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Algebraic number theory studies integer factorization, divisibility, prime ideals, and units after extending the rational numbers to a number field. Its basic objects are a number field $K$ and its ring of integers ${\mathcal{O}}_K$: the counterparts of $\mathbb{Q}$ and $\mathbb{Z}$. The field norm turns multiplication in $K$ into multiplication of rational numbers, and takes algebraic integers to integers. These connections let linear algebra and Euclidean geometry answer arithmetic questions.^[ref]

A first example is $K=\mathbb{Q}(\sqrt{2})$ with ring of integers $\mathbb{Z}[\sqrt{2}]$. The integer $7$ acquires the factorizationEach factor has field norm $7$. Neither is a unit, and each is an irreducible element: any factorization into two nonunits would factor the prime number $7$ into two integers of absolute value greater than one. Arithmetic in a larger ring thus reveals structure invisible in ordinary integer factorization.

There is also a complication: unique factorization of elements can fail, as in the failure of unique factorization in the ring of integers generated by square root of minus five. The general replacement is prime ideal factorization. The ideal class group measures the obstruction to making every ideal a principal ideal.

The historical motivation is visible in Richard Dedekind's original 1877 exposition: he begins with ordinary integer factorization, passes to Gaussian integers, and then uses a quadratic example to explain why ideals are needed. His later section establishes unique factorization of ideals in a number field. This is a direct record of the development of the theory, rather than a claim that all its ideas originated in one year.^[ref][ref]

An associated prime of an $R$-module $M$ is a prime ideal equal to the annihilator ${\mathrm{Ann}}_R(m)$ of some nonzero $m\in M$. The set of these primes is denoted ${\mathrm{Ass}}_R(M)$. Over a Noetherian ring, any maximal member among the annihilators of nonzero elements is prime: if $abm=0$ and $bm\ne 0$, maximality forces ${\mathrm{Ann}}_R(bm)={\mathrm{Ann}}_R(m)$, so $am=0$. Hence every nonzero module has an associated prime.

For a finite minimal primary decomposition $I={\bigcap }_i{Q}_i$, these are the prime ideals $\sqrt{Q_i}$. This terminology applies without a Noetherian hypothesis when the decomposition exists. Its invariance follows from the intrinsic description as those prime ideals of the form $\sqrt{(I:x)}$, where $x\notin I$ and $(I:x)$ is a colon ideal. Indeed, $(I:x)={\bigcap }_i(Q_i:x)$, and each factor is either the whole ring or a primary ideal with radical of an ideal $\sqrt{Q_i}$. A finite intersection of prime ideals can be prime only if it equals one of them: otherwise multiplying one element outside the intersection from each factor contradicts the prime ideal condition. Conversely, irredundancy supplies $x\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$, giving $\sqrt{(I:x)}=\sqrt{Q_i}$. In a Noetherian ring these agree with the associated primes of a module $R/I$. For arbitrary rings, the annihilator definition of an associated prime of a module must not be substituted without a further hypothesis.

Every ideal class of this real quadratic number field contains an integral ideal of ideal norm at most $\sqrt{14}<4$, by the diamond form of the Minkowski convex body theorem. The only possible norms are one, two and three. The sole prime ideal above two has norm two and is principal, generated by $4+\sqrt{14}$, whose field norm is two. The prime three is inert, since $X^2-14$ is irreducible modulo three, so there is no ideal of norm three. Every class therefore has a principal small representative, and the ideal class group is trivial.

This illustrates why the small-ideal bound can prove a class number without listing arbitrary ideals: only prime ideals of a few possible norms need to be tested. Real quadratic fields require the absolute field norm; the generator here happens to have positive norm.

For an irreducible closed subvariety $Y$ of an irreducible variety $X$, its codimension is $\dim X-\dim Y$. In an affine chart it also equals the height of the corresponding prime ideal, by the dimension formula for finite-type domains over a field.

A prime ideal of a number field splits completely in a finite extension of degree $n$ when it factors into $n$ distinct prime ideals, each of residue-field degree one. In a Galois extension, an unramified prime ideal splits completely exactly when its Frobenius conjugacy class is the identity class. The Chebotarev density theorem then gives these primes Dirichlet density $1/n$.

For a number field $K$ and a nonzero prime ideal $\mathfrak{p}\subset {\mathcal{O}}_K$, the completion $K_{\mathfrak{p}}$ is the completion for the $\mathfrak{p}$-adic absolute value. It is a finite extension of ${\mathbb{Q}}_p$, where $p\mathbb{Z}=\mathfrak{p}\cap \mathbb{Z}$.

The different exponent at a prime ideal $\mathfrak{P}$ is its exponent in the prime ideal factorization of the different ideal. It equals the local different exponent of the corresponding extension of completions. The different exponent and tame ramification theorem detects unramified, tame, and wild behavior without assuming a Galois extension.

For a normal variety with affine coordinate ring $A$ and a Weil divisor $D$, the fractional ideal of functions whose principal Weil divisor is at least $D$ isFor an effective prime divisor this is its height-one prime ideal. If the divisor is Cartier locally, this ideal is locally principal. A nonprincipal height-one prime ideal in a normal local ring can therefore detect a non-Cartier divisor.

The ring $k\times k$, for a field $k$, has two prime ideals, $0\times k$ and $k\times 0$. Their localizations at a prime ideal are both isomorphic to $k$. Nevertheless $(1,0)(0,1)=0$ is a product of nonzero elements, so $A$ is not an integral domain. Nonzero orthogonal idempotents separate the two components of its spectrum of a commutative ring.

The dual-number ring over a field $k$ consists of elements $a+b\epsilon$ with ${\epsilon }^2=0$. The element $\epsilon$ is a nonzero nilpotent element; its ideal is the ring's only prime ideal. Hence its affine scheme has the same one-point topology as $\mathrm{Spec}k$ while its structure sheaf differs.

A finite extension of number fields is unramified everywhere when every finite prime ideal is unramified. If infinite places are included in the convention, one also requires every real place to remain real.

For a prime ideal $P$, suppose the kernel of a ring homomorphism $K=\ker (R\to R_P)$ is a finitely generated ideal, say $(a_1,\dots ,a_t)$. Choose $s_i\notin P$ killing $a_i$ and let $f={\prod }_i{s}_i$. Then $f\notin P$ and $fK=0$. If a fraction $a/f^n$ maps to zero in $R_P$, then $a\in K$, so $fa=0$ and the fraction was already zero in $R_f$. Thus the canonical ring homomorphism $R_f\to R_P$ is injective. The empty product $f=1$ handles $K=0$. Every Noetherian ring satisfies the finite generation hypothesis.

In a Noetherian ring, every radical of an ideal is an intersection of finitely many prime ideals. A maximal counterexample ideal $J$ cannot be prime. Choose $a,b\notin J$ with $ab\in J$; the strictly larger ideals $J+(a)$ and $J+(b)$ satisfy the result, and $\sqrt{J}=\sqrt{J+(a)}\cap \sqrt{J+(b)}$. This contradicts maximality. The unit ideal uses an empty intersection.

If $\alpha =a+bi$ is a Gaussian integer and $P$ is a nonzero prime ideal of $R=\mathbb{Z}[\alpha ]$, then $R/P$ is a finite field. Complex conjugation preserves $R$, so a nonzero $\beta \in P$ gives the positive integer $\beta \bar{\beta }\in P$. Primality supplies a rational prime $p\in P$. Since $\alpha$ satisfies the monic polynomial $t^2-2at+a^2+b^2$, the ring $R/P$ is a quotient of ${\mathbb{F}}_p[t]/(\bar{F})$ and has at most $p^2$ elements. It is a finite integral domain, hence a field. The proof does not require $R$ to be a principal ideal domain.

In a finite Galois extension $L/K$, choose a prime ideal $\mathfrak{P}$ above an unramified $\mathfrak{p}$. The Frobenius automorphism at $\mathfrak{P}$ is characterized on its residue field by $x\mapsto x^{N\mathfrak{p}}$. Changing $\mathfrak{P}$ conjugates this element, so its conjugacy class is well-defined. It is the identity class exactly at a completely split prime.

Let $B$ be the integral closure of a Dedekind domain $R$ in a finite separable extension $L/K$ of degree $n$. For a nonzero prime ideal $\mathfrak{p}$ of $R$, writeand let $f_{\mathfrak{q}/\mathfrak{p}}=[B/\mathfrak{q}:R/\mathfrak{p}]$. Then

A generic point is a point whose closure is the whole space. An integral scheme has a unique generic point; on an affine scheme associated to an integral domain, it corresponds to the zero prime ideal. Its local ring is the fraction field of the domain. On every nonempty affine chart of an integral scheme these fields identify with the same function field.

Going up and going down concern lifting inclusions of prime ideals through a ring extension. Going-up theorem lifts upward from a chosen lower prime for an integral extension. Going-down theorem lifts downward from a chosen upper prime under additional hypotheses, including an integrally closed base domain for integral extensions of domains.

The height of a prime ideal is the supremum of lengths of strict prime chains ending at it, equivalently the Krull dimension of its localization at a prime ideal. The Krull height theorem bounds the height of a minimal prime over an $n$-generated ideal by $n$.