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Commutative algebra

Yoshifumi Tsuchimoto

\fbox{05. Length, Hilbert function, Samuel function}

LEMMA 05.1   Let

$\displaystyle 0 \to L \to M \to N\to 0
$

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

$\displaystyle l(L)+l(N)=l(M).
$

DEFINITION 05.2   Let $ A=\oplus_{i=0}^\infty A_i$ be a graded algebra. We assume
  1. $ l_{A_0}(A_0)<\infty$ (Length of $ A_0$ as an $ A_0$ module is finite.)
  2. $ A$ is generated by homogeneous elements $ x_1,x_2,\dots, x_r$ where $ \deg(x_i)=d_i$ .
Then for any graded finite $ A$ -module $ M$ , We define its Hilbert series as

$\displaystyle \varphi_M(t)=\sum_{j=0}^\infty l_{A_0}(M_j) t^j
$

PROPOSITION 05.3   Under the assumption of the definition above, The Hilbert series $ \varphi_M$ is a rational function on $ t$ . More precisely, we have

$\displaystyle \prod_{j=1}^r (1-t^{d_j}) \varphi_M(t) \in \mathbb{Q}[t]
$

PROPOSITION 05.4   If a graded algebra is generated by $ x_1,x_2,\dots, x_r$ of degree $ 1$ over a ring $ A_0$ with $ l_{A_0}(A_0)<\infty$ , there exists a polynomial $ p_M$ such that

% latex2html id marker 749
$\displaystyle l(M_k)=p_M(k) \quad (\forall k »0).
$

We call $ p_M$ the Hilbert polynomial of $ M$ .

COROLLARY 05.5   Let $ (A,\mathfrak{m})$ be a Noetherian local ring. Let $ I$ be an ideal of definition (That means, there exists $ n_0$ such that $ \mathfrak{m} \supset I\supset
\mathfrak{m}^{n_0}$ holds.) We put $ \chi^I_M(j)=l(M/I^j)$ . Then there exists a polynomial $ p$ such that $ p(j)=\chi^I_M(j)$ holds for $ j»0$ .

DEFINITION 05.6   Under the hypothesis of the Corollary above, we define the Samuel function of $ M$ as $ \chi_M^{\mathfrak{m}}(\bullet)$ .

THEOREM 05.7 (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

$\displaystyle M=\sum_{i=1}^t A m_i
$

holds. If an ideal $ I$ of $ A$ satisfies

% latex2html id marker 808
$\displaystyle I M=M \quad ($that is, $\displaystyle M/I M=0),
$

then there exists an element $ c\in I$ such that

% latex2html id marker 813
$\displaystyle c m= m \qquad (\forall m \in M)
$

holds. If furthermore $ I$ is contained in $ \cap_{\mathfrak{m}\in \operatorname{Spm}(A)} \mathfrak{m}$ (the Jacobson radical of $ A$ ), then we have $ M=0$ .

PROOF.. Since $ I M=M$ , there exists elements $ b_{i l}\in I$ such that

% latex2html id marker 830
$\displaystyle a_i= \sum_{l=1}^t b_{i l} a_l \qquad (1\leq i \leq t )
$

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

$\displaystyle v=B v
$

with $ v=^t (m_1,\dots,m_n)$ , $ B=(b_{i j})\in M_t (I)$ . Using the unit matrix $ 1_t\in M_t(A)$ one may also write :

$\displaystyle (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

$\displaystyle R(1_t-B)=(1_t-B)R=(\det(1_t-B)) 1_t.
$

Then we have

$\displaystyle \det(1_t -B)\cdot 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\in I$ . This $ c$ clearly satisfies

$\displaystyle v=c v.
$

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$ \qedsymbol$


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2012-05-25