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.

Excluding points of codimension zero and one

PDF p. 15: Codimension of the singular locus

Statement

Let $C$ be an integral scheme and $Z \subseteq C$ a subset containing no point of codimension zero or one. Then $Z$ has codimension at least two in $C$ .

The codimension of a point $x$ in $C$ is the supremum of lengths of strict chains of irreducible closed subsets of $C$ from $C$ to $\overline{{x}}$ ; the codimension of $Z$ is the infimum of the codimensions of its points, with infimum of the empty set defined as $+ \infty$ .

Facts & Assumptions

Given: An integral scheme $C$ and a subset $Z$ with no point of codimension zero or one.

Goal: Show that ${codim}_{C} Z \geq 2$ .

Proof

1.1given

For every $x \in Z$ , its codimension is a nonnegative integer or $+ \infty$ , being the supremum of lengths of finite chains. It is neither zero nor one by hypothesis, so it is at least two.

2.1step 1.1∎

Taking the infimum over $x \in Z$ proves the inequality; if $Z$ is empty, the stated convention gives $+ \infty$ .

Used by · 1

Smooth density and normal codimension over a perfect field