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

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

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

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

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

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

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

$$a_i= \sum_{l=1}^t b_{i l} a_l \qquad (1\leq i \leq t )$$

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

$$v=B v$$

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 :

$$(1_t -B) v=0.$$

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

$$\det(1_t -B)\cdot v = R (1_t -B) v=0.$$

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

$$v=c v.$$

$\qedsymbol$

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.