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

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associated graded polynomial surjection

Statement

Let $(R, m, k)$ be nonzero Noetherian local and let $x_{1}, \dots, x_{e}$ lift a basis of $m / m^{2}$ . There is a surjective graded $k$ -algebra map $ϕ : k [X_{1}, \dots, X_{e}] \to {gr}_{m} R$ , determined by $X_{i} \mapsto x_{i} + m^{2}$ , with every variable of degree one.

Facts & Assumptions

Given: The objects and hypotheses in the statement. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

embedding dimension is minimal maximal ideal generator number: For a nonzero Noetherian local ring $(R, m, k)$ , $edim R$ is the least number of generators of $m$ .

[F2]

The associated graded ring and associated graded module of an ideal-adic filtration: Let $R$ be a commutative ring, let $I \subset R$ be an ideal, and let $M$ be an $R$ -module. The associated graded ring of the $I$ -adic filtration is ${gr}_{I} (R) : = ⨁_{n \geq 0} I^{n} / I^{n + 1} .$ Multiplication is induced by multiplication in $R$ : $(a + I^{m + 1}) (b + I^{n + 1}) = a b + I^{m + n + 1} .$ The associated graded module is ${gr}_{I} (M) : = ⨁_{n \geq 0} I^{n} M / I^{n + 1} M,$ viewed as a graded ${gr}_{I} (R)$ -module by $(a + I^{m + 1}) (x + I^{n + 1} M) = a x + I^{m + n + 1} M .$

[F3]

Assuming the Axiom of Choice, Nakayama's lemma: If $I$ is contained in the Jacobson radical of a commutative ring and $M$ is finite with $I M = M$ , then $M = 0$ .

Proof

1.1

The degree-zero part is $R / m = k$ . On $m^{n} / m^{n + 1}$ the action of $R$ factors through $k$ , since $m m^{n} \subseteq m^{n + 1}$ . The graded multiplication therefore defines the displayed polynomial map. Altering a representative by $m^{n + 1}$ changes a product of degrees $n, j$ by $m^{n + j + 1}$ , so multiplication and the map are well-defined.

F2givenalgebra

2.1

Put $N = (x_{1}, \dots, x_{e}) \subseteq m$ . The basis hypothesis says $m = N + m^{2}$ , so the finite module $m / N$ satisfies $m (m / N) = m / N$ . Nakayama gives $m = N$ . Expanding products now shows that degree- $n$ monomials in the $x_{i}$ generate $m^{n}$ over $R$ ; reducing coefficients modulo $m$ spans the degree- $n$ quotient over $k$ . Thus every graded component is in the image. When $e = 0$ , the same Nakayama argument gives $m = 0$ and the map is the identity on $k$ .

F1F3step 1.1algebra∎

Depends on

Used by

associated graded polynomial map singular kernel Example
cusp local ring not regular Example
regular local graded surjection has zero kernel Lemma
associated graded ring of a regular local ring Theorem

Dependency tree · two levels

10 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

10.106.1, first proof paragraph (standard reference, not scraped)