Lecture. The Fundamental Theorem of Galois Theory [l13]

May 13, 2026

Example 1. some examples [22016]

Example 1.1. $\mathrm{Gal}\left( \mathbb{Q}(\zeta_{p}) / \mathbb{Q} \right)$ [22016A]

Let $F = \mathbb{Q}(\zeta_{p})$ where $\zeta_{p} = e^{ 2\pi 1 / p }$ is a primitive $p$-th root of unity.

Claim: $F / \mathbb{Q}$ is Galois.

$F$ is the splitting field of $x^{p} - 1$ over $\mathbb{Q}$ by Example 1505, so $F / \mathbb{Q}$ is normal by (4) of Proposition 2205. $\operatorname{ch}(\mathbb{Q}) = 0$, so $F / \mathbb{Q}$ is separable by (5) of Proposition 1704. Hence, $F / \mathbb{Q}$ is Galois.

Claim: $\mathrm{Gal}(F / \mathbb{Q}) \cong \left( \mathbb{Z} / p\mathbb{Z} \right)^{\times}$.

We showed that $m_{\zeta_{p}, \mathbb{Q}}(x) = \Phi_{p}(x) = \frac{x^{p} - 1}{x - 1} = \prod_{1 \leq j \leq p-1} (x - \zeta_{p}^{j})$ in Example 1505. For each integer $j$ coprime to $p$, $\zeta_{p}^{j}$ is a root of $\Phi_{p}(x)$. By Proposition 1709 and its proof, we know that $$ \mathbb{Q}(\zeta_{p}) \to \mathbb{Q}(\zeta_{p}^{j}) = \mathbb{Q}(\zeta_{p}), \qquad a(\zeta_{p}) \mapsto a(\zeta_{p}^{j}) \quad \forall a(x) \in \mathbb{Q}[x] $$ is a field isomorphism. By definition, we can check that $\sigma^{i} \circ \sigma^{j} = \sigma^{ij}$ and $\sigma^{j} = \operatorname{id}_{F}$ if and only if $j \equiv 1 \pmod{p}$. It follows that $$ \begin{align*} \left( \mathbb{Z} / p\mathbb{Z} \right)^{\times} &\to \mathrm{Gal}\left( \mathbb{Q}(\zeta_{p}) / \mathbb{Q} \right) \\ \overline{j} &\mapsto \sigma_{j} \end{align*} $$ is well defined and is an injective group homomorphism. On the other hand, $$ \lvert \mathrm{Gal}\left( \mathbb{Q}(\zeta_{p}) / \mathbb{Q} \right) \rvert = [\mathbb{Q}(\zeta_{p}) : \mathbb{Q}] = \deg \Phi_{p}(x) = p - 1 = \lvert \left( \mathbb{Z} / n\mathbb{Z} \right) ^{\times} \rvert . $$ Therefore, the group homomorphism is bijective and is a group isomorphism.

For general positive integer $n$, we will show that $\mathrm{Gal}\left( \mathbb{Q}(\zeta_{n}) / \mathbb{Q} \right) \cong \left( \mathbb{Z} / n\mathbb{Z} \right)^{\times}$.

Example 1.2. $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right)$ [22016B]

Let $\mathbb{F}_{p^{n}}$ be the unique (up to isomorphism) field extension of $\mathbb{F}_{p}$ of degree $n$.

Claim: $\mathbb{F}_{p^{n}} / \mathbb{F}_{p}$ is Galois.

By construction in Theorem 1801, $\mathbb{F}_{p^{n}}$ is the splitting field of $x^{p^{n}} - 1$ over $\mathbb{F}_{p}$. Therefore, $\mathbb{F}_{p^{n}} / \mathbb{F}_{p}$ is normal.

By (2) of Proposition 1902, every finite extension of $\mathbb{F}_{p}$ is separable. Hence $\mathbb{F}_{p^{n}} / \mathbb{F}_{p}$ is separable.

Therefore, $\mathbb{F}_{p^{n}} / \mathbb{F}_{p}$ is Galois.

Claim: $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) \cong \mathbb{Z} / n\mathbb{Z}$.

Denote by $\operatorname{Fr} : \mathbb{F}_{p^{n}} \to \mathbb{F}_{p^{n}}$ the Frobenius endomorphism. Since $\operatorname{Fr}$ is injective and $\mathbb{F}_{p^{n}}$ is finite, $\operatorname{Fr}$ is bijective and is an automorphism of $\mathbb{F}_{p^{n}}$. Theorem 1801 also shows that $\alpha^{p} = \alpha$ for all $\alpha \in \mathbb{F}_{p}$. Hence, $\operatorname{Fr}$ fixes $\mathbb{F}_{p}$. Then $$ \mathbb{Z} \to \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) , \qquad m \mapsto \operatorname{Fr}^{m} $$ is a group homomorphism.

It is not injective since the domain is infinite while the codomain is fintie. Its kernel equals $\mathbb{Z} / m\mathbb{Z}$ for some positive integer $m$. Then $\operatorname{Fr}^{m} = \operatorname{id}_{\mathbb{F}_{p^{n}}}$, so every $\alpha \in \mathbb{F}_{p^{n}}$ is a root of $x^{p^{m}} - x$. The polynomial $x^{p^{m}} - x$ has at most $p^{m}$ roots, so $p^{m} \geq p^{n}$. It follows that $m \geq n$. On the other hand, by Theorem 1801, every $\alpha \in \mathbb{F}_{p^{n}}$ satisfies $\alpha^{p^{n}} = \alpha$, so $\operatorname{Fr}^{n} = \operatorname{id}_{\mathbb{F}^{p^{n}}}, n \in m\mathbb{Z}$. Thus, $m = n$ and $\mathbb{Z} / n\mathbb{Z}$ is isomorphism to a subgroup of $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right)$. Then $$ \lvert \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) \rvert = [\mathbb{F}_{p^{n}} : \mathbb{F}_{p}] = n = \lvert \mathbb{Z} / n\mathbb{Z} \rvert $$ implies that $$ \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) \cong \mathbb{Z} / n\mathbb{Z}. $$ More generally, for all $n \mid m$, $\mathbb{F}_{p^{m}} / \mathbb{F}_{p^{n}}$ is a Galois extension with $\mathrm{Gal}\left( \mathbb{F}_{p^{m}} / \mathbb{F}_{p^{n}} \right) \cong \mathbb{Z} / (m / n)\mathbb{Z}$.

Example 1.3. characteristic 0 [22016C]

(D&F Example (2) at p.563) More generally, any quadratic extension $K$ of any field $F$ of characteristic different from $2$ is Galois.
(D&F Example (4) at p.563) The extension $\mathbb{Q}(\sqrt{ 2 }, \sqrt{ 3 })$ is Galois over $\mathbb{Q}$ because it is the splitting field of $(x^{2} - 2)(x^{2} - 3)$.
(D&F Example (6) at p.566) The field $\mathbb{Q}(\sqrt[4]{ 2 })$ is not Galois over $\mathbb{Q}$ while we have $\mathbb{Q} \subset \mathbb{Q}(\sqrt{ 2 }) \subset \mathbb{Q}(\sqrt[4]{ 2 })$ and both $\mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q}$ and $\mathbb{Q}(\sqrt[4]{ 2 }) / \mathbb{Q}(\sqrt{ 2 })$ are Galois.

Example 1.4. characteristic 2 [22016D]

(D&F Example (8) at p.566) The inseparable extension $\mathbb{F}_{2}(x)$ over $\mathbb{F}_{2}(t)$ where $x^{2} - t = 0$ is not Galois.

Proposition 2. $L / F$ is Galois $\implies$ $L / K$ is Galois [22017]

Let $K / F$ and $L / K$ be algebraic field extensions. Then $L / F$ is Galois $\implies$ $L / K$ is Galois.

Proof. sketch [22017A]

Suppose $L / F$ is Galois. Then by Proposition 17011, $L / K$ is separable.

We use (5) of Proposition 2205 to show that $L / K$ is normal. Fix an algebraic closure of $F$ such that $F \subseteq K \subseteq L \subseteq \overline{F}$. If $\varphi : L \to \overline{F}$ is a $K$-embeddings, then it is an $F$-embeddings. By Proposition 2205, $\varphi(L) = L$. Again by Proposition 2205, $\varphi(L) = L$ for all $K$-embedding $\varphi : L \to \overline{F}$ of $L$ $\implies$ $L / K$ is normal.

Remark 3. some remarks [22018]

$F \subseteq K \subseteq L$.

$L / F$ is Galois $\centernot\implies$ $K / F$ is Galois.
$K / F$ and $L / F$ are both Galois $\centernot{\implies}$ $L / F$ is Galois.

Definition 4. Galois closure [22019]

Let $K / F$ be a separable extension. A field extension $L / K$ is called the Galois closure of $K$ over $F$ if

$L / F$ is Galois, and
is minimal in the sense that $K \subseteq E \subseteq L$ and $E / K$ is Galois $\implies$ $E = L$.

Proposition 5. construction and uniqueness of Galois closure [22020]

Let $K / F$ be a separable extension and $S$ be a subset of $K$ such that $K = F(S)$. Fix an algebraic closure $\overline{K} = \overline{F}$. Let $S' \subseteq \overline{F}$ be the subset consisting of roots of $m_{\alpha, F}(x), \alpha \in S$.

$F(S')$ is Galois closure of $K$ over $F$.
If $L$ is a Galois closure of $K$ over $F$, then there exists a field isomorphism $\varphi : L \to F(S')$ such that $\varphi |_{K} = \operatorname{id}_{K}$. Moreover, if $L \subseteq \overline{F}$, then $L = F(S')$.
$$F(S') = \bigcap_{\begin{aligned}&K \subseteq L \subseteq \overline{F} \\ &L / F \text{ Galois}\end{aligned}} L.$$
Proof. sketch [22020A]
1. Given any $\alpha' \in S'$, by construction, $\alpha'$ is a root of $m_{\alpha, F}(x)$ for some $\alpha \in S$. And clearly $m_{\alpha', F}(x) = m_{\alpha, F}(x)$. Since $K = F(S) / F$ is separable, $m_{\alpha, F}(x)$ is separable over $F$ $\implies$ $m_{\alpha', F}(x)$ is separable over $F$ $\implies$ $F(S') / F$ is separable. Moreover, any other roots of $m_{\alpha', F}(x)$ lie in $S'$, hence in $F(S')$ $\implies$ $m_{\alpha', F}(x)$ splits completely over $F(S')$ $\implies$ $F(S') / F$ is normal. Therefore, $F(S') / F$ is Galois.
  
  Suppose $K \subseteq E \subseteq F(S')$ and $E / F$ is Galois. Then for all $\alpha \in S$, $m_{\alpha, F}(x)$ splits completely over $E$ $\implies$ all the roots of $m_{\alpha, F}(x)$ lie in $E$ $\implies$ $S' \subseteq E$ $\implies F(S') \subseteq E$ $\implies F(S') = E$.
  
  So $F(S')$ is a Galois closure of $K$ over $F$.
2. Let $\varphi : L \to \overline{K} = \overline{F}$ be a $K$-embedding such that $\varphi |_{K} = \operatorname{id}_{K}$. So $K \subseteq \varphi(L) \subseteq \overline{F}$. Since $\varphi$ is injective, we have $\varphi(L) \cong L$ and we can check that $\varphi(L)$ is also a Galois closure of $K$ over $F$. Then by the minimality of Galois closure, $\varphi(L) = F(S')$. Hence $L \cong F(S')$.
3. Since $K \subseteq F(S') \subseteq \overline{F}$ and $F(S') / F$ is Galois, we have $$F(S') \subseteq \bigcap_{\begin{aligned}K \subseteq L \subseteq \overline{F} \\ L / F \text{ Galois}\end{aligned}} L.$$ The other direction just follows from the construction of $S'$.

When $K / F$ is finite and separable, then the set $S$, hence $S'$ can be taken to be a finite set. Then we see that the Galois closure of $K$ over $F$ is a finite extension of $F$.

Homework:

$K / F$ finite. Then normal $\iff$ $K$ is the splitting field of some $f(x) \in F[x]$, and Galois $\iff$ $K$ is the splitting field of some separable $f(x) \in F[x]$.
Show that if $K = F(\alpha_{1}, \dots, \alpha_{n})$ and $\varphi_{1}, \dots, \varphi_{n}$ are all the $F$-embeddings from $K$ to $\overline{F}$, then $F(S)$ is a Galois closure of $K$ where $S = \{ \varphi_{i}(\alpha_{j}) \mid 1 \leq i, j \leq n \}$.
Show that $K_{1} / F$ and $K_{2} / F$ Galois $\implies$ $K_{1} \cap K_{2} / F$ Galois.

The Fundamental Theorem of Galois Theory

Theorem 6. “$[K : K^{G}] = \\lvert G \\rvert$” [2301]

Let $K / F$ be a finite field extension and $G \leq \mathrm{Aut}(K / F)$. Then $[K : K^{G}] = \lvert G \rvert$.

Proof. sketch (TODO) [2301A]