Spectrum is compact

This article gives a fact about the relation between ring-theoretic assumptions about a commutative unital ring and topological consequences for the spectrum
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Statement

The spectrum of a commutative unital ring is always a compact space.

Proof

We shall prove the statement using the finite intersection property formulation of compactness.

Given: $R$ is a commutative unital ring, and $X = S p e c (R)$ is its spectrum. Suppose $V_{i}$ is a collection of closed subsets (where $i \in I$ , an indexing set) such that any finite intersection of the $V_{i}$ s is nonempty.

To prove: The intersection of all the $V_{i}$ s is nonempty.

Proof: Let $I (V)$ denote the radical ideal corresponding to the closed set $V$ under the Galois correspondence between a ring and its spectrum.

Clearly, we have, by definition:

Failed to parse (unknown function "\bigplus"): {\displaystyle \mathcal{I}(\bigcap_i V_i) = \bigplus \mathcal{I}(V_i)}

In other words, a prime ideal contains all the $I (V_{i})$ s if and only if it contians their sum.

Suppose the intersection of the $V_{i}$ s was empty. Then the left side would be the whole ring $R$ , hence so would the right side. In other words, we'd have that $R$ is the sum of the $I (V_{i})$ s.

But this would imply that $1 \in R$ is in the sum of finitely many $I (V_{i})$ s, and that in turn would yield that the intersection of those finitely many $V_{i}$ s must be empty, contradicting the assumption of finite intersection property.