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NAK

THEOREM 10.5 (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

$\displaystyle I M=M \quad ($that is, $\displaystyle M/IM=0),
$

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

$\displaystyle c m= m \qquad (\forall m \in M)
$

holds. If furthermore $ I$ is contained in the nilradical of $ A$ , then we have $ M=0$ .

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

$\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.
$

$ \qedsymbol$

Let us interpret the claim of the above theorem in terms of a sheaf $ \mathcal{F}=\mathcal{O}_X \otimes_A M$ on $ X=\operatorname{Spec}(A)$ . $ M$ is assumed to be finitely generated over $ A$ . Note that this in particular means that every fiber of $ \mathcal{F}$ on a $ K$ -valued point (for each field $ K$ ) is finite dimensional $ K$ -vector space. In other words, it is ``a pretty little(=finite dimensional) vector spaces in a row.''

The next assumption simply means that $ \mathcal{F}$ restricted to $ V(I)$ is equal to zero. So $ \mathcal{F}$ sits somewhere other than $ V(I)$ .

The claim of the theorem (NAK) is that one may choose a regular function $ c$ which ``distinguishes $ V(I)$ and ``the support of $ \mathcal{F}$ ''. $ c$ is equal to 0 on $ V(I)$ and is equal to $ 1$ where $ \mathcal{F}$ sits.


next up previous
Next: Étale morphism Up: Unramified morphism Previous: Unramified morphism
2007-12-11