Let RR be a . A maximal ideal of RR is a proper ideal mR\mathfrak m\subsetneq R such that there is no ideal strictly between m\mathfrak m and RR; equivalently, R/mR/\mathfrak m is a .

The maximal spectrum of RR is the set

MaxSpec(R):={mRm is a maximal ideal}.\operatorname{MaxSpec}(R) := \{\mathfrak m \subset R \mid \mathfrak m \text{ is a maximal ideal}\}.

There is always an inclusion MaxSpec(R)Spec(R)\operatorname{MaxSpec}(R)\subseteq \operatorname{Spec}(R) (see ), since every maximal ideal is prime. One typically topologizes MaxSpec(R)\operatorname{MaxSpec}(R) by the subspace topology induced from the . Concretely, for an ideal IRI\subseteq R the corresponding closed subset of MaxSpec(R)\operatorname{MaxSpec}(R) is

V(I)MaxSpec(R)={mMaxSpec(R)Im}.V(I)\cap \operatorname{MaxSpec}(R)=\{\mathfrak m\in \operatorname{MaxSpec}(R)\mid I\subseteq \mathfrak m\}.

A point mMaxSpec(R)\mathfrak m\in \operatorname{MaxSpec}(R) has residue field R/mR/\mathfrak m, which agrees with the at m\mathfrak m.

Examples

  1. Local rings. If RR is a , it has a unique maximal ideal (see ), hence MaxSpec(R)\operatorname{MaxSpec}(R) is a single point.
  1. The integers. For R=ZR=\mathbb{Z}, the maximal ideals are exactly (p)(p) for primes pp. Thus
    MaxSpec(Z)={(p)p prime},\operatorname{MaxSpec}(\mathbb{Z})=\{(p)\mid p\ \text{prime}\},
    which is Spec(Z)\operatorname{Spec}(\mathbb{Z}) with the generic point (0)(0) removed.
  1. Polynomial rings over an algebraically closed field. If kk is algebraically closed and R=k[x1,,xn]R=k[x_1,\dots,x_n], then the identifies maximal ideals with points a=(a1,,an)kna=(a_1,\dots,a_n)\in k^n via
    a(x1a1,,xnan).a \longleftrightarrow (x_1-a_1,\dots,x_n-a_n).
    In this sense, MaxSpec(k[x1,,xn])\operatorname{MaxSpec}(k[x_1,\dots,x_n]) recovers affine nn-space over kk as a set, and its induced topology is the classical Zariski topology on knk^n.