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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-07
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The Zariski topology on an affine product is generally not the product topology

Statement

For affine varieties the product construction has the expected set-theoretic fibres, but its Zariski topology need not equal the product topology of the two Zariski topologies.

Proof

Given: The affine product Ak1×kAk1.

1.1

Its coordinate ring is k[x]kk[y]k[x,y], so the diagonal is the algebraic set V(xy) and is Zariski closed.

givenalgebra
2.1

Each factor has the cofinite Zariski topology. For a point (a,b) off the diagonal, every basic product neighbourhood U×V of it has UV, because both U and V are cofinite in the infinite field k. Choosing cUV gives (c,c)(U×V)D. Thus no product-topology neighbourhood of (a,b) is contained in the complement of the diagonal.

step 1.1
3.1

Hence the diagonal is not product-topology closed although it is Zariski closed. The projections still have fibres obtained by quotienting k[x,y] by the corresponding coordinate values, so the asserted fibre behaviour remains.

step 2.1algebra

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