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For a nonzero finite module and an ideal of definition, Hilbert-Samuel multiplicity is a positive integer

Statement

Assume the Axiom of Choice.

Let (R,m) be a Noetherian local ring, let M0 be a finite R-module, and let I be an ideal of definition for M. Then the Hilbert-Samuel multiplicity eI(M) is a positive integer.

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian local ring (R,m), a nonzero finite R-module M, and an ideal of definition I for M.

[L1]

The Hilbert-Samuel function agrees for large n with a polynomial written in binomial form χI,M(n)=j=0daj(n+jj) for integers aj (The Hilbert-Samuel function agrees eventually with a rational polynomial in binomial form).

[L2]

If a finite module over a local ring satisfies M/In+1M=0, then M=0, because the empty generating family lifts across the Jacobson-radical ideal In+1m (Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators).

[L3]

Hilbert-Samuel multiplicity is the factorial-scaled leading coefficient of the eventual Hilbert-Samuel polynomial (Hilbert-Samuel multiplicity as the factorial-scaled leading coefficient).

Proof

technique · direct
1.1

By [L1], there are integers a0,,ad and a polynomial PI,M(n)=j=0daj(n+jj) such that χI,M(n)=PI,M(n) for all sufficiently large n.

L1given
1.2

For every n0, the quotient M/In+1M is nonzero. Indeed, if M/In+1M=0, then M=In+1M; since In+1m=J(R), [L2] would force M=0, contradicting the hypothesis. Thus χI,M(n)=R(M/In+1M)>0 for every n.

L2givenalgebra
2.1

The polynomial PI,M therefore takes positive values for all sufficiently large integers, so its leading coefficient is positive. In the binomial expansion of step 1.1 the leading coefficient is ad/d!, hence ad>0.

step 1.1step 1.2algebra
3.1

By [L3], the Hilbert-Samuel multiplicity is eI(M)=d!(leading coefficient of PI,M)=ad. Since adZ>0, the multiplicity is a positive integer.

L3step 2.1algebra

Depends on

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