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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12
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Principal opens form a basis for the Zariski topology on an affine variety

Statement

Let X be a classical affine variety over an algebraically closed field k. Then the principal opens DX(f) form a basis for the Zariski topology on X. In particular, DX(f)DX(g)=DX(fg) for all f,gk[X].

Facts & Assumptions

Given: A classical affine variety X over an algebraically closed field k.

[L1]

For fk[X], the principal open is DX(f)={xX:f(x)0} (A principal open subset of a classical affine variety).

[L2]

Closed subsets of affine space are zero loci of sets of polynomials (Zero loci in affine space are the closed sets of the classical Zariski topology).

Proof

technique · direct
1.1

For any fk[X], the set DX(f) is open because it is the complement in X of the closed subset where f vanishes. Also a point lies in DX(f)DX(g) exactly when both f and g are nonzero there, equivalently when fg is nonzero there. Thus DX(f)DX(g)=DX(fg).

L1givenalgebra
1.2

Let UX be Zariski-open and xU. Then XU is closed in X, so there is a set S of polynomials on the ambient affine space with XU=XV(S) by [L2]. Since xV(S), choose fS with f(x)0. Then xDX(f)U. Hence every open set is a union of principal opens.

L1L2choose
2.1

Steps 1.1 and 1.2 are exactly the basis criterion.

step 1.1step 1.2

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