inductive step, we know $F _ {r-1}$ is contained in some radical Galois extension $L/F$ and that there is some $\alpha$ in $F _ r$ such that $\alpha^d = \theta \in F _ {r-1}$. Then we carefully check that a splitting field of $f _ {\text{Gal}(L/F) \cdot \theta}(t^d)$ gives the required radical extension $M$, carefully checking that: $M/F$ is radical, $M/F$ is Galois, and that $M$ actually contains $K$.

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

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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

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

and then $f _ {G \cdot \theta}$ has coefficients in $L^G$, and $L^G = F$ as $L/F$ is Galois. Thus

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

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

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

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

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

$M$ actually contains $K$:

Recall in the above notation $K = F _ r$. 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:

(most of the notes consider field extensions $K/F$ to mean $K \supseteq F$, but here we are using the “proper” definition that $K/F$ is a field extension if there is an embedding $F \hookrightarrow K$).

Since $\alpha^d = \theta$, we have $m _ {F _ {r-1}, \alpha} \mid t^d - \theta$ in $F _ {r-1}[t]$. But $t^d - \theta$ divides $g$ in $L[t]$, and so

\[m _ {F _ {r-1}, \alpha} \mid t^d - \theta \mid g\]

Thus there is some $q \in L[t]$ such that $g = m _ {F _ {r-1}, \alpha} q$.

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

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

by taking:

• Both $F$ and $\tilde F$ as $F _ {r-1}$
• $\alpha$ as $\alpha$ and $\tilde \alpha$ as $\beta$

we see that there is an embedding:

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

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

Suppose:

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

Then:

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

By the result that lets us enlarge radical extensions:

If:

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

Then:

• There exists a radical Galois 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 Galois extension
• $\alpha \in K$

Then:

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

it follows that it is also solvable: if

\[\text{Gal}(K/F) = G_0 \triangleright G_1 \triangleright \cdots \triangleright G_n = \lbrace e \rbrace\]

is a subnormal series with abelian quotients, we can apply $\varphi$ to obtain

\[\text{Gal} _ F(m _ {F, \alpha}) = \varphi(G_0) \triangleright \varphi(G_1) \triangleright \cdots \triangleright \varphi(G_n) = \lbrace e \rbrace\]

and note that each $\varphi(G _ i)$ is normal in $\varphi(G _ {i-1})$ and the quotients remain abelian.

Overall idea:

Induct on the degree of the splitting field $K/F$ and consider a prime-degree intermediate field $L$, which exists as $\text{Gal}(K/F)$ is solvable. Adjoin a $p$-th root of unity by consider the splitting field for $(t^p - 1)m _ {F, \alpha}(t)$. Then after some checking you can deduce $\text{Gal} _ {L(\varepsilon)}(m _ {L(\varepsilon), \alpha})$ is solvable and therefore by induction there is a radical extension of $L(\varepsilon)$ containing $\alpha$, say $R$.

But then you can check $F/F(\varepsilon)$ is radical, $L(\varepsilon) / F(\varepsilon)$ is radical and $R/L(\varepsilon)$ is radical, and so the overall extension $R/F$ is radical.

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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 (a justification for this fact is given at the end).

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 $M = K(\varepsilon)$ ($m _ {F, \alpha}$ splits in $K$, and $t^p - 1$ splits in $F(\varepsilon)$).

Then 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.

$M = K(\varepsilon)$ is Galois over $L(\varepsilon)$ as ($M$ is Galois over $F$, and $M/L(\varepsilon)/F$ is an intermediate field).

But then using the result that

If $K/F$ is a 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

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

Now recall the following result:

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.

Since $K(\varepsilon)/L$ is Galois and $K/L$ is Galois, viewing $K$ as an intermediate field $K(\varepsilon) / K / L$ implies that $K$ is $\text{Gal}(K(\varepsilon)/L)$-stable.

But then by this result:

If $K/F$ is a Galois extension, $L$ is an intermediate field, $L$ is $\text{Gal}(K/F)$-stable, then $r : \text{Gal}(K/F) \to \text{Gal}(L/F)$ is a well-defined surjective group homomorphism.

This gives a well-defined restriction map

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

(explicitly, we have the sequence of maps $\text{Gal}(K(\varepsilon) / L(\varepsilon) \hookrightarrow \text{Gal}(K(\varepsilon) / L) \to \text{Gal}(K/L)$). This restriction map is injective, since a $L(\varepsilon)$-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 $H = \text{Gal}(K/L)$.

To reiterate, we now have two maps:

• $r:\text{Gal}(K(\varepsilon)/L(\varepsilon)) \hookrightarrow \text{Gal}(K/L)$
• $\phi: \text{Gal}(K(\varepsilon) / L(\varepsilon)) \twoheadrightarrow \text{Gal} _ {L(\varepsilon)}(m _ {L(\varepsilon), \alpha})$

By the first isomorphism applied to $\phi$, we have

\[\text{Im }\phi = \text{Gal}_{L(\varepsilon)}(m_{L(\varepsilon), \alpha}) \cong \frac{\text{Gal}(K(\varepsilon)/L(\varepsilon))}{\ker \phi}\]

but because of the restriction map $r$, we know that:

• $r(\text{Gal}(K(\varepsilon)/L(\varepsilon))) \le \text{Gal}(K/L) \le \text{Gal}(K/F) = H$, and
• $r(\ker \phi) \trianglelefteq r(\text{Gal}(K(\varepsilon)/L(\varepsilon)))$. Thus:

\[\text{Gal}_{L(\varepsilon)}(m_{L(\varepsilon), \alpha}) \cong \frac{(\text{a subgroup of }H)}{(\text{a normal subgroup of that subgroup})}\]

In other words, $\text{Gal} _ {L(\varepsilon)}(m _ {L(\varepsilon), \alpha})$ is isomorphic to a subquotient of the finite solvable group $\text{Gal}(K/F)$.

But this means $\text{Gal} _ {L(\varepsilon)}(m _ {L(\varepsilon), \alpha})$ is solvable. Now look at the size of this group:

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

Therefore 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 $L$ 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 $F(\varepsilon)$ (since the intermediate extensions are Galois), 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, and we are done.

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Why may we find a subgroup of $G$ such that $[G : H]$ is prime?

Say we have the subnormal series

\[G \trianglerighteq H_{1} \trianglerighteq \cdots \trianglerighteq \{e\}\]

Then $G/H _ 1$ is abelian (in a solvable group, all factor groups are abelian), and thus $G/H _ 1$ contains a subgroup of order $p$ where $p$ is some prime dividing $ \vert G/H _ 1 \vert $ by Cauchy’s theorem. Thus there is a surjection $\phi : G/H _ 1 \to \mathbb Z / p\mathbb Z$ where the kernel is a subgroup of order $p$. But then compose with the projection to obtain the sequence

\[G \stackrel\pi\longrightarrow G/H_1 \stackrel\phi\longrightarrow \mathbb Z /p\mathbb Z\]

since $\pi$ and $\phi$ are surjections, the map $\psi = \phi \circ \pi$ is also a surjection. Call $\ker \psi = H$. Then by the first isomorphism theorem, $[G : H] = \vert G/H \vert = p$.

For any prime $p$, let $F := \mathbb F _ p(t)$ be the field of fractions of the polynomial ring $\mathbb F _ p[t]$ and let $K$ be a field extension of $F$ containing a root $\alpha$ of $f := x^p - t \in F[x]$.

This polynomial is irreducible and inseparable, and the minimal polynomial of $\alpha$. Note also that $K$ is a radical extension, as $K = F[\alpha]$ and $\alpha^p = t \in F$.

Suppose that $K$ were contained in some Galois extension $M$. By previous results, all Galois extensions are separable and so for all $\beta \in M$, $m _ {F, \beta}$ should be separable. But then $m _ {F, \alpha}$ is inseparable, a contradiction.