Proposition 10.59.5 (00K8)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.59: Noetherian local rings
Proposition 10.59.5 (cite)

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Proposition 10.59.5. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Let $I \subset R$ be an ideal of definition. The Hilbert function $\varphi _{I, M}$ and the function $\chi _{I, M}$ are numerical polynomials.

Proof. Consider the graded ring $S = R/I \oplus I/I^2 \oplus I^2/I^3 \oplus \ldots = \bigoplus _{d \geq 0} I^ d/I^{d + 1}$. Consider the graded $S$-module $N = M/IM \oplus IM/I^2M \oplus \ldots = \bigoplus _{d \geq 0} I^ dM/I^{d + 1}M$. This pair $(S, N)$ satisfies the hypotheses of Proposition 10.58.7. Hence the result for $\varphi _{I, M}$ follows from that proposition and Lemma 10.55.1. The result for $\chi _{I, M}$ follows from this and Lemma 10.58.5. $\square$

Comments (2)

Comment #3505 by Jonas Ehrhard on August 08, 2018 at 11:46

Lemma 00JD (10.54.1) needs $S$ to be Artinian. I don't see why this should be the case, as the ideals

$J_k = \bigoplus_{d \ge k} I^d / I^{d+1}$

seem to give a possibly infinite descending sequence. $S = J_0 \supset J_1 \supset J_2 \supset \cdots$ . This sequence stabilises iff $I^d = I^{d+1}$ for $d \gg 0$ , which must not be true. For example consider the localisation $k[x]_{(x)}$ of the polynomial ring at the maximal ideal $(x)$ and $I = (x)$ .

Comment #3547 by Johan on August 28, 2018 at 14:13

No, this is fine because Proposition 10.58.7 tells us we end up in $K'_0(R/I)$ and $R/I$ is Artinian because $I$ is an ideal of definition.

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