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An idempotent partitions the spectrum into complementary clopen subsets

Statement

Let $R$ be a commutative ring and let $e \in R$ satisfy $e^{2} = e$ . Then $Spec (R) = D (e) ⊔ D (1 - e) = V (1 - e) ⊔ V (e),$ and both $D (e)$ and $D (1 - e)$ are clopen.

Facts & Assumptions

Given: A commutative ring $R$ and an idempotent $e \in R$ .

[L1]

$D (f)$ is the set of prime ideals that do not contain $f$ , and $D (f g) = D (f) \cap D (g)$ (Principal distinguished subsets of the prime spectrum, Distinguished-subset identities).

[L2]

A prime ideal is proper and has the factor property $a b \in p \Rightarrow a \in p$ or $b \in p$ (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct

1.1L1L2given

Since $e (1 - e) = 0$ , every prime ideal $p$ contains $e$ or $1 - e$ by [L2]. It cannot contain both, because then $1 = e + (1 - e)$ would lie in $p$ , contradicting properness. Therefore each prime lies in exactly one of $D (e)$ and $D (1 - e)$ , so $Spec (R) = D (e) ⊔ D (1 - e)$ .

2.1L1step 1.1algebra

A prime ideal lies in $D (e)$ exactly when it does not contain $e$ , which by step 1.1 is equivalent to containing $1 - e$ . Hence $D (e) = V (1 - e)$ . Similarly $D (1 - e) = V (e)$ . Since vanishing sets are closed and distinguished opens are open, both subsets are clopen.

3.1step 1.1step 2.1∎

Thus an idempotent partitions the spectrum into complementary clopen subsets.

Depends on

Used by

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, Lemma 14.2 (standard reference, not scraped)
The Stacks Project, Section 10.22: Connected components of spectra (standard reference, not scraped)