Galois Theory¶

Job: control intermediate fields of $L/K$ via subgroups of $\operatorname{Gal}(L/K)$; decide solvability by radicals.

Core¶

Setup. $L/K$ finite field extension; $G=\operatorname{Aut}_K(L)$. Always $\|G\|\le[L:K]$, with equality iff $L/K$ is Galois (normal + separable). Then write $\operatorname{Gal}(L/K)=G$.

Fundamental theorem of Galois theory (FTGT)¶

Theorem (FTGT). Let $L/K$ be finite Galois, $G=\operatorname{Gal}(L/K)$. There is an inclusion-reversing bijection $$ {\text{subgroups }H\le G}\ \longleftrightarrow\ {\text{fields }K\subseteq M\subseteq L}, $$ $$ H\mapsto L^H={x\in L:\sigma x=x\ \forall\sigma\in H},\qquad M\mapsto\operatorname{Gal}(L/M). $$ Moreover:

Field / group condition	Corresponding
$[L:L^H]=\\|H\\|$, $[L^H:K]=[G:H]$	index–degree
$H\trianglelefteq G$	$L^H/K$ normal (hence Galois)
Then $\operatorname{Gal}(L^H/K)\cong G/H$	quotient of Galois groups
Composite / intersection	lattice dualities via $\langle H_1,H_2\rangle$, $H_1\cap H_2$

Cyclotomic extensions¶

$\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois with $$ \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^\times,\qquad \sigma_a:\zeta_n\mapsto\zeta_n^a,\ \gcd(a,n)=1. $$ $[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\varphi(n)$. Subfields correspond to subgroups of units mod $n$ (Gaussian periods, etc.).

Solvability by radicals¶

Definition. $f\in K[x]$ (char $0$, $K$ contains enough roots of unity for clean statements, or pass to cyclotomic closure) is solvable by radicals if the splitting field lies in a tower $K=K_0\subseteq K_1\subseteq\cdots\subseteq K_m$ with each $K_{i+1}=K_i(\sqrt[n_i]{a_i})$.

Theorem. Let $f\in K[x]$, char $0$, $L$ splitting field over $K$. Then $f$ is solvable by radicals iff $\operatorname{Gal}(L/K)$ is a solvable group (composition factors cyclic of prime order).

Corollary. General quintic: $\operatorname{Gal}(S_5)$ or $A_5$ for suitable $f$ (e.g. $x^5-6x+3$ over $\mathbb{Q}$ has Galois group $S_5$) $\Rightarrow$ not solvable by radicals.

Infinite Galois theory (pointer)¶

$\operatorname{Gal}(\overline{K}/K)=\varprojlim\operatorname{Gal}(L/K)$ over finite Galois $L/K$; Krull topology (basis: $\operatorname{Gal}(\overline{K}/L)$). Closed subgroups $\leftrightarrow$ intermediate fields. Absolute Galois group of $\mathbb{Q}$ is a central object in number theory.

Discriminant and embeddings¶

If $L=K(\alpha)$, $\min_K(\alpha)=f$, then number of $K$-embeddings $L\to\overline{K}$ equals number of distinct roots of $f$. Separable degree vs inseparable degree multiplies to $[L:K]$ in general char $p$.

See it¶

Cubic. $f=x^3+ax+b\in\mathbb{Q}[x]$ irreducible. Disc $\Delta=-4a^3-27b^2$. Gal$(f)\cong A_3$ iff $\Delta$ square in $\mathbb{Q}$, else $S_3$. Cardano’s formulas implement the radical tower corresponding to the solvable series of $S_3$ or $A_3$.

Kummer theory. If $\mu_n\subseteq K$ and char $\nmid n$, then cyclic degree-$n$ Galois extensions are $K(\sqrt[n]{a})/K$ for $a\in K^\times/(K^\times)^n$, with Gal generated by $\sqrt[n]{a}\mapsto\zeta\sqrt[n]{a}$.

Check¶

Why does FTGT need normality of $L/K$? Give a counterexample without it.

Take $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt[3]{2})$. Then $\operatorname{Aut}_K(L)=\{\mathrm{id}\}$ (other cube roots not real / not in $L$), but $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, so $\|G\|<[L:K]$; intermediate fields are only $K,L$, but the splitting field $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$ is degree $6$ with Gal $S_3$.

Sketch: if $G=\operatorname{Gal}(L/K)$ is solvable, why does a radical tower exist (char $0$)?

Refine to cyclic prime-order quotients; adjoin roots of unity (cyclotomic, abelian Gal); each cyclic extension is Kummer after roots of unity, hence radical. Track fixed fields down the subnormal series.

Pitfalls¶

Normal closure: intermediate $M$ may fail to be normal over $K$ even if $L/K$ is Galois — precisely when $\operatorname{Gal}(L/M)$ is not normal in $G$.
Separability in char $p$: purely inseparable extensions have trivial Aut; Artin–Schreier theory replaces Kummer for degree $p$.
**“Solvable polynomial” $\neq$ “solvable group of the polynomial’s Gal over $\mathbb{Q}$” without care about the base field and roots of unity — statements are cleanest after extending scalars.
Infinite extensions: only closed subgroups appear in the Krull FTGT; dense proper subgroups exist abstractly but are not of the form $\operatorname{Gal}(\overline{K}/M)$.

Refs¶

Lang, Algebra
Morandi, Field and Galois Theory
Neukirch, Algebraic Number Theory (infinite Galois, local/global)

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Group theory · Rings and fields · Modules · Abstract algebra · Number theory