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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 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 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.

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.

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 $n$.

The imaginary quadratic field $K=\mathbb{Q}(\sqrt{-163})$ has discriminant $-163$, and its Minkowski bound for ideal classes isThe polynomial $X^2-X+41$ defining its ring of integers of a quadratic field has no root modulo $2,3,5,$ or $7$, so these rational primes are inert. Factoring an integral ideal of norm at most eight into prime ideals then shows that its only possible nontrivial factor is the principal ideal $(2)$, of norm four. Every ideal class is therefore principal, and

A nonempty closed subset which, with its induced topology, is an irreducible topological space. In an affine algebraic set these subsets correspond, with reversed inclusion, to prime ideals of the coordinate ring.

For a nonnegatively graded ring $S={⨁}_{d\ge 0}{S}_d$, the irrelevant ideal isThe points of $\mathrm{Proj}S$ are precisely the homogeneous prime ideals that do not contain $S_{+}$.