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A DVR has Hilbert-Samuel polynomial $n + 1$ and multiplicity one

Example

Let $(V, (π))$ be a discrete valuation ring. Then

χ_{(π), V} (n) = ℓ_{V} (V / (π^{n + 1})) = n + 1

for every $n \geq 0$ . Hence the Hilbert-Samuel polynomial is exactly $n + 1$ , and the Hilbert-Samuel multiplicity is

e_{(π)} (V) = 1.

Facts & Assumptions

Given: A discrete valuation ring $V$ with maximal ideal $(π)$ .

[L1]

The quotient $V / (π^{n + 1})$ has length $n + 1$ for every $n \geq 0$ (Length and valuation in a DVR).

[L2]

The Hilbert-Samuel polynomial exists and multiplicity is its factorial scaled leading coefficient (The Hilbert-Samuel function agrees eventually with a rational polynomial in binomial form, Hilbert-Samuel multiplicity as the factorial-scaled leading coefficient).

Verification

technique · direct

1.1

By [L1], the Hilbert-Samuel function is $χ_{(π), V} (n) = ℓ_{V} (V / (π^{n + 1})) = n + 1$ for every $n \geq 0$ .

L1given

1.2

Therefore the eventual polynomial is already exactly $P (n) = n + 1$ . Its degree is $1$ and its leading coefficient is $1$ , so [L2] gives $e_{(π)} (V) = 1! \cdot 1 = 1.$

2.1

This computes both the Hilbert-Samuel polynomial and the multiplicity of a DVR.

algebra∎

Depends on

Used by

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Sources

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, §§21 and 23 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, discrete valuation rings (standard reference, not scraped)

A DVR has Hilbert-Samuel polynomial n+1n+1 and multiplicity one