Basics

Galois is defined as normal and separable.

An extension $L/k$ is simple iff $L = K( \alpha)$ for some $\alpha\in L$.

Every finite separable extension is simple.

${\mathbf{GF}}(p^n)$ is a simple extension over ${ \mathbf{F} }_p$.

A field extension $L/k$ is algebraic iff every $\alpha \in L$ is the root of some polynomial $f\in k[x]$.

Every finite extension is algebraic.

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 $f$.

Normal Extensions

Let $L/k$ be an extension. Then TFAE:

$L/k$ is normal.
Every irreducible polynomial $f\in k[x]$ that has one root in $L$ has all of its roots in $L$, and thus splits in $L[x]$.
- So if $\alpha\in L$ then every Galois conjugate $\alpha_k \in L$ as well.. Thus either $f$ splits in $L$ or $f$ has no roots in $L$.

Useful trick: if $[L: k] = 2$ then $L/k$ is automatically normal.
Useful trick: if $L/K/k$, then $K/k$ is normal iff ${ \mathsf{Gal}}(L/K) {~\trianglelefteq~}{ \mathsf{Gal}}(L/k)$.
$K \coloneqq{\mathbf{Q}}(2^{1\over 3})$ is not normal, since $K\subset {\mathbf{R}}$ but $(x^3-2) = \prod_k x-\zeta_3^k 2^{1\over 3}$ with $\zeta_3, \zeta_3^2 \in {\mathbf{C}}$.
- Another reason: an embedding $\sigma: K\to \overline{k}$ can send $2^{1\over 3}$ to any other root of $x^3-2$.
${\mathbf{Q}}(\sqrt 2, \sqrt 3)$ is normal over ${\mathbf{Q}}$, since it it is finite and splits $f(x) \coloneqq(x^2-2)(x^2-3)$, which is a separable polynomial.
$L \coloneqq{\mathbf{Q}}(2^{1\over 4})$ is not normal, since it is finite but not the splitting field of any polynomial.
${\mathbf{Q}}(\zeta_k)$ is normal for $\zeta_k$ any primitive $k$th root of unity.
A normal non-separable extension: ${ \mathbf{F} }_p(x, y) _{/ {{ \mathbf{F} }_p (x^p, y^p)}}$. This has finite degree $p^2$ but infinitely many subfields?

For $L/k$ algebraic: let $\overline{k}$ be an algebraic closure containing $L$, then $L/k$ is normal iff every $k{\hbox{-}}$embedding $\sigma: L\to \overline{k}$ satisfies $\operatorname{im}\sigma = L$, so $\sigma$ is a $k{\hbox{-}}$automorphism of $L$:

If $K/k$ is algebraic, then there is an extension $N_k/K$ such that $N_k/k$ is normal and $N_k/K/k$ is a tower. $N_k$ is referred to as the normal closure of $K/k$.

An extension $L/k$ is finite and normal $\iff L$ is the splitting field of some polynomial $f\in k[x]$.

$\implies$:

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 $[L(a): L] = 1$.

${\left\lvert {\mathop{\mathrm{Aut}}(L/k)} \right\rvert} \leq [L: k]$ with equality precisely when $L/k$ is normal.

Separable Extensions

A polynomial $f \in k[x]$ is separable iff $f$ has no repeated roots.

$x^2-1$ is separable over ${\mathbf{Q}}$, but inseparable over ${ \mathbf{F} }_2$ since it factors as $(x-1)^2$.
$(x^2-2)^2$ is inseparable over ${\mathbf{Q}}$
$x^2-t$ is inseparable over ${ \mathbf{F} }_2(t)$.
$f(x) \coloneqq x^n-1$ is inseparable over ${ \mathbf{F} }_p$ when $p\divides n$.
- Otherwise, $f' = nx^{n-1}$ has only $x=0$ as roots, whereas $0$ is not a root of $f$, so $f$ is separable.
$f(x) \coloneqq x^p-t$ is not separable over ${ \mathbf{F} }_p(t)$: it is irreducible by Eisenstein, but has only the single root $t^{1\over p}$.
$f(x) \coloneqq x^{p^n}-x$ is separable over ${ \mathbf{F} }_p$, since $f'(x) = -1$ has no roots at all.

Let $L/k$ be a field extension, $\alpha \in L$ be algebraic over $k$, and $f(x) \coloneqq\min(\alpha, k)$. The following are equivalent

$L/k$ is a separable extension.
Every element ${\alpha} \in L$ is separable over $k$, so $\alpha$ has separable minimal polynomial $m(x)$ in some splitting field of $m$.
Every finite subextension $L'/k$ is separable.

If $\alpha \in K/k$ is separable, then $\alpha$ is separable in any larger field $L/K/k$ since the minimal polynomial over the larger field will divide the minimal polynomial over the smaller field.

$f$ is separable iff $\gcd(f, f')=1$, so $f, f'$ share no common roots. Moreover, the multiple roots of $f$ are precisely the roots of $\gcd(f, f')$.

$\not\implies$: 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.

$f\in k[x]^{{\mathrm{irr}}}$ is inseparable (so $f$ has a repeated root) iff $f'(x) \equiv 0$.

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 $f' \equiv 0$.

For any field $k$, $f\in k[x]$ is separable $\iff f'\not\equiv 0 \in k[x]$.
For $\operatorname{ch}k = 0$, irreducible implies separable.
For $\operatorname{ch}k = p$, irreducibles $f(x)$ are inseparable iff $f(x) = g(x^p)$ for some $g\in k[x]$.

Thus for an irreducible polynomial $f$, \begin{align*} f\text{ separable} \iff \gcd(f, f')=1 \iff f'\not\equiv 0 \iff_{\operatorname{ch}k = p} f(x) = g(x^p) .\end{align*}

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 $f = a_0 + a_px^p + a_{2p}x^{2p} + \cdots$.

If $f\in k[x]^{{\mathrm{irr}}}$ and $p\coloneqq\operatorname{ch}k > 0$, then $f$ inseparable $\iff f(x) = q(x^{p^n})$ for some unique $n$.

$\implies$:

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 $f\in k[x]^{{\mathrm{irr}}}$, then $f$ is automatically separable.

Why this is true: assuming $f$ is irreducible, $\gcd(f, f') = 1$ or $f$. It can’t be $f$, since $f\divides f'$ would force $\deg f = \deg f' = 0$ and make $f$ a constant. So this $\gcd$ is 1.

Use that irreducible polynomial $f$ must have distinct roots, by the argument above. (In fact, it is the minimal polynomial of its roots.)
Toward a contradiction, suppose $f$ is irreducible but inseparable.
Then $f(x) = g(x^p)$ for some $g(x) \coloneqq\sum a_k x^k$.
Since Frobenius is bijective, write $a_k = b_k^p$ for some $b_k$, 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*} making $f$ reducible. $\contradiction$

A finite extension of a perfect field is automatically separable, and one only needs to show normality to show it’s Galois.

If $L/k$ is a finite extension, then, TFAE:

$L/k$ is separable.
$L = k( \alpha)$ for $\alpha$ a separable element.
$L = k( S )$ for $S$ some set of separable elements
$[L: K] = [L:K]_s$, i.e. the separable degree equals the actual degree. \begin{align*} [L: k] = \left\{{ L: k }\right\} \coloneqq{\sharp}\mathop{\mathrm{Aut}}_{{ \mathsf{Fields}_{k}}}(L) .\end{align*}

If $L/k$ is separable, then \begin{align*} [L: k] = \left\{{ L: k }\right\} .\end{align*} If $L/k$ is a splitting field, then \begin{align*} [L:K] = {\sharp}\mathop{\mathrm{Aut}}_{{ \mathsf{Fields}_{k}}}(L) \coloneqq{\sharp}{ \mathsf{Gal}}(L/k) .\end{align*}

Irreducible polynomials have distinct roots after passing to a splitting field.

If $\operatorname{ch}k = 0$ or $k$ is finite, then every algebraic extension $L/k$ is separable.

If $k$ is a perfect field, then every irreducible $f\in k[x]^{{\mathrm{irr}}}$ is automatically separable.

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. $\contradiction$.

The separable degree of an extension $L/k$ is defined by fixing an embedding $\sigma: k\hookrightarrow\overline{k}$ (the algebraic or separable closure) and letting $[L:k]_s$ be the number of embeddings $\sigma':L\to \overline{k}$:

If $L/K/k$, then $L/K$ is separable and $K/k$ is separable $\iff$ $L/k$ is separable:

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:

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.

$E/k$ and $F/k$ are separable $\iff$ $EF/k$ is separable.

$\impliedby$: 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 $EF/k$.

Galois Extensions

Let $L/k$ be a finite field extension. The following are equivalent:

$L/k$ is a Galois extension.
$L$ is normal, and separable.
The fixed field $L^H$ of $H\coloneqq\mathrm{Aut}(L/k)$ is exactly $k$.
$L$ is the splitting field of a separable polynomial $p\in K[x]$.
$L$ is a finite separable splitting field of an irreducible polynomial.
There is a numerical equality: \begin{align*} {\sharp}\mathop{\mathrm{Aut}}_{{ \mathsf{Fields}_{k}}} (L) = [L: k] = \left\{{ L: k}\right\} ,\end{align*} where $\left\{{E:F}\right\}$ is the number of isomorphisms to any field lifting $\operatorname{id}_F$:

figures/2021-08-09_22-29-40.png

In this case, we define the Galois group as \begin{align*} { \mathsf{Gal}}(L/k) \coloneqq\mathop{\mathrm{Aut}}_{{ \mathsf{Fields}_{k}}} (L/k) .\end{align*}

For $L/k$ algebraic and $\operatorname{ch}k = 0$, $L/k$ is Galois $\iff L/k$ is normal.

If $L/k$ is Galois, then $L/F$ is always Galois. Moreover, $F/k$ is Galois if and only if ${ \mathsf{Gal}}(L/F) {~\trianglelefteq~}{ \mathsf{Gal}}(L/k)$

In this case, \begin{align*} { \mathsf{Gal}}(F/k) \cong \frac{{ \mathsf{Gal}}(L/k)}{{ \mathsf{Gal}}(L/F)} .\end{align*}

Let $L/K/k$ with $L/k$ Galois. Then \begin{align*} K/k \text{ is Galois } \iff { \mathsf{Gal}}(L/K){~\trianglelefteq~}{ \mathsf{Gal}}(L/k) ,\end{align*} and moreover ${ \mathsf{Gal}}(K/k) = G$.

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.

${\mathbf{Q}}(\zeta_3, 2^{1/3})$ is normal but ${\mathbf{Q}}(2^{1/3})$ is not since the irreducible polynomial $x^3 - 2$ has only one root in it.
${\mathbf{Q}}(2^{1/3})$ is not Galois since its automorphism group is too small (only of size 1 instead of 3?).
${\mathbf{Q}}(2^{1/4})$ is not Galois since its automorphism group is too small (only of size 2 instead of 4). However, the intermediate extensions ${\mathbf{Q}}(2^{1/4}) / {\mathbf{Q}}(2^{1/2})$ and ${\mathbf{Q}}(\sqrt 2) / {\mathbf{Q}}$ are Galois since they are quadratic. Slogan: “Being Galois is not transitive in towers.”
A quadratic extension that is not Galois: $\operatorname{SF}(x^2 + y) \in { \mathbf{F} }_2(y)[x]$, which factors as $(x - \sqrt{y})^2$, making the extension not separable.

Fundamental Theorem of Galois Theory

Let $L/k$ be a Galois extension, then there is a correspondence: \begin{align*} \left\{{\substack{\text{Subgroups } H \leq { \mathsf{Gal}}(L/k)}}\right\} &\rightleftharpoons \left\{{\substack{\text{Fields } F \text{ such}\\ \text{that } L/F/k}}\right\} \\ H &\rightarrow \left\{{\substack{E^H \coloneqq~\text{The fixed field of $H$}}}\right\} \\ \left\{{\substack{{ \mathsf{Gal}}(L/F) \coloneqq\left\{{ \sigma \in { \mathsf{Gal}}(L/k) {~\mathrel{\Big\vert}~}\sigma(F) = F}\right\}}}\right\} &\leftarrow F \end{align*}

This is contravariant with respect to subgroups/subfields.
$[F: k] = [G: H]$, so degrees of extensions over the base field correspond to indices of subgroups.
$[K : F] = {\left\lvert {H} \right\rvert}$
$L/F$ is Galois and $Gal(K/F) = H$
$F/k$ is Galois $\iff H$ is normal, and ${ \mathsf{Gal}}(F/k) = { \mathsf{Gal}}(L/k)/H$.
The compositum $F_1 F_2$ corresponds to $H_1 \cap H_2$.
The subfield $F_1 \cap F_2$ corresponds to $H_1 H_2$.

A trick for remembering the degree/index correspondence:

If $\operatorname{ch}k = 0$ or $k$ is finite, then $k$ is perfect.
$k = {\mathbf{C}},{\mathbf{R}}, {\mathbf{Q}}, { \mathbf{F} }_p$ are perfect, so any finite normal extension is Galois.
Every splitting field of a polynomial over a perfect field is Galois.

If $F/k$ is finite and Galois and $L/k$ is arbitrary, then $FL/L$ is Galois and \begin{align*} { \mathsf{Gal}}(FL/L) = { \mathsf{Gal}}(F / F\cap L) \subset { \mathsf{Gal}}(F/k) .\end{align*}

Quadratic Extensions

If ${ \mathbf{F} }$ is a field with $\operatorname{ch}({ \mathbf{F} })\neq 2$ and $E_{/{ \mathbf{F} }}$ is a degree 2 extension, then $E$ is Galois and $E = F(\sqrt{a})$ for some squarefree $a\in { \mathbf{F} }$.

If $E_{/{\mathbf{Q}}}$ is a quadratic extension, $E = {\mathbf{Q}}(\sqrt q)$ for some $q\in {\mathbf{Q}}$ squarefree. Explicitly, use the primitive element theorem to write $E = {\mathbf{Q}}(\alpha)$, let $f$ be the minimal polynomial, then take $q=b^2-4ac$. One can do slightly better by writing $b^2-4ac = a/b$ so that $\sqrt{b^2-4ac}=\sqrt{ab}/b$ and taking $q=ab$.

For ${ \mathbf{F} }_p$ a finite field of prime order, all quadratic extensions $E/{ \mathbf{F} }_p$ are isomorphic.

Distinguished Classes

A collection of field extensions $\mathcal{S}$ is distinguished iff

(Transitive property) For any tower $L/K/k$, the extension $L/k \in \mathcal{S} \iff L/K \in {\mathcal{S}}$ (upper transitivity) and $K/k\in \mathcal{S}$ (lower transitivity):

(Lifting property) Lifts of distinguished extensions are distinguished: $K/k\in \mathcal{S}$ and $L/k$ any extension $\implies LK/L \in \mathcal{S}$:

(Compositing property) Whenever $L/k, K/k\in {\mathcal{S}}$, the amalgam $LK/k \in {\mathcal{S}}$ as well:

The following classes of extensions are distinguished:

Algebraic.
Finite.
Separable.
Purely inseparable.
Finitely generated.
Solvable.

Normal extensions are not distinguished, since they fail the forward implication for (lower) transitivity. However, they do have the (forward implication) upper transitive, lifting, and compositing properties.

As a consequence, Galois extensions are also not distinguished.

For $L/F/k$: $L/k$ normal/algebraic/Galois $\implies L/F$ normal/algebraic/Galois.

Algebraic Extensions

If $L/K/k$ (not necessarily finite) with $L/K$ and $K/k$ both algebraic, then $L/k$ is algebraic.

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.

If $L/K/k$ with $\alpha$ algebraic over $L$, then $\alpha$ is algebraic over $K$ and $\min_{\alpha, L}$ divides $\min_{\alpha, K}$ (so minimal polynomials only get smaller in extensions).

Normal Extensions

$E_1/k$ normal and $E_2/k$ normal $\implies E_1E_2/k$ normal and $E_1 \cap E_2 / k$ normal.

Issues with Normal Towers

One can similarly produce towers where the total extension is normal but the lower iterate is not normal: take \begin{align*} L/K/k \coloneqq{\mathbf{Q}}(2^{1\over 3}, \zeta_3) / {\mathbf{Q}}(2^{1\over 3}) / {\mathbf{Q}} .\end{align*} Now $K/k$ isn’t normal, since ${ \mathsf{Gal}}(L/k) = S_3$ but ${ \mathsf{Gal}}(L/K) = {\mathbf{Z}}/2 \not{~\trianglelefteq~}S_3$.

Another example: let $L/k$ be any algebraic extension that isn’t normal, and take $N_k$ to be the normal closure to get $N_k/L$. Concretely, $N_{\mathbf{Q}}/ {\mathbf{Q}}(2^{1\over 3})/{\mathbf{Q}}$ works.

One can produce towers of successively normal extensions whose total extension is not normal in a cheap way: take \begin{align*} L/K/k \coloneqq{\mathbf{Q}}(2^{1\over 4}) / {\mathbf{Q}}(2^{1\over 2}) / {\mathbf{Q}} .\end{align*} Each iterate is normal since it’s quadratic, but the overall extension misses complex roots and is thus not normal.

For $L/k$ finite,

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 $\operatorname{im}\sigma = L$:

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.

Field Theory: Extensions and Towers

Basics

Normal Extensions

Separable Extensions

Galois Extensions

Fundamental Theorem of Galois Theory

Quadratic Extensions

Distinguished Classes

Algebraic Extensions

Normal Extensions

Issues with Normal Towers