If $K/F$ and $[K:F] = n$, then pick any $\alpha \in K$ and consider $1, \alpha , \alpha ^2, ...$. This yields $n+1$ elements in an $n{\hbox{-}}$dimensional vector space, and thus there is a linear dependence $$\begin{align*} f(\alpha ) \coloneqq\sum_{j=1}^n c_j \alpha ^j = 0 .\end{align*}$$ But then $\alpha$ is the root of the polynomial $f$.

$\implies$:

• Write $L = k(a_1, \cdots, a_n)$ by finiteness.
• Let $m_i$ be the minimal polynomials of the $a_i$.
• By normality, the $m_i$ split in $L[x]$.
• Then $L$ is the splitting field of $f(x) \coloneqq\prod_i m_i(x)$.

$\impliedby$:

• Suppose $L/k = \operatorname{SF}(f)$, and pick any monic $m\in L[x]^{{\mathrm{irr}}}$ with a root $a\in L$, so that $m$ is the minimal polynomial of $a$.

• Toward showing $m$ splits in $L$: let $M = \operatorname{SF}(m)$, we’ll show $M=L$.

• To show that for any root $b\in M$ we have $b\in L$, it suffices to show $[L(b): L] = 1$.

  The strategy: use that $[L(a):L] = 1$ since $a\in L$ by assumption, and try to relate the two degrees.

• We have $L/k$, and a number of towers to work with: $$\begin{align*} [L(a):k] &= [L(a): k(a)] [k(a): k] &= [L(a) : L] [L: k] \\ \\ [L(b):k] &= [L(b): k(b)] [k(b): k] &= [L(b) : L] [L: k] .\end{align*}$$

• In the first set of equalities, note that $k(a)_{/ {k}} \cong k(b)_{/ {k}}$ since $a,b$ are conjugate roots over $k$. Moreover $L(a)_{/ {k(a)}} \cong L(b) _{/ {k(b)}}$ since both are splitting fields for $f$.

• Thus $[L(a):k] = [L(b): k]$, which forces $[L(a): L] = [L(b): L]$ after dividing by $[L:k]$. But $[L(a): L] = 1$.

$\not\implies$: Suppose $f$ has a repeated root $r_i$, so its multiplicity is at least 2. Then $$\begin{align*} f(x) = (x-r)^2 g(x) \implies f'(x) = 2(x-r)g(x) + (x-r)^2g'(x) ,\end{align*}$$ so $r$ is a root of $f'$.

$\not\impliedby$: Suppose $r$ is a root of $f, f'$. Write $f(x) = (x-r)p(x)$ and $f'(x) = (x-r)p'(x) + p(x)$. Rearranging, $f'(x) - (x-r)p'(x) = p(x)$, and since $r$ is a root of the LHS it’s a root of the RHS. So $p(x) = (x-r) q(x)$ and $f(x) = (x-r)^2 q(x)$, making $r$ a repeated root.

Assume $f$ is monic, then $f$ is inseparable iff $f, f'$ have a common root $a$. So $(x-a)\divides q\coloneqq\gcd(f, f')$, and since $f$ is irreducible, it must be the minimal polynomial of $a$. Since $f'(a) = 0$, this forces $f'\divides f$, and since $\deg f' = \deg f - 1 < \deg f$ this forces $f' \equiv 0$.

• First part:
  • $\not A\implies \not B$:
    • Let $f$ be irreducible, and suppose $f$ is separable. If $d(x) \coloneqq\gcd(f, f') \neq 1$, then $f'$ can not divide $f$ since $f$ is irreducible, so $f$ divides $f'$. But $\deg f' < f$ and $f\divides f'$ forces $f'\equiv 0$.
  • $\not B \implies \not A$:
    • If $f'\equiv 0$, then $d(x) \coloneqq\gcd(f, f') = \gcd(f, 0) = f \neq 1$ and $f$ is not separable.
• Second part:
  • If $\operatorname{ch}k = 0$ and $f$ is irreducible, then $\deg f \geq 2$ and $\deg f' \geq 1$ so $f' \neq 0$ and $f$ is thus separable.
• Third part:
  • $\impliedby$: If $f(x) = g(x^p)$ then $f'(x) = g'(x^p)\cdot px^{p-1}\equiv 0$.
  • $\implies$: Let $f$ be irreducible and inseparable, so $f' \equiv 0 \in k[x]$. Then $f(x) \coloneqq\sum_{k=0}^n a_k x^k$ implies $f'(x) \coloneqq\sum_{k=1}^{n}ka_k x^{k-1}$, which is zero iff $ka_k \equiv 0$ so $p$ divides $ka_k$. So $a_k\not\equiv 0$ forces $p\divides k$, so $f = a_0 + a_px^p + a_{2p}x^{2p} + \cdots$.

$\implies$:

Use that $f$ is inseparable iff $f' \equiv 0$. The claim is that $f' \equiv 0$ in characteristic $p$ iff all exponents present in $f$ are divisible by $p$. If $f'\equiv 0$, write $$\begin{align*} f(x) &= a_nx^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \\ \implies f'(x) &= na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 \\ &\equiv 0 ,\end{align*}$$ which forces $i a_i = 0$ for all $i$. For any $a_i\neq 0$, this forces $i\equiv 0 \operatorname{mod}p$, so $a_i$ can only be nonzero when $p\divides i$, so $i=kp$ for some $k$. So reindex to write $$\begin{align*} f(x) = a_0 + a_1x^p + a_2x^{2p} + \cdots + a_n x^{np} = \qty{b_0 + b_1 x + b_2 x^{2} + \cdots + b_nx^{n}}^p \in \overline{k}[x] ,\end{align*}$$ using $(c+d)^p = c^p + d^p$ in characteristic $p$, and taking $b_i \coloneqq a_i^{1\over p} \in \overline{k}$ So $f' \equiv 0\implies f(x) = q(x^p)$ where $q(t) \coloneqq\sum b_i t^i$.

$\impliedby$: If $f(x) = q(x^p)$ for some $q$, the previous calculation shows $q$ has multiple roots, thus so does $f$, so $f$ is inseparable.

If $\operatorname{ch}k = 0$ and $f$ is irreducible, then since $\deg f' < \deg f$ and $f$ is irreducible we must have $\gcd(f, f')=1$ and $f$ is separable.

If $\operatorname{ch}k = p>0$, then if $f$ is irreducible and inseparable then $f(x) = g(x^p)$ for some $g$. Write $g(x) = \sum a_k x^k$, and since $k$ is perfect, write $b_k \coloneqq a_k^{1\over p}$, then $$\begin{align*} f(x) = \sum a_k x^{pk} = \sum b_k^p x^{pk} = \qty{\sum b_k x^k}^p ,\end{align*}$$ so $f$ is reducible. $\contradiction$.

Use that $L/k$ is separable $\iff [L:k] = [L:k]_s$.

$\impliedby$:

• By definition, every $\alpha \in L$ is separable over $k$.
• $K/k$ is separable:
  • Since $K \subseteq L$, any $\alpha \in K$ is also separable over $k$.
• $L/K$ is separable:
  • If $\alpha \in L$, then $\min_{\alpha, k}(x)$ is a separable polynomial over some splitting field.
  • Use that $L/k$ implies $\min_{\alpha, L}(x)$ divides $\min_{\alpha, k}(x)$, so the former is separable, done.

$\implies$:

• Now use that the separable degree is multiplicative in towers.

• If all extensions in sight are finite, this direction is immediate: $$\begin{align*} [L:k]_s = [L:K]_s [K:k]_s = [L:K][K:k] = [L:K] .\end{align*}$$

• For the infinite case, want to show every $\alpha\in L$ is separable over $k$. It suffices to show $\alpha$ is contained in some finite separable subextension. The strategy:

Link to Diagram

• Let $f(x) \coloneqq\min_{\alpha, K}(x)$ be the minimal polynomial of $\alpha$ over the intermediate extension $K$, which by assumption is separable since $L/K$ is separable.
  • So $f\in K[x]$, and letting $S$ be the finite set of coefficients of $f$, $S \subseteq K$.
  • Note that each coefficient $s\in S$ is separable over $k$ since $K/k$ is separable by assumption.
• Set $F\coloneqq k(\alpha, S) \cap K$. Note $K/k$ is separable and $F \subseteq K$, so $F/k$ is separable.
• Moreover $k(\alpha, S)/F$ is separable, since the minimal polynomial of $\alpha$ over $F$ is still $f$.
• Now $k(\alpha, S) / F /K$ is a tower of finite extensions where $k(\alpha, S)/F$ and $F/k$ are separable, so this reduces to the finite case.

$\impliedby$: Separability always descends to subfields, and $E \leq EF, F\leq EF$.

$\implies$:

• Write $E = k(S)$ for some finite set $S$. Then $EF = F(S)$.
• Use that $k(S)/k$ is separable iff $s\in S$ is a separable element for all $s$.
  • Since $E/k$ is separable, each $s\in S$ is separable over $k$.
• Since $F/k$ is separable, each $s\in S$ is separable over $F$.
• So $F(S)/F$ is separable.
• Now use the tower $F(S)/F/k$ to obtain $F(S)/k$ separable, which is $EF/k$.

• Note separability is distinguished, so $K/k$ is separable.
• $K/k$ is Galois $\iff$ $F/k$ is normal (since we already have separability).
• $\iff \sigma(K) = K$ for all $\sigma\in G$
• $\iff \sigma H \sigma^{-1}= H$ for all $\sigma \in G$.

So $H$ is normal and $G/H$ is a group. For the isomorphism, take $$\begin{align*} \rho: { \mathsf{Gal}}(L/k) &\to { \mathsf{Gal}}(K/k) \\ \rho &\mapsto { \left.{{\rho}} \right|_{{K}} } .\end{align*}$$ This is well-defined since by normality $\sigma(K) = K$. Any $f\in \ker \rho$ is the identity on $K$, so $f\in { \mathsf{Gal}}(L/K)$ and $\ker \phi = H$. Since $L/K$ is Galois, every $f\in { \mathsf{Gal}}(K/k)$ lifts to ${ \mathsf{Gal}}(L/k)$, making $\rho$ surjective.

• We want to show every $\alpha \in L$ is algebraic over $k$, and it suffices to show $\alpha$ is algebraic over some finite subextension $k(S)$.
• Pick $\alpha\in L$, then $\alpha$ is algebraic over $K$ by assumption, so it is a root of some $f\in K[x]$.
• Let $S$ be the finitely many coefficients of $f$, then $\alpha$ is algebraic over $k(S)$.
• Note that $k(S)/k$ is finite and thus algebraic, and $k(S,\alpha)/k(s)$ is finite and also algebraic, so we’re reduced to the finite case.
• It suffices to show $k(S, \alpha)/k(s)/k$ is finite, which follows from multiplicativity of degrees.

• Use the fact that for finite extensions, $L/k$ is normal and separable $\iff L$ is the splitting field of a separable polynomial $f\in k[x]$.
• Now regard $f$ as a polynomial in $K[x]$; then $L$ is still the splitting field of $f$ over $K$, done.

Alternatively,

• Let $\alpha\in L$ be a root of $f\in K[x]$ with $f$ irreducible, it suffices to show all roots of $f$ are in $L$.
• Let $m\in K[x]$ be the minimal polynomial of $\alpha$ over $K$, and let $m'\in k[x]$ be the minimal polynomial of $\alpha$ over $k$.
• Since $L/k$ is normal and $\alpha\in L$, $m'$ splits in $L$.
• Minimal polynomials are divisible in towers, so $m$ divides $m'$. Since $m'$ splits in $L$, so must $m$.

• Suppose $L/K/k$ with $L/k$ normal, we want to show $L/K$ is normal.
• Use the embedding characterization, it suffices to show that every embedding $\sigma: L\hookrightarrow\overline{K}$ satisfies $\operatorname{im}\sigma = L$:

Link to Diagram

• Now just use the fact that $\overline{k} = \overline{K}$, and since $k\subseteq K$, any $K{\hbox{-}}$morphism is also a $k{\hbox{-}}$morphism.
• Since $L/k$ is normal, $\sigma(L) = L$ and $L/K$ is thus normal.