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The closed points of the prime spectrum are exactly the maximal ideals

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $p \in Spec (R)$ . Then the singleton ${p}$ is closed in $Spec (R)$ if and only if $p$ is a maximal ideal.

Facts & Assumptions

Given: A commutative ring $R$ , a prime ideal $p \subseteq R$ , and the Axiom of Choice.

[L1]

The closure of ${p}$ is $V (p)$ (The closure of a prime is its vanishing set).

[L2]

A maximal ideal is a proper ideal contained in no strictly larger proper ideal (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct

1.1L1

The point $p$ is closed exactly when $\overline{{p}} = {p}$ . By [L1], this is equivalent to $V (p) = {p}$ .

2.1L2step 1.1

If $p$ is maximal, then every prime ideal containing $p$ equals $p$ by [L2]. Hence $V (p) = {p}$ , so $p$ is a closed point.

2.2L2step 1.1

Conversely, if ${p}$ is closed, then step 1.1 gives $V (p) = {p}$ . If $p \subseteq q$ for a prime ideal $q$ , then $q \in V (p)$ and therefore $q = p$ . Thus no strictly larger proper ideal can contain $p$ , so [L2] shows that $p$ is maximal.

3.1step 2.1step 2.2∎

Therefore the closed points of $Spec (R)$ are exactly the maximal ideals.

The closed points of the prime spectrum are exactly the maximal ideals

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources