ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-09-24 (gpt-6-sol)

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.

The polynomial ring and a homogeneous quotient have the expected Hilbert series and Hilbert polynomial

Example

Let $k$ be a field and give $k [x, y]$ the standard grading. Then ${HS}_{k [x, y]} (t) = \frac{1}{(1 - t)^{2}},$ because the degree- $n$ piece has basis $x^{n}, x^{n - 1} y, \dots, x y^{n - 1}, y^{n}$ and therefore dimension $n + 1$ .

For the homogeneous quotient $A : = k [x, y] / (y^{2}),$ the degree- $0$ piece has dimension $1$ and each degree- $n \geq 1$ piece has basis $x^{n}, x^{n - 1} y$ . Hence ${HS}_{A} (t) = 1 + 2 t + 2 t^{2} + \dots = \frac{1 + t}{1 - t},$ so the Hilbert polynomial of $A$ is the constant polynomial $2$ .

Facts & Assumptions

Given: A field $k$ , the standard grading on $k [x, y]$ , and the quotient $A = k [x, y] / (y^{2})$ .

[L1]

The geometric-series identity $\sum_{n \geq 0} t^{n} = (1 - t)^{- 1}$ and its formal derivative give $\sum_{n \geq 0} (n + 1) t^{n} = (1 - t)^{- 2}$ .

Verification

technique · direct

1.1L1givenalgebra

In $k [x, y]$ , the degree- $n$ monomials are exactly $x^{n - i} y^{i}$ for $0 \leq i \leq n$ , so the degree- $n$ piece has dimension $n + 1$ . Therefore ${HS}_{k [x, y]} (t) = \sum_{n \geq 0} (n + 1) t^{n} = \frac{1}{(1 - t)^{2}} .$

1.2givenalgebra

In the quotient by $(y^{2})$ , every monomial containing $y^{2}$ vanishes. So for $n \geq 1$ the degree- $n$ piece is spanned by $x^{n}$ and $x^{n - 1} y$ , and these two classes are linearly independent. Hence ${HS}_{A} (t) = 1 + \sum_{n \geq 1} 2 t^{n} = \frac{1 + t}{1 - t} .$

2.1step 1.2

The eventual coefficient sequence of ${HS}_{A} (t)$ is constant equal to $2$ , so its Hilbert polynomial is $2$ .

3.1step 1.1step 1.2step 2.1∎

Thus both the polynomial ring and this homogeneous quotient realize the expected Hilbert series and Hilbert polynomial.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Stacks Project, Example 10.58.9 (standard reference, not scraped)
Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, §20 (standard reference, not scraped)