Maximal spectrum - Knowlpedia

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

The maximal spectrum of $R$ is the set

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

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

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

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

Examples

Local rings. If $R$ is a local ring, it has a unique maximal ideal (see the characterization of local rings by a unique maximal ideal), hence $\operatorname{MaxSpec}(R)$ is a single point.

The integers. For $R=\mathbb{Z}$ , the maximal ideals are exactly $(p)$ for primes $p$ . Thus $\operatorname{MaxSpec}(\mathbb{Z})=\{(p)\mid p\ \text{prime}\},$ which is $\operatorname{Spec}(\mathbb{Z})$ with the generic point $(0)$ removed.

Polynomial rings over an algebraically closed field. If $k$ is algebraically closed and $R=k[x_1,\dots,x_n]$ , then the Nullstellensatz identifies maximal ideals with points $a=(a_1,\dots,a_n)\in k^n$ via $a \longleftrightarrow (x_1-a_1,\dots,x_n-a_n).$ In this sense, $\operatorname{MaxSpec}(k[x_1,\dots,x_n])$ recovers affine $n$ -space over $k$ as a set, and its induced topology is the classical Zariski topology on $k^n$ .