When $S = \{ \alpha_{1}, \dots, \alpha_{n} \}$, we write $F(S)$ as $F(\alpha_{1}, \dots, \alpha_{n})$.

TODO

By the construction of $F(S)$ from elements of $F \cup S$ in Proposition 1302, obviously $\varphi = id_{F(S)}$.

(1) $\iff$ (2) is trivial.

(5) $\implies$ (1): Since $F(\alpha) = F[\alpha]$, there exists $c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \in F[\alpha]$ with at least one nonzero coefficient such that it equals $\alpha ^{-1} \in F(\alpha)$. Hence $$ \begin{align*} &\alpha ^{-1} = c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \\ \iff &1 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha \\ \iff &0 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha - 1 \in F[\alpha] \end{align*} $$ Therefore, $\alpha$ is algebraic over $F$.

(2) $\implies$ (3)(4)(5): Since $\varphi$ is not injective, $\ker \varphi$ is non-trivial. By the First Isomorphism Theorem, $F[x] / \ker\varphi \cong \mathrm{Im} \varphi$. Here, $\mathrm{Im}\varphi$ is a subring of $K$ containing $1$. Now $K$ is a field $\implies$ $\mathrm{Im} \varphi$ is an ID $\implies$ $F[x] / \ker \varphi$ is an ID $\implies$ $\ker\varphi$ is a PID, i.e., $\ker \varphi = \left( p(x) \right)$ for some $p(x) \in F[x]$ $\implies$ $p(x)$ is prime in $F[x]$ $\implies$ $p(x)$ is irreducible in $F[x]$. By Theorem 1206, $F[x] / \ker \varphi$ is a field and $$ F[x] / \ker\varphi = F + F\overline{x} + \cdots + F\overline{x}^{n-1}. $$ With $x \mapsto \alpha$ and $p(\alpha) = 0$ for $p(x) \in \ker\varphi$, we obtain $$ \mathrm{Im}\varphi = F + F\alpha + \cdots + F\alpha^{n-1}. $$ So $\mathrm{Im}\varphi$ is field extension of $F$ and $[\mathrm{Im}\varphi : F] < \infty$. By definition of $\varphi$, we have $F[\alpha] \subseteq \mathrm{Im}\varphi$. The above implies $\mathrm{Im}\varphi \subseteq F[\alpha]$. So $\mathrm{Im}\varphi = F[\alpha]$. Now $\mathrm{Im}\varphi$ is a field $\implies$ $F[\alpha] = F(\alpha)$. Therefore, $[F(\alpha) : F] < \infty$ and $[F[\alpha] : F] < \infty$.

(4) $\implies$ (1): $[F[\alpha] : F] < \infty$ implies that $F[\alpha] = F + F\alpha + \dots + F \alpha^{n-1}$ for some $n \in \mathbb{Z}_{\ge 0}$. Hence $1 \in F[\alpha]$ can be written as $1 = c_{0} + c_{1}\alpha + \cdots + c_{n-1}\alpha^{n-1}$, i.e., $c_{0} - 1 + c_{1}\alpha + \dots + c_{n-1}\alpha^{n-1} = 0$. So $\alpha$ is algebraic over $F$.

(3) $\implies$ (4): $F(\alpha) \subseteq F[\alpha] \subseteq F$. Then use Theorem 1207.

Existence: let $p(x)$ be the very polynomial in the proof of Proposition 1403. Let $a$ be the leading coefficient of $p(x)$. Then $m_{\alpha, F}(x) = a^{-1} p(x)$. It is obviously monic, and the irreducibility is inherited from $p(x)$.

Uniqueness: use the same notation as in the proof of Proposition 1403. Then $\ker \varphi = \left( p(x) \right) = \left( m_{\alpha, F}(x) \right)$. Suppose there is another monic irreducible polynomial $m_{\alpha, F}'(x) \in F[x]$ such that $m_{\alpha, F}'(\alpha) = 0$. Then $m_{\alpha, F}'(x) \in \left( m_{\alpha, F}(x) \right)$, hence $m_{\alpha, F}'(x) = m_{\alpha, F}(x) q(x)$ for some $q(x) \in F[x]$. Since $m_{\alpha, F}'(x), m_{\alpha, F}(x)$ are monic, $q(x)$ should also be monic. If $\deg q > 0$, then it contradicts irreducibility of $m_{\alpha, F}'(x)$. So $q(x) = 1$, which implies $m_{\alpha, F}' = m_{\alpha, F}$.

$\implies$: Just use the same argument in the proof of Proposition 1403. We have $f(x) \in \ker \varphi = \left( m_{\alpha, F}(x) \right)$. So $m_{\alpha, F}(x) \mid f(x)$ in $F[x]$.

$\impliedby$: Similarly, $m_{\alpha, F}(x) \mid f(x) \implies f(x) \in \left( m_{\alpha, F}(x) \right) = \ker\varphi$ and therefore $f(\alpha) = 0$.

1. By the proof of Proposition 1405, the kernel of the following ring homomorphism $$\varphi : F[x] \to K, \qquad a(x) \mapsto a(\alpha)$$ is $\left( m_{\alpha, F}(x) \right)$. By the First Isomorphism Theorem, $$F[x] / \left( m_{\alpha, F}(x) \right) \to \mathrm{Im}\varphi, \qquad \overline{a(x)} \mapsto a(\alpha)$$ is a well-defined isomorphism. Moreover, in the proof of Proposition 1403, we have shown $\mathrm{Im}\varphi = F(\alpha)$. And by Theorem 1206, $[F(\alpha) : F] = \deg m_{\alpha, F}(x)$.

2. Suppose $\frac{a(x)}{b(x)} = \frac{c(x)}{d(x)}$, then $$a(x)d(x) = b(x)c(x) \implies a(\alpha)d(\alpha) = b(\alpha)c(\alpha) \implies a(\alpha)b(\alpha)^{-1} = c(\alpha)d(\alpha)^{-1}.$$ So the map is well-defined. Also, it is clearly a field homomorphism. By definition of $F(\alpha)$, it is surjective. Denote the map by $\phi$. If $\phi$ is not injective, then there exists $\frac{a(x)}{b(x)}\in F(x)$ such that $a(\alpha) b(\alpha)^{-1} = 0 \implies a(\alpha) = 0$. Then $\alpha$ is algebraic, contradicting $\alpha$ is transcendental. So $\phi$ is an isomorphism.

   Now, if $[F(\alpha) : F] < \infty$, then by Proposition 1403, $\alpha$ is algebraic over $F$, a contradiction. So $[F(\alpha) : F] = \infty$.

WLOG, assume $p_{1}(x)$ and $p_{2}(x)$ are monic. Then by Proposition 1407, $$ \begin{align*} &\varphi_{1} : F_{1}[x] / \left( p_{1}(x) \right) \to F_{1}(\alpha), \qquad a(x) \mapsto a(\alpha), \\ &\varphi_{2} : F_{2}[x] / \left( p_{2}(x) \right) \to F_{2}(\beta), \qquad a(x) \mapsto a(\beta) \end{align*} $$ are field isomorphisms. Also, define $$ \sigma : F_{1}[x] / \left( p_{1}(x) \right) \to F_{2}[x] / \left( p_{2}(x) \right) , \quad c_{m}x^{m} + \cdots + c_{0} \mapsto \varphi(c_{m}) x^{m} + \cdots + \varphi(c_{0}), $$ which is well-defined and is an isomorphism induced by the isomorphism $\varphi$.

Then $\varphi_{2} \circ \sigma \circ \varphi_{1} : F_{1}(\alpha) \to F_{2}(\beta)$ is a field isomorphism and we have $\Psi = \varphi_{2} \circ \sigma \circ \varphi_{1}^{-1}$.

1. $\implies$: Suppose $[K : F] = n < \infty$. Let $\alpha_{1}, \dots, \alpha_{n}$ be a basis of $K$ as an $F$-vector space. Then $K = F(\alpha_{1}, \dots, \alpha_{n})$. Moreover, $F(\alpha_{i}) \subseteq K$ implies $[F(\alpha_{i}) : F] \leq [K : F] < \infty$, then by Proposition 1403, $\alpha_{j}$ is algebraic over $F$.

   $\impliedby$: By Theorem 1207, $$[K : F] = [F(\alpha_{1}, \dots, \alpha_{n}) : F] \leq [F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] [F(\alpha_{1}, \dots, \alpha_{n-1}) : F].$$ By HW2 Prob 1, $[F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] \leq [F(\alpha_{n}) : F]$. So $$[K : F] \leq [F(\alpha_{n}) : F][F(\alpha_{1}, \dots, \alpha_{n-1}) : F]$$ Therefore, by induction we have $[K : F] \leq [F(\alpha_{n}) : F][F(\alpha_{n-1}) : F] \cdots [F(\alpha_{1}) : F]$, in which every $[F(\alpha_{i}) : F] < \infty$.

2. $\implies$: Take $S = K$.

   $\impliedby$: Given $\beta \in K = F(S)$, $\beta \in F(\alpha_{1}, \dots, \alpha_{n})$ for some $\alpha_{i} \in S$. The $\alpha_{i}$’s are algebraic, so $F(\alpha_{1}, \dots, \alpha_{n}) / F$ is finite. Hence $[F(\beta) : F]$ is finite since $F(\beta) \subseteq F(\alpha_{1}, \dots, \alpha_{n})$. Then by Proposition 1403, $\beta$ is algebraic over $F$. Therefore $K / F$ is algebraic.

Let $\alpha, \beta$ be any elements of $E$, it suffices to show that $\alpha + \beta, -\alpha, \alpha \beta, \alpha ^{-1} \in E$. Since they are all in $F(\alpha, \beta)$, it suffices to show that $F(\alpha, \beta) \subseteq E$. By Proposition 1410, $F(\alpha, \beta) / F$ is finite and hence is algebraic. Therefore, $F(\alpha, \beta) \subseteq E$.

$\implies$: Given $\alpha \in L$, it is algebraic over $K$ since $L / K$ is algebraic. So there exist $a_{0}, \dots, a_{n-1} \in K$ such that $$ a(\alpha) = \alpha^{n} + a_{n-1} \alpha^{n-1} + \cdots + a_{0} = 0. $$ Hence $\alpha$ is algebraic over $F(a_{0}, \dots, a_{n-1})$, and $$ [F(\alpha, a_{0}, \dots, a_{n-1}) : F(a_{0}, \dots, a_{n-1})] < \infty. $$ Since $K / F$ is algebraic, each $a_{i}$ is algebraic over $F$, so $[F(a_{i}) : F] < \infty$ for all $i$. Thus $$ [F(a_{0}, \dots, a_{n-1}) : F] < \infty. $$ By Theorem 1207, we write $$ [F(\alpha, a_{0}, \dots, a_{n-1}) : F] = [F(\alpha, a_{0}, \dots, a_{n-1}) : F(a_{0}, \dots, a_{n-1})][F(a_{0}, \dots, a_{n-1}) : F], $$ then we can conclude that $[F(\alpha, a_{0}, \dots, a_{n-1}) : F] < \infty$. Since $F \subseteq F(\alpha) \subseteq F(\alpha, a_{0}, \dots, a_{n-1})$, it follows that $[F(\alpha) : F] < \infty$. Then by Proposition 1403, $\alpha$ is algebraic over $F$. Since $\alpha$ is arbitrary, $L / F$ is algebraic.

$\impliedby$: Since $L / F$ is algebraic and $K \subseteq L$, $K / F$ is also algebraic. Moreover, every $\alpha \in L$, there exist $a_{0}, \dots, a_{n-1} \in F$ such that $$ a(\alpha) = \alpha^{n} + a_{n-1}\alpha^{n-1} + \cdots + a_{0} = 0. $$ Since $F \subseteq K$, each $a_{i} \in K$, hence $a(x) \in K[x]$, so $\alpha$ is algebraic over $K$. Therefore, $L / K$ is algebraic.

Clearly, $f(x)$ splits completely in $F(\alpha_{1}, \dots, \alpha_{n})$.

Claim: $F \subseteq E \subseteq K$ and $f(x)$ splits completely over $E$ $\implies$ $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$.

Proof of the Claim: Suppose $f(x) = b(x - \beta_{1})\cdots(x - \beta_{n})$ with $b, \beta_{1}, \dots, \beta_{n} \in E$. Then we have $$ a(x - \alpha_{1}) \cdots (x - \alpha_{n}) = b(x - \beta_{1}) \cdots (x - \beta_{n}) $$ in $K[x]$. Since $K[x]$ is a UFD, after reordering we have $\alpha_{i} = \beta_{i}$ for all $1 \leq i \leq n$. And $F(\beta_{1}, \dots, \beta_{n}) \subseteq E$. Hence $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$.

1. Suppose $F \subseteq E \subseteq F(\alpha_{1}, \dots, \alpha_{n})$ and $f(x)$ splits completely over $E$, then by our claim, $E = F(\alpha_{1}, \dots, \alpha_{n})$. So $F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x)$.
2. Since $f(x)$ splits completely over $E$, by our claim, $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$. Moreover, $E$ is a splitting field of $f(x)$, then by definition, $E = F(\alpha_{1}, \dots, \alpha_{n})$.

Let $\deg f = n$. We prove by induction on $n$.

$n = 1$: it is clear that $F[x]$ is a splitting field of $f(x)$.

Assume it holds for $\leq n - 1$. Suppose $f(x) = p(x)a(x)$ for some irreducible $p(x) \in F[x]$ with $\deg p > 0$ and $a(x) \in F[x]$. Then $p(x)$ is prime in $F[x]$, and by Theorem 1206, let $E_{1} = F[x] / \left( p(x) \right)$, then $E_{1}$ is a field extension of $F$ and $\theta = \overline{x} \in E_{1}$ is a root of $p(x)$. Then $\theta$ is also a root of $f(x)$ since $f(x) \in \left( p(x) \right)$. So $f(x) = (x - \theta)f_{1}(x)$ for some $f_{1}(x) \in E_{1}[x]$ and $\deg f_{1} = n - 1$. By induction, there exists a field extension $E / E_{1}$ such that $f_{1}(x) = a(x - \alpha_{1})\cdots(x - \alpha_{n-1})$ with $\alpha_{1}, \dots, \alpha_{n-1} \in E$. Then $f(x) = a(x - \theta)(x - \alpha_{1}) \cdots (x - \alpha_{n-1})$ in $E[x]$, i.e., splits completely in $E[x]$.

By Proposition 1503, $K = F(\theta, \alpha_{1}, \dots, \alpha_{n-1})$ is a splitting field of $f(x)$ in $F[x]$.

By Proposition 1410, $F(\theta, \alpha_{1}, \dots, \alpha_{n-1})$ is a finite extension of $F$.

The splitting field for $x^{2} - 2 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt{ 2 })$. $\pm \sqrt{ 2 } \in \mathbb{Q}(\sqrt{ 2 })$. In general, the splitting of $x^{2} + bx + c \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt{ b^{2} - 4c })$.

The splitting field of $x^{3} - 2 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$.

Exegesis 5.1.2.3. subfields of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ [1505BC]

We have the diagram

Actually, these are all the subfields of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$. We will show it later.

The splitting field of $x^{n} - 1 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\zeta_{n})$.

The polynomial splits completely in $\mathbb{C}$. Let $\zeta_{n} = e^{ 2\pi i / n } = \cos \left( \frac{2\pi}{n} \right) + i \sin \left( \frac{2\pi}{n} \right) \in \mathbb{C}$. Then $\zeta_{n}$ is a root of $x^{n} - 1$ and $\zeta_{n}^{k}, 0 \leq k \leq n -1$ are distinct roots of $x^{n} - 1$. It follows that $$ (x - \zeta_{n}^{0})(x - \zeta_{n}^{1}) \cdots (x - \zeta_{n}^{n-1}) \mid x^{n} - 1 $$ in $\mathbb{C}[x]$. By comparing degrees, we know that $$ x^{n} - 1 = (x - \zeta_{n}^{0})(x - \zeta_{n}^{1}) \cdots (x - \zeta_{n}^{n-1}). $$ Therefore, a splitting field of $x^{n} - 1$ is $\mathbb{Q}(\zeta_{n}, \dots, \zeta_{n}^{n-1}) = \mathbb{Q}(\zeta_{n}) = \mathbb{Q}[\zeta_{n}]$.

If $k$ is coprime to $n$, then $\mathbb{Q}(\zeta_{n}) = \mathbb{Q}(\zeta_{n}^{k})$.

When $n = p$ is prime, we have shown that $\Phi_{p}(x) = \frac{x^{p} - 1}{x - 1} = x^{p-1} + \cdots + x + 1$ is irreducible in $\mathbb{Q}[x]$ in MATH 111B. And $\Phi_{p}(\zeta_{p}) = 0$ and $\Phi_{p}(x)$ is monic, so $m_{\zeta_{p}, \mathbb{Q}}(x) = \Phi_{p}(x)$. Therefore, $[\mathbb{Q}(\zeta_{p}) : \mathbb{Q}] = \deg \Phi_{p}(x) = p-1$.

Later, we will show that $[\mathbb{Q}(\zeta_{n}) : \mathbb{Q}] = \lvert (\mathbb{Z} / n\mathbb{Z})^{\times} \rvert$.

1. Proposition 1503 show that all splitting fields of $f(x)$ contained in a common field extension $K$ of $F$ are the same.
2. We prove by induction on $\deg f_{1}(x) = n$.

   1. Case $n = 1$: in this case, $E_{1} = F_{1}$ and $E_{2} = F_{2}$. The isomorphism $\sigma$ is simply $\varphi$.

   2. Case $n$: assume the theorem holds for all $\deg f_{1}(x) < n$.

      Let $\alpha$ be a root of $f_{1}(x)$ over $E_{1}$, we first find its corresponding root $\beta$ of $f_{2}(x)$ over $E_{2}$ such that $\sigma(\alpha) = \beta$. After that, we establish $F_{1}(\alpha) \cong F_{2}(\beta)$ and write $f_{1}(x) = (x - \alpha)g_{1}(x)$ and $f_{2}(x) = (x - \beta)g_{2}(x)$ with $\deg g_{1}(x) < n$ and $\deg g_{2}(x) < n$, and then use assumption on $g_{1}, g_{2}$ over $F_{1}(\alpha), F_{2}(\beta)$ to get $F_{1}(\alpha)(\alpha_{1}, \dots, \alpha_{n-1}) \cong F_{2}(\beta)(\beta_{1}, \dots, \beta_{n-1})$ where $\alpha_{1}, \dots, \alpha_{n-1}$ are the roots of $g_{1}(x)$ and $\beta_{1}, \dots, \beta_{n-1}$ are the roots of $g_{2}(x)$. And finally claim that $E_{1} = F_{1}(\alpha, \alpha_{1}, \dots, \alpha_{n-1}) = F_{1}(\alpha, \dots, \alpha_{n-1})$ and the same for $E_{2}$.

Let $F$ be the base field and $f(x) \in F[x]$ be the polynomial with $\deg f(x) = n$. Let $\alpha_{1}, \dots, \alpha_{n}$ be the roots of $f(x)$ and $E = F(\alpha_{1}, \dots, \alpha_{n})$ be the splitting field of $f(x)$.

Then $[F(\alpha_{1}) : F]$ has degree at most $n$ since $1, \alpha_{1}, \dots, \alpha_{1}^{n}$ is not linearly independent ($f(\alpha_{1}) = 0$). Moreover, by Homework 2, $[F(\alpha_{1}, \alpha_{2}) : F(\alpha_{1})] < [F(\alpha_{1}) : F] \leq n \implies [F(\alpha_{1}, \alpha_{2}): F(\alpha_{1})] \leq n-1$. If we keep doing so, we will finally arrive at $$ \begin{align*} [E : F] &= [F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] \cdots [F(\alpha_{1}, \alpha_{2}) : F(\alpha_{1})][F(\alpha_{1}) : F] \\ &\leq 1 \cdots \cdot (n-1)n \\ &= n! \end{align*} $$

1 $\iff$ 2 $\iff$ 3 is obvious.

3 $\implies$ 4: Take $\alpha \in K$ and consider $m_{\alpha, \Omega}(x)$. It must be $m_{\alpha, \Omega}(x) = x - \alpha \in \Omega[x]$ since it is irreducible and monic, hence $\alpha \in \Omega$. So we conclude that $K = \Omega$.

4 $\implies$ 5: Obvious because of Proposition 14010.

5 $\implies$ 1: Take any $f(x) \in \Omega[x]$ and let $K$ be its splitting field. Then by Proposition 1507, $[K : \Omega]$ is at most $n!$ where $n = \deg f(x)$, hence it is finite. So $K = \Omega$. Then we can conclude that every non-constant polynomial in $\Omega[x]$ splits completely in $\Omega$.

Case 1: $K / F$ is finite.

We prove by induction on $[K / F] = n$. The initial case $n = 1$ implies $K = F$ and hence $\varphi = \psi$. Assume the theorem holds for $n-1$. Pick any $\alpha \in K - F$, we consider $m_{\alpha, F}(x) \in F[x]$. Our strategy is to use Theorem 1506 to deduce it from $n$ to $n-1$ and then use the assumption.

Clearly $\varphi(m_{\alpha, F}(x)) \in \Omega[x]$. Since $\Omega$ is algebraically closed, $\varphi(m_{\alpha, F}(x))$ has a root $\beta \in \Omega$. Moreover, $\varphi(m_{\alpha, F(x)})$ is monic and irreducible because $m_{\alpha, F}(x)$ is monic and irreducible. So $m_{\beta, \Omega}(x) = \varphi(m_{\alpha, F}(x))$. Now we can apply Theorem 1506 and obtain an isomorphism $\varphi' : F(\alpha) \to \Omega$such that $\varphi'|_{F} = \varphi$. Then, by our induction hypothesis, we can extend $\varphi'$ to $\psi : K \to \Omega$ since $$ [K : F(\alpha)] = \frac{[K : F]}{[F(\alpha) : F]} $$ and we have $[F(\alpha) : F] > 1$ (because $\alpha \not\in F$).

Case 2: $K / F$ is infinite.

In this case, we use Zorn’s Lemma and the proved case that $K / F$ is finite for assistance.

Let $S$ be the set of pairs $(L, \phi)$ where $L$ is a subfield of $K$ containing $F$ with $\phi : L \to \Omega$ a field embedding with $\phi|_{F} = \varphi$. First, $S$ is not empty since $(F, \operatorname{id}_{F}) \in S$. Second, we define an order on $S$ by $$ (L, \phi) \leq (L', \phi') \iff L \subset L' \text{ and } \phi'|_{L} = \phi. $$ Suppose $C \subseteq S$ is a chain. We show that $C$ has an upper bound in $S$. Let $$ E = \bigcup_{(L, \phi) \in C}L. $$ $E$ is indeed a subfield of $K$. Given $\alpha \in E$, there exists $(L, \phi) \in C$ such that $\alpha \in L$. Set $\sigma(\alpha) = \phi(\alpha)$. We show that $\sigma(\alpha)$ does not depend on the choice of $(L, \phi)$. If $(L', \phi')$ is another element in $C$ satisfying $\alpha \in L'$, then $(L', \phi') \leq (L, \phi)$ or $(L, \phi) \leq (L', \phi')$ because $C$ is a chain. If it is the former case, then $L' \subseteq L$ and $\phi |_{L'} = \phi'$, which implies $\phi'(\alpha) = \phi(\alpha)$. The latter case is almost the same. Therefore, for every $\alpha \in E$, we have defined a unique $\sigma(\alpha) \in \Omega$. So $\sigma : E \to \Omega, \alpha \mapsto \phi(\alpha)$ is a well-defined map. Moreover, $\sigma |_{F} = \varphi$ and $(E, \sigma) \in S$ and is an upper bound of $C$. By Zorn’s Lemma, $S$ has a maximal element $(L, \phi)$.

We then show that $L = K$ by contradiction. If $L \subsetneq K$, pick any $\alpha \in K - L$, then $L(\alpha) / L$ is finite. By the case of finite extension proved above, $\phi : L \to \Omega$ extends to $\phi': L(\alpha) \to \Omega$. Then $(L, \phi) \leq (L(\alpha), \varphi')$ and $(L, \phi) \neq (L(\alpha), \phi')$. This contradicts the maximality of $(L, \phi)$. Therefore, $L = K$ and $\phi$ is the desired extension of $\varphi$.

$\impliedby$: Trivial.

$\implies$: Take any non-constant $f(x) \in \Omega[x]$. By Theorem 1504, $f(x)$ has a splitting field $\Omega'$ and $\Omega' / \Omega$ is finite thus algebraic by Proposition 1404. Then by Proposition 14012, since $\Omega' / \Omega$ and $\Omega / F$ are algebraic, $\Omega' / F$ is algebraic too. Hence for any $\alpha \in \Omega'$, there exists $a(x) \in F[x]$ such that $a(\alpha) = 0$. Now, since $\Omega$ is an algebraic closure of $F$, $a(x)$ splits completely in $\Omega$, which implies $\alpha \in \Omega$. Therefore, $\Omega' = \Omega$. So every non-constant $f(x) \in \Omega[x]$ splits completely over $\Omega$, $\Omega$ is algebraically closed.

1. Clearly $F' / F$ is algebraic. Take any $f(x) \in F[x]$, $f(x)$ splits completely over $\Omega$. By Proposition 1503, $F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x)$, where $f(x) = a(x - \alpha_{1})\cdots(x - \alpha_{n}), a \in F, \alpha_{i} \in \Omega$. Hence each $\alpha_{i}$ is algebraic over $F$ and belongs to $F'$, which implies that $f(x)$ splits completely over $F'$. Therefore, $F'$ is an algebraic closure of $F$.
2. $\alpha \in \Omega$ is algebraic over $F$ implies it is algebraic over $K$, so $\alpha \in K'$, which means $F' \subseteq K'$. By Proposition 14012, $K' / F$ is algebraic. So every $\beta \in K'$ is algebraic over $F$, hence $\beta \in F'$, it follows that $K' \subseteq F'$. So $K' = F'$.

1. TODO
2. Apply Proposition 1603 to $L / F$ and the natural field embedding $F \to L'$, then there exists a field embedding $\varphi : L \to L'$ such that $\varphi |_{F} = \operatorname{id}_{F}$. By Proposition 1605, $L$ and $L'$ are both algebraically closed. Then $L$ is algebraically closed $\implies$ $\varphi(L)$ is algebraically closed since $L \cong \varphi(L)$ (recall that a field homomorphism is either $0$ or its kernel is $0$). By Proposition 14012, $L' / F$ is algebraic $\implies$ $L' / \varphi(L)$ is algebraic. By part 4 of Proposition 1601, $L' = \varphi(L)$ since $L'$ is algebraically closed.

We often denote an algebraic closure by $\overline{F}$.

WLOG, assume $f(x)$ is monic in $F[x]$.

Let $\alpha, \beta \in \overline{F}$ be roots of $f(x)$ with multiplicities $a, b$ respectively. Suppose $a < b$. Since $f(x)$ is monic and irreducible in $F[x]$, $m_{\alpha, F}(x) = m_{\beta, F}(x) = f(x)$. Therefore, by Theorem 1408, there exists a field isomorphism $$ \varphi : F(\alpha) \to F(\beta) $$ such that $\varphi(\alpha) = \beta$ and $\varphi(r) = r$ for all $r \in F$.

Over $F(\alpha)$, $f(x) = m_{\alpha, F}(x) = (x - \alpha)^{a}g(x)$ for some $g(x) \in F(\alpha)[x]$ such that $(x - \alpha) \nmid g(x)$. Over $F(\beta)$, $f(x) = m_{\beta, F}(x) = (x - \beta)^{b}h(x)$ for some $h(x) \in F(\beta)[x]$ such that $(x - \beta) \nmid h(x)$. Then $$ \begin{align*} \varphi \left( f(x) \right) &= (x - \varphi(\alpha))^{a} \varphi \left( g(x) \right) \\ &= (x- \beta)^{a} \varphi \left( g(x) \right) \\ &= (x - \beta)^{b}h(x). \end{align*} $$ It follows that $\varphi \left( g(x) \right) = (x - \beta)^{b-a}h(x)$. Since $b > a$, $(x - \beta) \mid \varphi \left( g(x) \right)$. Apply $\varphi ^{-1}$ on $\varphi \left( g(x) \right)$, it turns out that $$ g(x) = \varphi ^{-1} \varphi \left( g(x) \right) = (x - \alpha)^{b-a} \varphi ^{-1} \left( h(x) \right). $$ Hence $(x - \alpha) \mid g(x)$ in $F(\alpha)$, contradicting the fact that $(x - \alpha) \nmid g(x)$. So we must have $a = b$, i.e. the multiplicities of $\alpha$ and $\beta$ are equal. Because $\alpha, \beta$ are arbitrary, all roots of $f(x)$ in $\overline{F}$ have the same multiplicity.

1. Omitted.

2. Let $d(x) = \gcd(f(x), f'(x))$.

   Claim: $\alpha \in \overline{F}$ is a common root of $f(x)$ and $f'(x)$ if and only if $\alpha$ is a root of $d(x)$. Then, just follow part 1.

   Let’s prove the claim.

   $\impliedby$: It is clear that $d(\alpha) = 0 \implies f(\alpha) = f'(\alpha) = 0$.

   $\implies$: Since $F[x]$ is a PID, there exists $a(x), b(x) \in F[x]$ such that $a(x)f(x) + b(x)f'(x) = d(x)$. Then $d(\alpha) = a(\alpha)f(\alpha) + b(\alpha)f'(\alpha) = 0$.

3. A direct proposition of part 2.

4. By part 3.

5. $\operatorname{ch}(F) = 0 \implies f'(x) \neq 0$, then use part 3. For every $K / F$ and $\alpha \in K$, by part 4, $m_{\alpha, F}(x)$ is separable, so $\alpha$ is separable over $F$, hence $K / F$ is separable.

We prove by induction on $[K : F]$.

Trivial case: $[K : F] = 1$. In this case, $K = F$, the proposition is obviously true.

Assume the statement holds for all field extensions of degree $\leq n - 1$. Given a field extension $K / F$ with $[K : F] = n$, take $\alpha \in K - F$, by Proposition 17010, $$ \begin{align*} &\# \{ F\text{- embeddings } \varphi : K \to \overline{F} \} \\ = &\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \cdot \#\{ F(\alpha)\text{-embeddings }\varphi : K \to \overline{F(\alpha)} = \overline{F} \}. \end{align*} $$ By Proposition 1709, $$ \#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \} \leq [F(\alpha) : F] $$ $[K : F(\alpha)] < n$, so by the induction hypothesis, $$ \# \{ F(\alpha)\text{-embeddings } \varphi : K \to \overline{F} \} \leq [K : F(\alpha)]. $$ Therefore, by Theorem 1207, $$ \#\{ F\text{-embeddings } \varphi : K \to \overline{F} \} \leq [F(\alpha) : F][K : F(\alpha)] = [K : F]. $$ The equality holds if and only if the equality holds for $$ \#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \} \leq [F(\alpha) : F]. $$ So only if $$ \begin{align*} &\# \{ \text{roots of }m_{\alpha, F}(x) \text{ in } \overline{F} \} = [F(\alpha) : F] = \deg m_{\alpha, F}(x) \\ \iff &\text{all roots of }m_{\alpha, F}(x)\text{ are simple} \\ \iff &\alpha\text{ is separable over }F. \end{align*} $$ Since $\alpha \in K - F$ is arbitrarily chosen, it implies that the equality holds if and only if all $\alpha \in K - F$ is separable over $F$, so if and only if $K / F$ is separable.

1. The map $$\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \to \overline{F}, \qquad \varphi \mapsto \varphi(\alpha)$$ is injective since the image of $\varphi$ is completely determined by $\varphi(\alpha)$. To associate it with the roots of $m_{\alpha, F}(x)$ in $\overline{F}$, suppose $\varphi(\alpha) = \beta \in \overline{F}$. Then since $\varphi |_{F} = \operatorname{id}_{F}$ and $m_{\alpha, F}(x) \in F[x]$, we have $\varphi(m_{\alpha, F}(\alpha)) = m_{\alpha, F}(\beta) = \varphi(0) = 0$, hence $\beta$ is a root of $m_{\alpha, F}(x)$ in $\overline{F}$. Thus $\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \leq \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \}$.

   Conversely, given a root $\gamma$ of $m_{\alpha, F}(x)$ in $\overline{F}$, then the kernel of the field homomorphism $F[x] \to \overline{F}, a(x) \mapsto a(\gamma)$ is $\left( m_{\alpha, F}(x) \right)$. Thus we obtain a field embedding $$\tau : F[x] / \left( m_{\alpha, F}(x) \right) \to \overline{F}, \qquad \overline{a(x)} \mapsto a(\gamma).$$ Moreover, by Proposition 1407, $$\sigma : F[x] / \left( m_{\alpha, F}(x) \right) \to F(\alpha), \qquad \overline{a(x)} \mapsto a(\alpha)$$ is a field isomorphism. Then we can check that the field homomorphism $\tau \circ \sigma ^{-1} |_{F} = \operatorname{id}_{F}$ and $\tau \circ \sigma ^{-1} (\alpha) = \gamma$, so $\tau \circ \sigma ^{-1}$ is the desired field embedding. Thus $\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \geq \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \}$.

2. By $\# \{ \text{roots of }m_{\alpha, F}(x) \text{ in } \overline{F} \} \leq \deg m_{\alpha, F}(x) = [F(\alpha) : F]$, and the equality holds if and only if $\alpha$ is separable over $F$.

$\implies$: Suppose $L / F$ is separable. Given $\alpha \in K$, we have that $\alpha \in L$ is hence separable over $F$, so $K / F$ is separable. To show that $L / K$ is separable, consider $m_{\alpha, F}(x)$. Since $\alpha$ is separable over $F$, all roots of $m_{\alpha, F}(x)$ are simple. Moreover, $m_{\alpha, K}(x) \mid m(\alpha, F)(x)$ in $K[x]$, so all roots of $m_{\alpha, K}(x)$ are simple as well $\implies$ $\alpha$ is separable over $K$. Hence $L$ is separable over $K$.

$\impliedby$: Suppose that $K / F$ and $L / K$ are both separable.

Case 1: $[L : F] < \infty$. By Proposition 1707, $$ \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} = [K : F], \, \#\{ K\text{-embeddings } \varphi : L \to \overline{K} \} = [L : K]. $$ Then Proposition 17010 implies $$ \#\{ F\text{-embeddings }\varphi : L \to \overline{F} \} = [K : F][L : K] = [L : F]. $$ By Proposition 1707, this implies that $L / F$ is separable.

Case 2: $[L : F] = \infty$.

$\implies$ : Simply take $S = K$.

$\impliedby$ : We divide it into two cases.

Case 1: $[K : F] < \infty$. Use induction on $[K : F]$. The inertial case $[K : F] = 1$ is obviously true. Assume the statement holds for all field extensions of degree $\leq n - 1$. Suppose $[K : F] = n, K = F(S)$ and every element in $S$ is separable over $F$. Take any $\alpha \in S \setminus F$, $\alpha$ is separable over $F$ $\implies$ $\#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = [F(\alpha) : F]$ by Proposition 17010. Moreover, by induction hypothesis, $[K : F(\alpha)] < [K : F] = n$ (bc $\alpha \not\in F$), it follows that $K / F(\alpha)$ is separable. So by Proposition 1707, $\#\{ F(\alpha)\text{-embeddings} \varphi : K \to F(\alpha) \} = [K : F(\alpha)]$. Thus by Proposition 17010, $$ \begin{align*} &\#\{ F\text{-embeddings } \varphi : K \to \overline{F} \} \\ = &\#\{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \cdot \#\{ F(\alpha)\text{-embeddings } \varphi : K \to \overline{F(\alpha)} = \overline{F} \} \\ = &[F(\alpha) : F][K : F(\alpha)] \\ = &[K : F]. \end{align*} $$ So again, by Proposition 1707, $K / F$ is separable.

Case 2: $[K : F] = \infty$. Since $K = F(S)$, given any $\beta \in K$, there exists $\alpha_{1}, \dots, \alpha_{n} \in S$ such that $\beta \in F(\alpha_{1}, \dots, \alpha_{n})$. Now Case 1 $\implies$ $F(\alpha_{1}, \dots, \alpha_{n}) / F$ is separable, hence $\beta$ is separable over $F$. Therefore, $K / F$ is separable.

1. Let $S \subseteq \overline{\mathbb{F}}_{p}$ be the set of roots of $x^{p^{n}} - x$. We first show that $S$ is a subfield of $\overline{\mathbb{F}}_{p}$, then we show that $\lvert S \rvert = p^{n} = \deg (x^{p^{n}} - x)$, this suffices to show that $x^{p^{n}} - x$ is separable.

   First, $0 \in S$. Given $\alpha, \beta \in S$, we have $(\alpha - \beta)^{p^{n}} - (\alpha - \beta) = \alpha^{p^{n}} - \beta^{p^{n}} - (\alpha - \beta) = 0$. Hence $\alpha - \beta \in S \implies (S, +)$ is a subgroup of $(\overline{\mathbb{F}}_{p}, +)$. Next, $(\alpha\beta)^{p^{n}} - \alpha\beta = \alpha^{p^{n}}\beta^{p^{n}} - \alpha\beta = 0 \implies \alpha\beta \in S$ and $(\alpha ^{-1})^{p^{n}} - \alpha ^{-1} = (\alpha^{p^{n}})^{-1} - \alpha ^{-1} = 0 \implies \alpha ^{-1} \in S$ shows that $S$ is a subfield of $\overline{\mathbb{F}}_{p}$. Moreover, $\mathbb{F}_{p^{n}} = \mathbb{F}_{p}[S] = S$. This is because any $\alpha \in \mathbb{F}_{p}$ satisfies $\alpha^{p^{n}} - \alpha = \alpha^{p^{n} - 1}\alpha - \alpha = 1 \cdot \alpha - \alpha = 0$ by Fermat’s little theorem. Hence $\alpha \in S$ $\implies$ $\mathbb{F}_{p}[S] = S$.

   Now since $(x^{p^{n}} - x)' = p^{n}x^{p^{n} - 1} - 1 = -1$, we have $\gcd(x^{p^{n}} - x, (x^{p^{n}} - x)') = 1$. By Proposition 1704, $x^{p^{n}} - x$ is separable over $\mathbb{F}_{p}$, hence $\lvert S \rvert = p^{n}$.

   Then we prove the uniqueness of $\mathbb{F}_{p^{n}}$. Suppose that $F$ is a subfield of $\overline{\mathbb{F}}_{p}$ with $\lvert F \rvert = p^{n}$. Then the multiplicative group $F^{\times}$ is abelian of order $p^{n} - 1$. By Lagrange’s Theorem, we know that $\alpha^{p^{n} - 1} = 1$ for all $\alpha \in F^{\times}$. Hence every element in $F$ is a root of $x^{p^{n}} - x$, hence $F \subseteq \mathbb{F}_{p^{n}}$. Since they have the same cardinality, $F = \mathbb{F}_{p^{n}}$.

2. Let $F_{0}$ be $F$’s prime field. By the definition of $[F : F_{0}]$, which is $\dim_{F_{0}} F$, we have $\lvert F \rvert = \lvert F_{0} \rvert^{[F:F_{0}]}$. Hence $\lvert F_{0} \rvert$ is a prime that divides $p^{n}$ $\implies$ $\lvert F_{0} \rvert = p \implies F_{0} \cong \mathbb{F}_{p}$. Let $\varphi : F_{0} \to \overline{\mathbb{F}}_{p}$ be defined by $F_{0} \xrightarrow{\cong} \mathbb{F}_{p} \hookrightarrow \overline{\mathbb{F}}_{p}$. By [Proposition 1603], $\varphi$ extends to a field embedding $\psi : F \to \overline{\mathbb{F}}_{p}$. Then $\mathrm{Im}\psi \cong F$ is a subfield of $\overline{\mathbb{F}}_{p}$ and $\lvert \mathrm{Im}\psi \rvert = \lvert F \rvert = p^{n}$. By (1), we know that $\mathrm{Im}\psi = \mathbb{F}_{p^{n}}$. Therefore, $F \cong \mathbb{F}_{p^{n}}$.

1. Let $F$ be a field of characteristic $0$. Given finite field extension $K / F$ and $\alpha \in K$, let $m_{\alpha, F}(x) \in F[x]$ be the minimal polynomial of $\alpha$ over $F$. Then $m_{\alpha, F}(x)$ is irreducible. By (3) of Proposition 1704, $m_{\alpha, F}(x)$ is separable. Hence $\alpha$ is separable over $F$. So $K / F$ is separable $\implies$ $F$ is perfect.
2. Let $F$ be a finite field with $\operatorname{ch}(F) = p$ and $\lvert F \rvert = p^{n}$ for some integer $n$. By the proof of Theorem 1801 ,$\alpha = \alpha^{p^{n}} = (\alpha^{p^{n-1}})^{p}$ for every $\alpha \in F$. In particular, every element in $F$ is a $p$-th power in $F$. By (3), $F$ is perfect.
3. Suppose that $F = F^{p}$. Let $K / F$ be a finite extension and $\alpha \in K$. Let $m_{\alpha, F}(x)$ be the minimal polynomial of $\alpha$ over $F$. We show that $m_{\alpha, F}(x)$ is separable by contradiction. Suppose the opposite. Then by (2) of Proposition 1704, $\gcd \left( m_{\alpha, F}(x), m_{\alpha, F}'(x) \right) \neq 1$, then $\gcd \left( m_{\alpha, F}(x), m_{\alpha, F}'(x) \right) = m_{\alpha, F}(x)$ because $m_{\alpha, F}(x)$ is irreducible. It follows that $m_{\alpha, F}(x) \mid m'_{\alpha, F}(x)$. We deduce that $m_{\alpha, F}'(x) = 0$. Otherwise, $\deg m_{\alpha, F}'(x) < \deg m_{\alpha, F}(x)$, so $m_{\alpha, F}(x)$ cannot divide $m_{\alpha, F}'(x)$. It follows from the definition of derivative of polynomials that$$m_{\alpha, F}'(x) = 0 \implies m_{\alpha, F}(x) = a_{m}x^{p^{m}} + a_{m-1}x^{p^{m-1}} + \cdots + a_{1}x^{p} + a_{0}.$$The condition $F = F^{p}$ implies that there exists $b_{0}, \dots, b_{m} \in F$ such that $a_{j} = b_{j}^{p}$. Then$$\begin{align*} m_{\alpha, F}(x) &= b_{m}^{p}x^{p^{m}} + b_{m-1}^{p}x^{p^{m-1}} + \cdots + b_{1}^{p}x^{p} + b_{0}^{p} \\ &= \left( b_{m}x^{p^{m-1}} + b_{m-1}x^{p^{m-1}} + \cdots + b_{1}x + b_{0} \right)^{p},\end{align*}$$and $m_{\alpha, F}(x)$ is not irreducible. We get a contradiction. Hence, $m_{\alpha, F}(x)$ is separable, and $\alpha$ is separable over $F$, so $K / F$ is separable and $F$ is perfect.

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt{ 2 })$. One can check that $$ \tau : \mathbb{Q}(\sqrt{ 2 }) \to \mathbb{Q}(\sqrt{ 2 }), \qquad \tau(a + b\sqrt{ 2 }) = a - b\sqrt{ 2 }, \, a, b \in \mathbb{Q} $$ is an automorphism in $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$. If $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$, then $\sigma(\sqrt{ 2 })^{2} = \sigma(\sqrt{ 2 }^{2}) = \sigma(2) = 2$. Hence, $\sigma(\sqrt{ 2 }) = \pm \sqrt{ 2 }$. If $\sigma(\sqrt{ 2 }) = \sqrt{ 2 }$, then $\sigma = \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}$, then $\sigma = \tau$. Thus, $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right) = \{ \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}, \tau \} \cong \mathbb{Z} / 2\mathbb{Z}$ ($\tau \circ \tau = \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}$).

Let $K / F$ be an algebraic field extension. If $\alpha \in K$ with $m_{\alpha, F}(x)$ and $\sigma \in \mathrm{Aut}(K / F)$, then $\varphi(\alpha)$ is a root of $m_{\alpha, F}(x)$. Therefore, we have the map $$ \mathrm{Aut}\left( F(\alpha) / F \right) \to \{ \text{root of }m_{\alpha, F}(x) \text{ in }F(\alpha) \}, \qquad \sigma \mapsto \sigma(\alpha). $$ The map is injective because an automorphism $\sigma \in \mathrm{Aut}\left( F(\alpha) / F \right)$ is completely determined by $\sigma(\alpha)$. Next, we show that it is also surjective. If $\beta \in F(\alpha)$ is a root of $m_{\alpha, F}(x)$, then $F(\beta) = F(\alpha)$ ($[F(\alpha) : F(\beta)] = \frac{[F(\alpha) : F]}{F(\beta) : F} = \frac{\deg m_{\alpha, F}(x)}{\deg m_{\beta, F}(x)} = 1$), and $$ \sigma: F(\alpha) \to F(\beta) = F(\alpha), \qquad a(\alpha) \mapsto a(\beta) \, \forall a(x) \in F[x] $$ defines an automorphism of $F(\alpha)$ fixing $F$ by Theorem 1408. Thus, $\beta = \sigma(\alpha)$ belongs to the image of the map. This shows that it is surjective.

Example 2106 (1) can be viewed as a special case of this. The minimal polynomial of $\sqrt{ 2 }$ over $\mathbb{Q}$ is $x^{2} - 2$, and the set $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$ is in bijection with the set of roots of $x^{2} - 2$ in $\mathbb{Q}(\sqrt{ 2 })$ which is $\{ \pm \sqrt{ 2 } \}$.

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt[3]{ 2 })$. The minimal polynomial of $\sqrt[3]{ 2 }$ over $\mathbb{Q}$ is $x^{3} - 2$. $x^{3} - 2$ has three roots $$ \theta_{1} = \sqrt[3]{ 2 }, \quad \theta_{2} = \sqrt[3]{ 2 }e^{ 2 \pi i / 3 } = \sqrt[3]{ 2 } \frac{-1 + \sqrt{ 3 }}{2}, \quad \theta_{3} = \sqrt[3]{ 2 } e^{ 4 \pi i / 3 } = -\frac{\sqrt[3]{ 2 }(1 + \sqrt{ 3 })}{2} $$ in $\overline{\mathbb{Q}}$, but only $\sqrt[3]{ 2 }$ belongs to $\mathbb{Q}(\sqrt[3]{ 2 })$. $x^{3} - 2$ has only one root in $\mathbb{Q}(\sqrt[3]{ 2 })$. By Example 2106 (2), $$ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q} \right) = \{ 1 \}, \qquad \text{the trivial group}. $$

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$, which is the splitting field of $x^{3} - 2$ over $\mathbb{Q}$ with $[\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}] = 6$ (part 2 of Example 1505). We show that $\mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \cong S_{3}$.

Keep the notation in Example 2106 (4). We have $$ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) = \left< \sigma_{1}, \sigma_{2} \right> \cong S_{3}, $$ with $$ \begin{align*} \sigma_1 : && \sqrt[3]{2} &\longmapsto \sqrt[3]{2}\frac{-1 + \sqrt{-3}}{2}, & \sigma_2 : && \sqrt[3]{2} &\longmapsto \sqrt[3]{2}, \\ && \frac{-1 + \sqrt{-3}}{2} &\longmapsto \frac{-1 + \sqrt{-3}}{2}, & && \frac{-1 + \sqrt{-3}}{2} &\longmapsto \frac{-1 - \sqrt{-3}}{2}. \end{align*} $$ One can check that $$ \begin{align*} \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}(\sqrt[3]{ 2 }) \right) &= \left< \sigma_{2} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2} \right) \right) & = \left< \sigma_{2} \sigma_{1} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 - \sqrt{ -3 }}{2} \right) \right) & = \left< \sigma_{2} \sigma_{1}^{2} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}(\sqrt{ -3 }) \right) & = \left< \sigma_{1} \right> \quad (\cong \mathbb{Z} / 3\mathbb{Z}), \end{align*} $$ and $$ \begin{align*} \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{2} \right> } &= \mathbb{Q}(\sqrt[3]{ 2 }), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{2} \sigma_{1} \right> } &= \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2} \right), \\ \mathbb{Q}(\sqrt[3]{ 2 }. \sqrt{ -3 })^{\left< \sigma_{2}\sigma_{1}^{2} \right> } &= \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 - \sqrt{ -3 }}{2} \right), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{1} \right> } &= \mathbb{Q}(\sqrt{ -3 }), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{1}, \sigma_{2} \right> } &= \mathbb{Q}. \end{align*} $$

1. One can check that $F(S)$ is a splitting field for $A$ over $F$, where $S \subseteq \overline{F}$ is the subset consisting of roots of the polynomials in $A$. (cf. Proposition 1503)
2. It suffices to show that if $K$ is a splitting field for $A$ over $F$, then there exists a field isomorphism $\varphi : K \to F(S)$ with $\varphi |_{F} = \operatorname{id}_{F}$. Apply Proposition 1603 to

   

   then we obtain a field embedding $\varphi : K \to \overline{F}$. Every polynomial splits completely over $K$ $\implies$ $S \subseteq \varphi(K)$ $\implies$ $F(S) \subseteq \varphi(K)$ $\implies$ $\varphi ^{-1}(F(S)) \subseteq K$. By (2) of Definition 2202, since $K$ is a splitting field for $A$ over $F$, it follows that $\varphi ^{-1}(F(S)) = K$ $\implies$ $F(S) = \varphi(K)$. Moreover, $\varphi$ is injective and $\varphi |_{F} = \operatorname{id}_{F}$. So $\varphi$ is a field isomorphism that maps $K$ to $F(S)$ with $\varphi |_{F} = \operatorname{id}_{F}$.

Let $\alpha \in \overline{\mathbb{F}_{p}}$ be a root of $x^{p} - x + a$. To show $x^{p} - x + a$ is irreducible, it suffices to show that $m_{\alpha, \mathbb{F}_{p}}(x) = x^{p} - x + a$. Moreover, since $m_{\alpha, \mathbb{F}_{p}}(x) \mid (x^{p} - x + a)$, it suffices to show that $\deg m_{\alpha, \mathbb{F}_{p}}(x) \geq p$. We also know that $[\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}] = \deg m_{\alpha, \mathbb{F}_{p}}(x)$, so we only need to prove that $[\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}] \geq p$. This can be done by showing that $$ p \leq \lvert \mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}) \rvert \leq \#\{ \mathbb{F}_{p}\text{-embeddings }\varphi : \mathbb{F}_{p}(\alpha) \to \overline{\mathbb{F}_{p}} \} \leq [\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}]. $$ $\operatorname{Fr}$ is a good instance of $\mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p})$. If $\operatorname{id}_{\mathbb{F}_{p}(\alpha)}, \operatorname{Fr}, \dots, \operatorname{Fr}^{p-1}$ are different, then of course $\lvert \mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}) \rvert \geq p$ for that $\mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p})$ is a group. Let us check it on $\alpha$. First, since $\alpha$ is a root of $x^{p} - x + a$, we have $$ \operatorname{Fr}(\alpha) = \alpha^{p} = \alpha - a. $$ Then $$ \operatorname{Fr}^{2}(\alpha) = \operatorname{Fr}(\alpha - a) = \operatorname{Fr}(\alpha) - \operatorname{Fr}(a) = \alpha - a - a^{p} = \alpha - 2a. $$ Using induction on $n$, we can prove that $\operatorname{Fr}^{n}(\alpha) = \alpha - na$. Therefore, $\operatorname{id}_{\mathbb{F}_{p}(\alpha)}(\alpha), \operatorname{Fr}(\alpha), \dots, \operatorname{Fr}^{p-1}(\alpha)$ are differentiable, so are those automorphisms.

$\implies$: Since $K / F$ is Galois, it is normal and separable. By (2) of Proposition 2207, $\lvert \mathrm{Aut}(K / F) \rvert = \# \{ F\text{-embeddings }\varphi : K \to \overline{F} \}$. By Proposition 1707, $\#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} = [K : F]$. Hence $\lvert \mathrm{Aut}(K / F) \rvert = [K : F]$.

$\impliedby$: By (2) of Proposition 2207, $\lvert \mathrm{Aut}(K / F)\rvert = [K : F] \leq \# \{ F\text{-embeddings }\varphi : K \to \overline{F} \}$. By Proposition 1707, $[K : F] \geq \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \}$. Hence $[K : F] = \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} \implies$ $K / F$ is separable. At the same time, $\lvert \mathrm{Aut}(K / F) \rvert = \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} \implies$ $K / F$ is normal. Therefore, $K / F$ is Galois.

Fix $\overline{F} = \overline{K}$. Then $F \subseteq K \subseteq \overline{F}$.

$\implies$: Let $A = \{ m_{\alpha, F}(x) \in F[x] \mid \alpha \in K \}$. First, we show that $K$ is the splitting field of $A \subseteq F[x]$. Let $S = \{ \text{roots of } \overline{F} \text{ of polynomials in } A \}$. Then $F(S)$ is the splitting field for $A$. We need to show $K = F(S)$. Given $\alpha \in K$, by the definition of $A$, $m_{\alpha, F}(x) \in A$ and by the definition of $S$, $\alpha$ is a root of $m_{\alpha, F}(x) \in A$, so $\alpha \in S$. Thus, $K \subseteq S \subseteq F(S)$. Conversely, if $\beta \in S$, then $\beta$ is a root of $m_{\alpha, F}(x)$ for some $\alpha \in K$. Since $K / F$ is normal, $K$ contains all the roots of $m_{\alpha, F}(x)$ in $\overline{F}$, so $K$ contains $\beta$. This shows that $S \subseteq K$, so $F(S) \subseteq K$. We have proved $K = F(S)$, the splitting field for $A \subseteq F[x]$.

Also, because $K / F$ is separable, we know that every $\alpha \in K$ is separable over $F$ and $m_{\alpha, F}(x)$ is separable. Hence $A$ is a subset of separable polynomials in $F[x]$.

$\impliedby$: Suppose $A$ is the subset of separable polynomials in $F[x]$ and $K$ is the splitting field for $A \subseteq F[x]$. We show that $K / F$ is both normal and separable. Proposition 2205 $\implies$ $K / F$ is normal. By the proof of Proposition 2204, $K \cong F(S)$ where $S \subseteq \overline{F}$ is the set of roots of polynomials in $A$. The polynomials in $A$ are separable $\implies$ every element in $S$ is separable over $F$. Then Proposition 17012 $\implies$ $K = F(S)$ is a separable extension of $F$.

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 $\mathrm{Gal}\left( \mathbb{Q}(\zeta_{n}) / \mathbb{Q} \right) \cong \left( \mathbb{Z} / n\mathbb{Z} \right)^{\times}$.

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

1. (D&F Example (2) at p.563) More generally, any quadratic extension $K$ of any field $F$ of characteristic different from $2$ is Galois.
2. (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 $(x^{2} - 2)(x^{2} - 3)$.
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.

(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.

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.

In (2) of Example 1505, we have seen that $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right)$ is the splitting field of $x^{3} - 2$ over $\mathbb{Q}$, so $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right)/ \mathbb{Q}$ is Galois. By (5) of Example 2106, we know that $\mathrm{Gal}\left( \mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q} \right) = \left< \sigma_{1}, \sigma_{2} \right> \cong S_{3}$ with $$\begin{aligned} \sigma_1 : \sqrt[3]{2} &\longmapsto \sqrt[3]{2}\frac{-1 + \sqrt{-3}}{2}, &\qquad \sigma_2 : \sqrt[3]{2} &\longmapsto \sqrt[3]{2}, \\ \sqrt{-3} &\longmapsto \sqrt{-3}, &\qquad \sqrt{-3} &\longmapsto -\sqrt{-3}. \end{aligned}$$ The nontrivial proper subgroups of $\mathrm{Gal}\left( \mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q} \right)$ are $\left< \sigma_{1} \right>, \left< \sigma_{2} \right>, \left< \sigma_{1}\sigma_{2} \right>, \left< \sigma_{1}^{2}\sigma_{2} \right>$, and $\left< \sigma_{1} \right>$ is normal of order $3$, $\left< \sigma_{2} \right>, \left< \sigma_{1} \sigma_{2} \right>, \left< \sigma_{1}^{2} \sigma_{2} \right>$ are not normal and of order $2$. $$\begin{aligned} \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1 \rangle} = \mathbb{Q}(\sqrt{-3}), &\qquad \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_2 \rangle} = \mathbb{Q}(\sqrt[3]{2}), \\ \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1\sigma_2 \rangle} = \mathbb{Q}\left(\sqrt[3]{2}\frac{-1-\sqrt{-3}}{2}\right), &\qquad \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1\sigma_2^2 \rangle} = \left(\sqrt[3]{2}\frac{-1+\sqrt{-3}}{2}\right), \end{aligned}$$ and these are all the proper nontrivial sub-extensions of $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q}$.

(D&F p.576) Since every subgroup of the Klein-4 group is normal, all the subfields of $\mathbb{Q}(\sqrt{ 2 }, \sqrt{ 3 })$ are Galois extensions of $\mathbb{Q}$.

(D&F Example p.577-581)

Consider $\mathbb{F}_{p^{n}}$, the unique subfield of $\overline{\mathbb{F}}_{p}$ of cardinality $p^{n}$.

By Example 22016, $$ \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) = \left< \operatorname{Fr} \right> \cong \mathbb{Z} / n\mathbb{Z}. $$ The subgroups of $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right)$ are $\left< \operatorname{Fr}^{d} \right> \cong d\mathbb{Z} / n\mathbb{Z}, d \mid n, d \geq 1$. (They are all normal subgroups since $\mathbb{Z} / n\mathbb{Z}$ is abelian.) $$ \mathbb{F}_{p^{n}}^{\left< \operatorname{Fr}^{d} \right> } = \{ \alpha \in \mathbb{F}_{p^{n}} \mid \alpha^{p^{d}} = \alpha \} = \mathbb{F}_{p^{d}}. $$ By the Fundamental Theorem, we have $$ \mathrm{Gal}\left( \mathbb{F}_{p^{d}} / \mathbb{F}_{p} \right) = \mathrm{Gal}\left( \mathbb{F}_{p^{n}}^{\left< \operatorname{Fr}^{d} \right> } / \mathbb{F}_{p} \right) \cong \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) / \left< \operatorname{Fr}^{d} \right> = \left< \operatorname{Fr} \right> / \left< \operatorname{Fr}^{d} \right> \cong \mathbb{Z} / d\mathbb{Z}, $$ compatible with what we have proved in Example 22016: $\mathrm{Gal}\left( \mathbb{F}_{p^{d}} / \mathbb{F}_{p} \right) \cong \mathbb{Z} / d\mathbb{Z}$.

$KE / E$ is Galois: Use Proposition 17012 and (3) of Proposition 2205 and the premise that $K / F$ is Galois.

$\mathrm{Gal}(KE / E) \cong \mathrm{Gal}(K / K \cap E)$: The map $$ \mathrm{Gal}(KE / E) \to \mathrm{Gal}(K / K \cap E), \qquad \sigma \mapsto \sigma |_{K} $$ is a group homomorphism. The kernel is $\{ \operatorname{id}_{KE} \}$. Let $H$ be its image. Then $$ \begin{aligned} \alpha \in K^{H} &\iff \alpha \in K \text{ and } \sigma(\alpha) = \alpha \text{ for all } \sigma \in \mathrm{Gal}(KE / E) \\ &\iff \alpha \in K \cap \left( KE \right)^{\mathrm{Gal}(KE / E)}. \end{aligned} $$ $KE / K$ is Galois $\implies (KE)^{\mathrm{Gal}(KE / E)} = E$. Hence, $$ \alpha \in K^{H} \iff \alpha \in K \cap E. $$ It follows that $K^{H} = K \cap E$. Then again by the Fundamental Theorem of Galois Theory, we have $$ H = \mathrm{Gal}(K / K^{H}) = \mathrm{Gal}(K / K \cap E). $$

1. What we want to show is $$\begin{aligned}&[K_{1}K_{2} : F] \leq \frac{[K_{1} : F][K_{2} : F]}{[K_{1} \cap K_{2} : F]} \\\iff &[K_{1}K_{2} : K_{1} \cap K_{2}][K_{1} \cap K_{2} : F] \leq [K_{1} : F][K_{2} : K_{1} \cap K_{2}] \\\iff & \frac{[K_{1}K_{2} : K_{1} \cap K_{2}]}{[K_{2} : K_{1} \cap K_{2}]} \leq \frac{[K_{1} : F]}{[K_{1} \cap K_{2} : F]} \\\iff & [K_{1} K_{2} : K_{2}] \leq [K_{1} : K_{1} \cap K_{2}].\end{aligned}$$ Let $E = K_{1} \cap K_{2}$. Suppose $[K_{1} : E] = n$ and $K_{1} = E\alpha_{1} + \cdots + E\alpha_{n}$ where $\alpha_{1}, \dots, \alpha_{n} \in K_{1}$. Then $K_{1} = E(\alpha_{1}, \dots, \alpha_{n}) \implies K_{1}K_{2} = K_{2}(\alpha_{1}, \dots, \alpha_{n})$. Now, $[K_{1} : E] < \infty$ implies that $\alpha_{1}, \dots, \alpha_{n}$ are algebraic over $E$, hence algebraic over $K_{2}$. Then we have $K_{1}K_{2} = K_{2}[\alpha_{1}, \dots, \alpha_{n}]$.

   To show that $[K_{1}K_{2} : K_{2}] < n$, it suffices to show that $K_{1}K_{2} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$. Now every element of $K_{1}K_{2}$ is of the form of linear combinations of $K_{2}\alpha_{1}^{r_{1}}\cdots\alpha_{n}^{r_{n}}$.Since each $\alpha_{i} \in K_{1}$, their multiple is also in $K_{1}$, i.e. in $E\alpha_{1} + \dots + E\alpha_{n} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$. So is the combinations, i.e. the element of $K_{1}K_{2}$. Hence we have shown $K_{1}K_{2} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$.

2. Omitted

3. $K_{1} / F$ Galois $\implies K_{1} / K_{1} \cap K_{2}$ Galois. Then use Proposition 2404.