Algebraic Number Theory¶

Job: study finite extensions $K/\mathbb{Q}$ (and global fields) via rings of integers, ideals, units, and Galois actions — arithmetic of $\mathbb{Z}$ upgraded to $\mathcal{O}_K$.

Core¶

Unique factorization of elements fails in general; it is restored at the level of ideals in Dedekind domains. The class group measures failure of PID; the unit group is described by Dirichlet’s unit theorem. Abelian extensions are controlled by class field theory (Artin reciprocity).

Page	Focus
Foundations	number fields, $\mathcal{O}_K$, discriminants, UFD failure, Dedekind domains, $\mathrm{Cl}_K$, unit theorem
Class field theory	abelian extensions, Artin map overview, local–global

Theorem spine¶

$\mathcal{O}_K$ is a Dedekind domain: every nonzero ideal factors uniquely into primes.
$\mathrm{Cl}_K=\mathrm{Pic}(\mathcal{O}_K)$ is finite; $h_K=|\mathrm{Cl}_K|$.
$\mathcal{O}_K^\times\cong\mu_K\times\mathbb{Z}^{r_1+r_2-1}$ (Dirichlet).
Global class field theory: $\mathrm{Gal}(K^{\mathrm{ab}}/K)\cong C_K/\overline{K^\times}$ (idèle class group quotient) via the Artin map.

Interfaces¶

Number theory: $\mathbb{Z}$, modular arithmetic, elementary primes
Abstract algebra: rings, modules, Galois theory
Analytic number theory: $\zeta_K$, Hecke $L$, class number formula
Complex analysis: Minkowski geometry of numbers (embeddings $K\otimes\mathbb{R}$)

See it¶

$\mathbb{Z}[\sqrt{-5}]$. $6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$ with irreducibles not associates — not a UFD. As ideals, $$ (2,1+\sqrt{-5})^2=(2),\quad (3,1+\sqrt{-5})(3,1-\sqrt{-5})=(3), $$ unique factorization into prime ideals; class group $\mathrm{Cl}_{\mathbb{Q}(\sqrt{-5})}\cong C_2$.

Check¶

Why does unique factorization of ideals not imply $\mathcal{O}_K$ is a UFD?

UFD concerns elements. Ideal factorization says every nonzero ideal is a product of prime ideals; principal ideals $(\pi)$ may still factor into non-principal ideals, so $\pi$ need not factor uniquely into element-irreducibles. PID $\Leftrightarrow$ UFD for Dedekind domains ($h_K=1$).

What is the unit rank of a totally real cubic field?

$r_1=3$, $r_2=0$ $\Rightarrow$ rank $r_1+r_2-1=2$. So $\mathcal{O}_K^\times\cong\mu_K\times\mathbb{Z}^2$ with $\mu_K=\{\pm 1\}$ typically.

Pitfalls¶

Order vs maximal order: $\mathbb{Z}[\sqrt{d}]$ need not be $\mathcal{O}_K$ (e.g. $d\equiv 1\pmod 4$); invertibility of ideals needs the full ring of integers (or careful conductor theory).
Ramification vs splitting vs inertia are local-global notions at each prime — do not conflate residue degree $f$ with ramification index $e$.
Class number formula mixes algebraic ($h,R,\mu$) and analytic ($\zeta_K$ residue) data — effectivity is subtle (Siegel zeros).

Refs¶

Neukirch, Algebraic Number Theory
Marcus, Number Fields
Cassels–Fröhlich, Algebraic Number Theory
Washington, Introduction to Cyclotomic Fields

Next¶

Foundations · Class field theory · Number theory · Analytic number theory · Galois theory