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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 .
Its coordinate ring is , so the diagonal is the algebraic set and is Zariski closed.
Each factor has the cofinite Zariski topology. For a point off the diagonal, every basic product neighbourhood of it has , because both and are cofinite in the infinite field . Choosing gives . Thus no product-topology neighbourhood of is contained in the complement of the diagonal.
Hence the diagonal is not product-topology closed although it is Zariski closed. The projections still have fibres obtained by quotienting by the corresponding coordinate values, so the asserted fibre behaviour remains.
Depends on
Used by
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Sources
- MIT 18.725 Algebraic Geometry, Lecture 7, Remarks 10 and 12 (standard reference, not scraped)