COMMUTATIVE ALGEBRA

YOSHIFUMI TSUCHIMOTO

05. Length, Hilbert function, Samuel function

Lemma 5.1. Let

0 → L → M → N → 0

be an exact sequence of A-modules. Then we have

l(L) + l(N) = l(M).

Definition 5.2. Let A = ⊕_{i=0}^{∞}A_{i} be a graded algebra. We assume (1) l_{A_{0}}(A_{0}) < ∞ (Length of A_{0} as an A_{0} module is finite.)

(2) A is generated by homogeneous elements x_{1}, x_{2}, …, x_{r} where

deg(x_{i}) = d_{i}. Then for any graded finite A-module M, We define its Hilbert series as

φ_{M}(t) = ∑_{j=0}^{∞} l_{A_{0}}(M_{j})t^{j}

Proposition 5.3. Under the assumption of the definition above, The Hilbert series φ_{M} is a rational function on t. More precisely, we have

∏_{j=1}^{r}(1 − t^{d_{j}})φ_{M}(t) ∈ ℚ[t]

Proposition 5.4. If a graded algebra is generated by x_{1}, x_{2}, …, x_{r} of degree 1 over a ring A_{0} with l_{A_{0}}(A_{0}) < ∞, there exists a polynomial p_{M} such that

l(M_{k}) = p_{M}(k) (∀k >> 0).

We call p_{M} the Hilbert polynomial of M.

Corollary 5.5. Let (A, 𝔪) be a Noetherian local ring. Let I be an ideal of definition (That means, there exists n_{0} such that 𝔪 ⊃ I ⊃ 𝔪^{n_{0}} holds.) We put χ_{M}^{I}(j) = l(M/I^{j}). Then there exists a polynomial p such that p(j) = χ_{M}^{I}(j) holds for j >> 0.

Definition 5.6. Under the hypothesis of the Corollary above, we define the Samuel function of M as χ_{M}^{𝔪}(•).

Theorem 5.7 (Nakayama’s lemma, or NAK). Let A be a commutative ring. Let M be an A-module. We assume that M is finitely generated (as a module) over A. That means, there exists a finite set of elements {m_{i}}_{i=1}^{t} such that

M = ∑_{i=1}^{t} Am_{i}

holds. If an ideal I of A satisfies

IM = M (that is, M/IM = 0),

then there exists an element c ∈ I such that

cm = m (∀m ∈ M)

YOSHIFUMI TSUCHIMOTO

holds. If furthermore I is contained in ∩_{𝔪∈Spm(A)}𝔪 (the Jacobson radical of A), then we have M = 0.

Proof. Since IM = M, there exists elements b_{il} ∈ I such that

a_{i} = ∑_{l=1}^{t} b_{il}a_{l} (1 ≤ i ≤ t)

holds. In a matrix notation, this may be rewritten as

v = Bv

with v =^{t} (m_{1}, …, m_{n}), B = (b_{ij}) ∈ M_{t}(I). Using the unit matrix 1_{t} ∈ M_{t}(A) one may also write :

(1_{t} − B)v = 0.

Now let R be the adjugate matrix of 1_{t} − B. In other words, it is a matrix which satisfies

R(1_{t} − B) = (1_{t} − B)R = (det(1_{t} − B))1_{t}.

Then we have

det(1_{t} − B) · v = R(1_{t} − B)v = 0.

On the other hand, since 1_{t} − B = 1_{t} modulo I, we have det(1_{t} − B) = 1 − c for some c ∈ I. This c clearly satisfies

v = cv.
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