ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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.

An internal shift reverses the published commutative twist parameter

Example

Let $k$ be a field and let $A = k [x]$ be graded by $\deg x = 1$ , so that $A_{j} = k x^{j}$ for $j \geq 0$ and $A_{j} = 0$ for $j < 0$ . The internal shift of Associative graded algebras, bimodules, and internal shifts and the twist of the published commutative convention (Nonnegatively graded rings and modules, homogeneous elements, and twists) move degrees in opposite directions:

module	$d$ -th homogeneous piece	degree of $x^{j}$	degree of $1$
$A {2}$	$A {2}_{d} = A_{d - 2}$	$j + 2$	$2$
$A (2)$	$A (2)_{d} = A_{d + 2}$	$j - 2$	$- 2$

Thus $A {2} = A (- 2)$ as graded modules, and the internal shift is nothing but the published twist with the sign of the parameter reversed. Shifting is a relabelling of degrees: the multiplication $A \times A \to A$ , the action of $A$ on the module and every degree-zero map are the same underlying maps in $A {2}$ and in $A (- 2)$ , and no sign enters.

Facts & Assumptions

Given: A field $k$ , the graded $k$ -algebra $A = k [x]$ with $\deg x = 1$ , and the internal shift $A {2}$ and published twist $A (2)$ .

[L1]

The internal shift has pieces $M {r}_{d} = M_{d - r}$ , is again graded with the same scalar action, and satisfies $M {r} = M (- r)$ in the published convention (Associative graded algebras, bimodules, and internal shifts).

[L2]

The published twist of a graded module over a nonnegatively graded commutative ring has pieces $M (a)_{d} = M_{d + a}$ , so $A_{j} = k x^{j}$ for the standard grading of $k [x]$ (Nonnegatively graded rings and modules, homogeneous elements, and twists).

Verification

1.1

In $A {2}$ the homogeneous piece of degree $d$ is $A {2}_{d} = A_{d - 2}$ , so the monomial $x^{j} \in A_{j}$ is a homogeneous element of degree $j + 2$ in $A {2}$ , and the element $1 \in A_{0}$ has degree $2$ .

1.2

In the published twist, $A (2)_{d} = A_{d + 2}$ , so $x^{j} \in A_{j}$ has degree $j - 2$ in $A (2)$ , while $A (- 2)_{d} = A_{d - 2}$ gives $x^{j}$ degree $j + 2$ in $A (- 2)$ ; hence $A {2}_{d} = A_{d - 2} = A (- 2)_{d}$ for every $d$ .

L1L2

2.1

The equality of piecewise $k$ -modules in step 1.2 holds for every degree, and both modules carry the same scalar action inherited from $A$ ; hence $A {2} = A (- 2)$ as graded $A$ -modules. The twist $A (2)$ is a genuinely different grading, not the same one written with the opposite sign: $A (2)_{0} = A_{2} = k x^{2} \neq 0$ , while $A (- 2)_{0} = A_{- 2} = 0$ , so $A (2)$ and $A (- 2)$ have different degree-zero pieces.

step 1.1step 1.2L1

3.1

Shifting changes no underlying map. The multiplication of $A$ , the action of $A$ on these graded modules, and every degree-zero $A$ -linear map are the same functions before and after the shift; only the degree attached to a homogeneous element changes, by the fixed amount $r$ . In particular the shift of a complex would move each homogeneous piece to the $r$ -fold shifted degree without introducing a sign in any differential. Since $A$ carries no differential, no sign is introduced here at all.

step 2.1L1

4.1

The computation exhibits both claims: $A {2} = A (- 2)$ as graded modules, so the internal shift reverses the published twist parameter, and the shift does not alter multiplication or differential signs. ∎

step 2.1step 3.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

Stacks Project, Algebra, §10.56, tag 00JL (standard reference, not scraped)
Alexander Kleshchev, Representation Theory of Symmetric Groups and Related Hecke Algebras (2009), §2.2, printed pp. 6-7 (standard reference, not scraped)