Galois Theory HT25, Galois groups and Galois extensions

The group of all $F$-linear field automorphisms of $K$.

A bijective map $\theta : K \to K$ such that for all $k _ 1, k _ 2 \in K$, and $f \in F$:

\[\begin{aligned} \theta(k _ 1 k _ 2) &= \theta(k _ 1) \theta(k _ 2) \\\\ \theta(k _ 1 + k _ 2) &= \theta(k _ 1) + \theta(k _ 2) \\\\ \theta(f k _ 1) &= f \theta(k _ 1) \end{aligned}\]

\[K^H = \\{x \in K \mid \forall \sigma \in H,\text{ }\sigma(x) = x \\}\]

$K \supseteq F$ is said to be a splitting field of $f$ if:

• $f$ splits completely in $K[t]$, and
• $K$ is generated as a field by $F$ together with the roots of $f$.

If $f \in F[t]$ and $K$ is a splitting field of $f$, then $[K : F] < \infty$.

Since $f$ is irreducible in $F[t]$, $\langle f \rangle$ must be a maximal ideal in $F[t]$ and hence $K := \frac{F[t]}{\langle f\rangle}$ is a field. Then $\alpha := t + \langle f \rangle$ is a root of $f$ in $K$.

Since $\alpha$ generates $K$ as a ring together with $F$, $K = F[\alpha]$.

Since $\alpha$ is algebraic over $F$, it follows

\[F(\alpha) = F[\alpha] = K\]

and $K / F$ is the required extension.

Induct on $d := \deg f$.

Base case. $d = 1$. Then $F$ is already a splitting field.

Inductive step. $d > 1$. Suppose that $g$ is an irreducible factor of $f$. By the result that says:

If $g \in F[t]$ is irreducible, there exists a simple extension $F(\alpha)$ where $\alpha$ is a root of $g$.

we can construct $L := F(\alpha)$. Since $g \mid f$, $f(\alpha) = 0$ and hence $(t - \alpha) \mid f$.

Define $h := \frac{g}{(t - \alpha)} \in L[t]$. Since $\deg h < d$, by induction there exists a splitting field $K$ of $h$.

Since $a \in L \subset K$, $f = (t - \alpha) \cdot h$ splits completely in $K[t]$.

Since the roots of $f$ in $K$ generate $K$ together with $F$, it follows that $K$ is a splitting field of $f$.

Yes.

Take $\varphi = \text{id} _ K$. Then since $[K : F] \ge 1$, there is at least one isomorphism $\varphi^\ast$ extending $\varphi$ to $K \to \tilde K$, and so they are isomorphic.

We induct on the degree of $f$.

If $\deg f = 1$, then there is nothing to prove.

If $\deg f > 1$, pick some root $a \in K$ of $f$ in $K$ and let $m _ {F, \alpha}$ be its minimal polynomial. Then $m _ {F, \alpha}$ splits in $K$ and in $K’$ (since it divides $P(x)$).

Let $\alpha _ 1$ be a root of $m _ {F, \alpha}$ in $K$. Notice that $K/F(\alpha)$ is a splitting extension of $f(x)/(x - \alpha) \in F(\alpha)$.

Similarly, if $K/F(\alpha _ 1)$ is a splitting extension of $f(x)/(x - \alpha _ 1)$ in $F(\alpha _ 1)$.

Let $J := F[x]/\langle m _ {F,\alpha}(x)\rangle$. The ring $J$ is a field, since $m _ {F, \alpha}$ is irreducible and furthermore, there are natural isomorphisms $J \cong F(\alpha)$ and $J \cong F(\alpha _ 1)$.

Consider the two $F$-extensions $K/J$ and $K’/J$ arising from these isomorphisms. By the inductive hypothesis, these two $J$-extensions are isomorphic. By construction, an isomorphism $K \cong K’$ of $J$-extensions is also an isomorphism of $F$-extensions, so we are done.

---

Alternatively, apply the result that says

If:

• $\varphi : F \to \tilde F$ is an isomorphism between two fields
• $K / F$ and $\tilde K / \tilde F$ 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 $\varphi^\ast$ where

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

such that $\varphi^\ast (\alpha) = \tilde \alpha$.

To each of the roots in the splitting field in turn.

If there are no such homomorphisms, the statement is empty.

Suppose there is such a map $\varphi : K \to L$. Since $K$ is generated by $F$ by the roots of $f$, the image of $\varphi$ is generated over $F$ by the images of these roots in $L$ under $\varphi$. But the images are the roots of $f(x)$ in $L$.

To see this, let $\alpha _ 1, \ldots, \alpha _ d$ be the (possibly duplicate) roots of $f(x)$ in $K$. Then we have

\[f(x) = x^d - \sigma_1 (\alpha_1, \ldots, \alpha_d) x^{d-1} + \cdots + (-1)^d \sigma_d(\alpha_1, \ldots, \alpha_d)\]

Thus the elements $\varphi(\alpha _ 1), \cdots, \varphi(\alpha _ d)$ are roots of

\[\begin{aligned} &x^d - \sigma_d(\varphi(\alpha_1), \ldots, \varphi(\alpha_d)) x^{d-1} + \cdots + (-1)^d \sigma_d(\varphi(\alpha_1), \cdots, \varphi(\alpha_d)) \\\\ &=x^d - \varphi(\sigma(\alpha_1, \ldots, \alpha_d))x^{d-1} + \cdots + (-1)^d \varphi(\sigma_d(\alpha_1, \ldots, \alpha_d)) \\\\ &= f(x) \end{aligned}\]

since the coefficients of $f(x)$ lie in $F$.

Now the set of roots of $f(x)$ in $L$ does not depend on $\varphi$, hence the assertion.

Since any two splitting fields of a polynomial are isomorphic, this follows from showing that $\frac{K[x]}{\langle f \rangle}$ is a splitting field for $f$. Clearly $\frac{K[x]}{\langle f \rangle}$ contains a root $\alpha := \langle f \rangle$ of $f$, and note that if $u \in \mathbb F _ {125}^\times$, then $u\alpha$ is also a root (since $u^{124} = 1$). Since $ \vert \mathbb F _ {125}^\times \vert = 124$ and $u \in K$, it follows $\frac{K[x]}{\langle f \rangle}$ contains all the roots and is generated by them, so it is a splitting field.

@example~ @exam~

There is a separable polynomial $f \in F[x]$ such that $K/F$ is the splitting field of $f$.

The following are equivalent:

• $K/F$ is Galois (i.e. $K$ is the splitting field of a separable polynomial in $F[x]$)
• $ \vert G \vert = [K : F]$
• $F = K^G$
• $K/F$ is normal and separable.

\[F = K^G\]

This is a consequence of three other results:

• $G = \text{Gal}(K/K^G)$
• $ \vert G \vert \le [K : K^G]$ (applying the upper bound on the size of Galois groups to $\text{Gal}(K/K^G)$)
• $[K : K^G] \le [K : F]$ by the tower law, since $K/K^G/F$.

Then:

\[|G| \le [K : K^G] \le [K : F] = |G|\]

and so $[K : K^G] = [K : F]$ and hence $[K^G : F] = 1$, and so $K^G = F$.

@important~ (main theorem)

$K/F$ is Galois (i.e. $K$ is the splitting field of a separable polynomial in $F[x]$).

Let $\{z _ 1, \ldots, z _ n\}$ be an $F$-basis for $K$. Since $G$ is finite (by the result that says finite field extensions lead to finite Galois groups), it follows that the set

\[X := \bigcup^n _ {i = 1} G\cdot z _ i\]

is finite as well as $G$-stable.

Then we can construct the polynomial $f _ X$ where

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

which is a separable polynomial with coefficients in $K^G$ (this is not immediate, but follows from considering the fact the coefficients are symmetric functions). Since $K^G = F$, it follows $f _ X \in F[t]$.

Since $f _ X$ splits completely over $K$ ($G \cdot z _ i \subseteq K$) and since $K$ is generated by the roots of $f _ X$ in $K$, it is the splitting field of $f _ X$ over $F$.

@important~

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

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Useful for:

• Property 1 is used in showing that $t^{p^n} - t$ is separable, and hence that $F$ is a splitting field of $t^{p^n} - t$ iff it has order $p^n$.
• Both properties are used in showing that the minimal polynomial of $\alpha$ over $K/F$ is the product of $(x - g(\alpha))$ for each $g \in \text{Gal}(K/F)$.
• Both properties are used in showing that $F = K^G$ implies $K/F$ is Galois.
• Both properties are used in showing $F=K^G$ implies that $K/F$ is normal and separable.
• Both properties are used in showing that the $n$-th cyclotomic polynomial lives in $\mathbb Q[t]$.

Separable: Any factor (and hence any irreducible factor) of $f _ X$ must have the form $f _ Y$ for $Y \subseteq X$, and so it suffices to just check that $D(f _ X) \ne 0$. Note that

\[D(f _ X) = \sum _ {y \in X} \prod _ {z \in X \setminus \\{y\\} }(t - z)\]

Choose any value $u \in X$. If $y \ne u$, then $\prod _ {z \in X \setminus \{y\} } (u - z) = 0$ since the product includes the factor $u - z$ with $z = u$. Therefore

\[D(f _ X)(u) = \prod _ {z \in X \setminus \\{u\\} } (u - z) \ne 0\]

which is non-zero, and so $f _ X$ is separable.

---

If $X$ is $H$-stable, then $f _ X$ has coefficients in $L^H$: $H$ acts on $L[t]$ by applying each automorphism to the coefficients and fixing $t$.

Under the assumption $X$ is $H$-stable, the set of linear polynomials

\[\\{t - x : x \in X\\}\]

is $H$-stable. Since the $H$-action respects multiplication in $L[t]$, it follows that the product of these linear polynomials is fixed by $H$:

\[f _ X \in L[t]^H = L^H[t]\]

and so $f _ X$ has coefficients in $L^H$.

(Alternatively, can consider that the coefficients are given by symmetric functions of the roots).

\[\text{Gal}(K / \mathbb Q)\]

is an abelian group.

Naturally isomorphic to a subgroup of $S _ n$:

We show there is an injection $\rho : G \to S _ n$. 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 let

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

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

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

If $f$ irreducible, then this subgroup is transitive:

If $f$ is irreducible, then $f = m _ {F, \alpha}$ where $\alpha \in V(f)$. Then applying the result

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

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

such that $\varphi^\ast (\alpha) = \tilde \alpha$.

where $\varphi$ is the identity and $\tilde \alpha$ is any other root. This gives at least one automorphism of $K$ that swaps $\alpha$ and $\tilde \alpha$, so $G$ is transitive.

Suppose:

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

Then:

• $m _ {F, \alpha}$ is separable
• $m _ {F, \alpha} = \prod _ {\psi \in \text{Gal}(K/F) \cdot \alpha} (x - \psi)$
• There is a surjective group homomorphism $\text{Gal}(K/F) \to \text{Gal} _ F(m _ {F, \alpha})$

(This result is useful mainly because it comes up in the proof that field extensions with solvable Galois groups are radical. Also, since $K/F$ is defined to be a separable extension whenever the minimal polynomial of every $\alpha \in K$ is separable, this also shows that all Galois extensions are separable).

• $m _ {F, \alpha}$ is irreducible over $F$ and has a zero in $K$ (namely $\alpha$)
• Since $K/F$ is Galois, it is particular normal.
• Therefore $m _ {F, \alpha}$ splits completely in $K$, and so $K$ contains a splitting field of $m _ {F, \alpha}$, say $L$.
• By another result, $m _ {F, \alpha} = m _ {K^G, \alpha} = f _ {G \cdot \alpha} = \prod _ {\beta \in G\cdot\alpha}(t - \beta)$.

• Since $f _ X$ is always separable, it follows $m _ {F, \alpha}$ is always separable.

• Therefore $L/F$ is Galois (it is the splitting field of some separable polynomial).
• $\text{Gal} _ F(m _ {F, \alpha}) = \text{Gal}(L/F)$ and so the restriction map $\text{Gal}(K/F) \to \text{Gal}(L/F)$ is a surjective group homomorphism.

• $L/F$ is also Galois
• There is an injective homomorphism $\phi : \text{Gal}(L/F) \to \text{Gal}(L _ 1/F) \times \text{Gal}(L _ 2 / F)$
• If $L _ 1 \cap L _ 2 = F$ (“linearly disjoint”), then this homomorphism is an isomorphism

@exam~

$L/F$ is Galois:

$L _ 1$ and $L _ 2$ are the splitting fields of separable polynomials $f _ 1, f _ 2 \in F[x]$. Then $F$ is the splitting field of $f _ 1 f _ 2$, which is also separable. Hence $L/F$ is also Galois.

---

There is an injective homomorphism $\phi : \text{Gal}(L/F) \to \text{Gal}(L _ 1/F) \times \text{Gal}(L _ 2 / F)$:

Let $G := \text{Gal}(L/F)$ and $H := \text{Gal}(L _ 1/F) \times \text{Gal}(L _ 2 / F)$. Let $\phi : G \to H$ be the map which sends $\sigma \to (\sigma \vert _ {L _ 1}, \sigma \vert _ {L _ 2})$. This map is well-defined as $L _ 1 / F$ and $L _ 2/F$ are both Galois extensions contained in $K$.

To see it is injective, suppose $\phi(\sigma) = (\text{id} _ {L _ 1}, \text{id} _ {L _ 2})$. Since $L$ is generated by the union of the generators for $L _ 1$ and $L _ 2$, it must also be the identity on $L$.

---

If $L _ 1 \cap L _ 2 = F$, then this homomorphism is an isomorphism:

It suffices to show that $ \vert G \vert = \vert H \vert $. Let $\iota$ be the restriction map

\[\iota : \text{Gal}(L / L _ 2) \to \text{Gal}(L _ 1 / F)\]

which sends $\sigma$ to $\sigma \vert _ {L _ 1}$. Since $L _ 2$ contains $F$, the image of $\iota$ fixes $F$. Since $L _ 1$ is Galois, $\iota$ is well-defined and it is injective since any element of its kernel must be the identity on both $L _ 1$ and $L _ 2$.

To see $\iota$ is also surjective, we show that

\[[F : L _ 2] = |S| = |\text{Gal}(L _ 1 / K)| = [L _ 1 : K]\]

as then the size of the domain and range are equal. Consider the subgroup $\iota(\text{Gal}(L/L _ 2))$. Then

\[[L _ 1 : L _ 1^\iota(\text{Gal}(L/L _ 2))] = |\text{Gal}(L/L _ 2)|\]

as the action of $\iota(\text{Gal}(L/L _ 2))$ on $L _ 1$ is faithful. If we can show $L _ 1^{\iota(\text{Gal}(L/L _ 2))} = L _ 1 \cap L _ 2$, then we will be done as $L _ 1 \cap L _ 2 = F$.

Note

\[\\{x \in L : g(x) = x \text{ } \forall g \in \text{Gal}(L/L _ 2)\\} = L _ 2\]

by the Galois correspondence on $L/F$. Therefore

\[L _ 1^{\iota(\text{Gal}(L/L _ 2))} = \\{x \in L _ 1 : g(x) = x \text{ } \forall g \in \text{Gal}(F/L _ 2)\\} = L _ 1 \cap L _ 2\]

and hence

\[|H| = |\text{Gal}(L _ 1 / F)| \times |\text{Gal}(L _ 2 / F)| = [L _ 1 : F][L _ 2 : F] = [L : L _ 2][L _ 2 : F] = [L : F] = |G|\]

as required.

@exam~

Suppose:

• $f$ is a degree $p$ polynomial where $p$ prime
• $f$ has exactly $p - 2$ real roots

Then:

• $\text{Gal} _ \mathbb Q(f) \cong S _ p$

• The Galois group must contain a transposition corresponding to complex conjugation
• The Galois group must also contain a $p$-cycle by Cauchy’s theorem
• Therefore the Galois group contains the whole symmetric group

Consider $\mathbb Q[\alpha] / \mathbb Q$ where $\alpha = \sqrt[3]{2}$. Then $m _ \mathbb Q(x) = x^3 - 2$, but $m(x)$ does not split completely, since it has two roots which are not in $\mathbb Q[\alpha]$.

Consider $F := \mathbb F _ p(t)$ be the field of fractions over the finite field of size $p$, and let $K$ be a field extension of $F$ containing a root $\alpha$ of $f(x) := x^p - t$. But then $K / F$ is not Galois as it is not separable, since $m _ {F, \alpha}$ is not separable.

• Let $F = \mathbb Q$, $K = \mathbb Q(\sqrt 2)$, $K’ = \mathbb Q(2^{1/4})$.
• Then:
  • $K/F$ is Galois (it’s the splitting field of $x^2 - 2$)
  • $K’/F$ is Galois (it’s the splitting field of $x^2 - \sqrt{2}$)
  • $K’/F$ is not Galois, since $m _ {\mathbb Q, 2^{1/4} } = x^4 - 2$, which does not split over $K’ \subseteq \mathbb R$.

@exam~

Pick any $\alpha \in L \setminus K$. Then $m _ {K, \alpha}(x) = x^2 + bx + c$ for some $b, c \in K$ and this polynomial is irreducible. Since $m _ {K, \alpha}’(x) = 2x + b$, it follows that $m _ {K, \alpha}$ is separable.

We aim to show that $L$ is a splitting field for $m _ {K, \alpha}$. By the quadratic formula,

\[\begin{aligned} m _ {K, \alpha}(x) &= \left(x - \frac{-b +\sqrt{b^2 - 4c} }{2}\right) \left(x - \frac{-b - \sqrt{b^2 - 4c} }{2}\right) \\\\ &= (x - \alpha)(x - (\alpha - b)) \end{aligned}\]

Since $\alpha, \alpha - b \in L$ and $L = F(\alpha)$, so $L$ is the splitting field.

@exam~ @example~