LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Residue fields and normalization point degrees

PDF p. 7: Smooth point generator; PDF p. 13: Normalization pushforward

Statement

Let $k$ be an algebraically closed field. Every closed point of a finite-type $k$ -scheme has residue field canonically equal to $k$ via its structure map. If $f : Y \to X$ is a finite surjective morphism between finite-type $k$ -schemes, then the fiber over each closed point $x \in X$ is a nonempty finite set of closed points of $Y$ . Each such point $y$ has residue field $k$ , and its zero-cycle pushforward is $f_{*} [y] = [x]$ .

Here a zero-cycle is a finite integral linear combination of closed points. For a finite morphism, the pushforward of a closed point is defined by $f_{*} [y] = [κ (y) : κ (f (y))] [f (y)]$ ; the field degree is finite. A finite morphism is one whose inverse image of every affine open $Spec A$ is affine, say $Spec B$ , with $B$ a finite $A$ -module.

Statement references

Over an algebraically closed field, every maximal ideal is an evaluation ideal

Facts & Assumptions

[F1]

For algebraically closed $k$ , every maximal ideal of $k [t_{1}, \dots, t_{n}]$ is an evaluation ideal and has quotient $k$ (Over an algebraically closed field, every maximal ideal is an evaluation ideal).

[F2]

For a finite family of pairwise comaximal ideals, the quotient map to the product of the quotient rings is surjective (Chinese remainder theorem for pairwise comaximal ideals).

Proof

Given: An algebraically closed field $k$ and the finite-type schemes and finite surjective morphism in the statement.

Goal: Prove the residue-field assertion and the stated description and pushforward of each closed fiber.

1.1givenF1choose

Let $z$ be a closed point of a finite-type $k$ -scheme and choose an affine open neighborhood $Spec A$ of $z$ . The point is closed in that open, so it corresponds to a maximal ideal $m$ of $A$ . Since $A$ is a finitely generated $k$ -algebra, write $A = k [t_{1}, \dots, t_{n}] / I$ . The inverse image of $m$ is maximal in the polynomial ring. By [F1], its quotient is $k$ , with the identification induced by the structure map. Hence $κ (z) = A / m$ is canonically $k$ .

2.1givenstep 1.1choosealgebra

Let $x \in X$ be closed and choose an affine neighborhood $U = Spec A$ , with $x$ represented by $m$ . By finiteness, $f^{- 1} (U) = Spec B$ for a finite $A$ -algebra $B$ . The fiber over $x$ is $Spec C$ , where $C = B \otimes_{A} κ (x) = B / m B$ is a finite-dimensional $k$ -algebra by step 1.1. Surjectivity makes this fiber nonempty, so $C \neq 0$ . If $p$ is a prime of $C$ , then $C / p$ is a finite-dimensional domain over $k$ ; multiplication by any nonzero element is an injective, hence surjective, linear map, so $C / p$ is a field. Thus every prime of $C$ is maximal.

3.1F2step 2.1step 1.1algebra

Distinct maximal ideals of $C$ are pairwise comaximal. For any finite collection of them, the Chinese remainder map of [F2] from $C$ onto the product of their nonzero residue fields is surjective. Consequently the number in any such collection is at most $\dim_{k} C$ , so $Spec C$ is finite. Every point $y$ in this fiber is closed in the fiber because its prime is maximal, and the fiber is closed in $Y$ because $x$ is closed in $X$ ; hence $y$ is closed in $Y$ . Step 1.1 gives $κ (y) = k$ compatibly with $κ (x) = k$ .

4.1givenstep 3.1algebra∎

The fiber is nonempty by surjectivity and finite by step 3.1. For each of its points, the induced extension $κ (x) \to κ (y)$ is identified with the identity of $k$ , so its degree is one. The definition of zero-cycle pushforward therefore gives $f_{*} [y] = [x]$ .

Depends on

Used by · 2

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