Lemma 6

Let M be a free $\mathbf{Z}$-module of finite type, $u$ an endomorphism of M, and $v$ the endomorphism $u\otimes 1$ of $M\otimes_{\mathbf{Z}}\mathbf{Q}$. Assume that $(^v_n)(M)\subset M$ for all $n\in \mathbf{N}$. Then $u$ is diagonalizable.

a) For any polynomial $P\in \mathbf{Q}[T]$ such that $P(\mathbf{Z})\subset \mathbf{Z}$, we have $P(v)(M)\subset M$ (no. 4, Cor. of Prop. 2), so det $P(v)\in \mathbf{Z}$.

b) Denote by $\chi_v(t) =t^d+\alpha_1t^{d-1}+\cdots$ the characteristic polynomial of $v$. Let $k\in \mathbf{Z}, n\in \mathbf{N}$. Applying a) to the polynomial $(^{T-k}_n)$, we see that the number

$a_n=$ det $(v-k)=1_d$ det($v-k$) det($v-k-1$)$. .$. det($v-k-n+ 1$)

$n(n$!)

$(n$!)

is an integer. Take $k-1<-\alpha_1/d$. Then

and

hence, if $a_n\not= 0$ for all $n\in \mathbf{N}$, the sequence of the $|a_n|$ is strictly decreasing for $n$ sufficiently large, which is absurd. It follows that $v$ has an integer eigenvalue $\lambda$. Put $M'=$ Ker($u-\lambda .1$) and $M''= M/M'$. Then $M'$ is the intersection with M of a vector subspace of $M\otimes_{\mathbf{Z}}\mathbf{Q}$, so the $\mathbf{Z}$-module $M''$ is torsion-free of finite type, and consequently free of rank $< d$. Arguing by induction on $d$ and applying the induction hypothesis to the endomorphism of $M''$ induced by $u$, we conclude that all the eigenvalues of $v$ in an algebraically closed extension of $\mathbf{Q}$ are integers.

c) We show that $v$ is diagonalizable. Let $\lambda$ be an eigenvalue of $v$ and let $x\in M\otimes_{\mathbf{Z}}\mathbf{Q}$ be such that $(v-\lambda )^2x= 0$. We have $v(vx-\lambda x) =\lambda (vx-\lambda x)$, so

$n1$!$(v-\lambda -n+ 1)(v-\lambda -n+ 2). . .(v-\lambda -1)(v-\lambda )x$

By a), this implies that $vx-\lambda x\in nM$ for all $n\in \mathbf{N}$, so $(v-\lambda )x= 0$.

d) Let $\lambda$ be an eigenvalue of $v$ and let $\lambda -a, \lambda +b$ be an interval in $\mathbf{Z}$ containing all the eigenvalues of $v$. Consider the polynomial

$b$!

$a$!

We have $P(\mathbf{Z})\subset \mathbf{Z},P(\lambda ) = 1,P(\mu) = 0$ for $\mu\in \mathbf{Z}\cap (\lambda -a, \lambda +b)$ and $\mu\not=\lambda$. By $a), P(v)(M)\subset M$. By $c), P(v)$ is a projection of $M\otimes_{\mathbf{Z}}\mathbf{Q}$ onto the eigenspace corresponding to $\lambda$. Q.E.D.