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The prime-spectrum construction is a contravariant functor to topological spaces

Statement

For every ring homomorphism $φ : R \to A$ , contraction defines a continuous map $Spec (φ) : Spec (A) ⟶ Spec (R) .$ These maps satisfy $Spec ({id}_{R}) = {id}_{Spec (R)} and Spec (ψ \circ φ) = Spec (φ) \circ Spec (ψ),$ so $Spec$ is a contravariant functor from commutative rings to topological spaces.

Facts & Assumptions

Given: Ring homomorphisms $φ : R \to A$ and $ψ : A \to B$ of commutative rings.

[L1]

For every ideal $I ⊴ R$ , $Spec (φ)^{- 1} (V (I)) = V (I A),$ so contraction pulls back vanishing sets to vanishing sets (A ring map induces a contraction map on prime spectra).

[L2]

The Zariski-closed subsets are exactly the vanishing sets (The vanishing sets define the Zariski topology on the prime spectrum).

Proof

technique · direct

1.1L1L2

Let $Z \subseteq Spec (R)$ be closed. By [L2], $Z = V (I)$ for some ideal $I ⊴ R$ . Then [L1] gives $Spec (φ)^{- 1} (Z) = Spec (φ)^{- 1} (V (I)) = V (I A)$ , which is closed in $Spec (A)$ by [L2]. Therefore $Spec (φ)$ is continuous.

1.2givenalgebra

For a prime ideal $p \subseteq R$ one has $Spec ({id}_{R}) (p) = {id}_{R}^{- 1} (p) = p$ , so $Spec ({id}_{R}) = {id}_{Spec (R)}$ .

1.3givenalgebra

For a prime ideal $q \subseteq B$ , one has $Spec (ψ \circ φ) (q) = (ψ \circ φ)^{- 1} (q) = φ^{- 1} (ψ^{- 1} (q)) = Spec (φ) (Spec (ψ) (q))$ . Hence $Spec (ψ \circ φ) = Spec (φ) \circ Spec (ψ)$ .

2.1step 1.1step 1.2step 1.3∎

Steps 1.1, 1.2, and 1.3 prove that $Spec$ is a contravariant functor to topological spaces.

The prime-spectrum construction is a contravariant functor to topological spaces

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources