LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The affine diagonal is cut out by coordinate differences

Statement

If $X \subseteq A^{n}$ is an affine variety, then the diagonal in $X \times X$ is cut out by $x_{i} \otimes 1 - 1 \otimes x_{i}$ for $1 \leq i \leq n$ .

Proof

Given: $k [X] \otimes_{k} k [X]$ as the coordinate ring of $X \times X$ .

1.1

The multiplication map $m : k [X] \otimes k [X] \to k [X]$ , $a \otimes b \mapsto a b$ , is the pullback of the diagonal map. It kills every $x_{i} \otimes 1 - 1 \otimes x_{i}$ .

givenalgebra

2.1

Quotienting by these differences identifies the two copies of every coordinate class, hence identifies the quotient with $k [X]$ . Therefore their ideal is $\ker m$ .

step 1.1algebra

3.1

The closed set of this kernel consists exactly of pairs whose coordinate values agree, namely the diagonal.

step 2.1∎

Depends on

The product of affine varieties has coordinate ring k[X] tensor_k k[Y]

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Geometry, §§5h--5i (standard reference, not scraped)