Lemma 9.3.32EC38CE665F6

Proof. We represent elements of $L$ by multiplication matrices over $K$ and evaluate these at points of $\mathbb C^ d$; integrality bounds their eigenvalues, and lemma 9.2.9 turns bounded elements of $K$ into elements of $\mathcal O_ d$.

We first recall the determinant criterion for integrality. Let $R$ be a subring of an integral domain $D$, let $b\in D$, and suppose that multiplication by $b$ maps an $R$-submodule of $D$ generated by elements $v_1,\ldots ,v_ N$, not all zero, into itself: $bv_ i=\sum _ ja_{ij}v_ j$ with $a_{ij}\in R$. Multiplying the matrix $bI-(a_{ij})$ by its adjugate shows that the monic polynomial $\det (bI-(a_{ij}))$ in $b$ annihilates every $v_ i$; since $D$ is a domain, this determinant is zero, and $b$ is integral over $R$. In particular, every element of $B$ is integral over $\mathcal O_ d$; take for the $v_ i$ generators of the $\mathcal O_ d$-module $B$. Conversely, if $x_1,\ldots ,x_ k\in D$ are integral over $R$, then $R[x_1,\ldots ,x_ k]$ is a finitely generated $R$-module, since powers of each $x_ i$ can be reduced with its monic equation; by the criterion, sums and products of integral elements are integral.

The ring $KB=\{ b/a:b\in B,\ a\in \mathcal O_ d\setminus \{ 0\} \} $ is a domain, and it is spanned over $K$ by finitely many module generators of $B$, so it is a finite-dimensional vector space over $K$. Multiplication by a nonzero element is an injective linear map of $KB$, hence surjective; so $KB$ is a field, and it equals $L$. Choose a basis $e_1=1,e_2,\ldots ,e_ q$ of $L$ over $K$ consisting of elements of $B$; it can be extracted from $1$ and the module generators of $B$. For $x\in L$ let $[x]\in K^{q\times q}$ be the matrix of the $K$-linear map $y\mapsto xy$ in this basis. The map $x\mapsto [x]$ is an injective homomorphism of $K$-algebras, because the first column of $[x]$ is the coordinate vector of $xe_1=x$. Let $\operatorname {Tr}(x)=\operatorname {tr}[x]$, which does not depend on the basis, and let $\chi _ x(T)=\det (TI-[x])\in K[T]$, a monic polynomial of degree $q$; by the Cayley–Hamilton theorem $[\chi _ x(x)]=\chi _ x([x])=0$, so $\chi _ x(x)=0$. The minimal polynomial $\mu _ x$ of $x$ exists, because $1,x,\ldots ,x^ q$ are linearly dependent over $K$, and $\deg \mu _ x\leq q$. By Euclidean division in $K[T]$, $\mu _ x$ divides every polynomial in $K[T]$ that vanishes at $x$.

Values of elements of $K$. Let $U$ be a connected neighborhood of $0$ in $\mathbb C^ d$. An element of $K$ is defined at $z'\in U$ if it can be written as $a/h$ with $a,h$ holomorphic on $U$ and $h(z')\neq 0$; its value at $z'$ is then $a(z')/h(z')$. This value does not depend on the representation, since two representations $a/h=a'/h'$ satisfy $ah'=a'h$ near $0$, hence on $U$ by the identity theorem. If finitely many elements of $K$ are defined at $z'$, so are their sums and products, and their quotients by those with nonzero value at $z'$, with the corresponding values. Finitely many elements of $K$ can be written with a common denominator $h\neq 0$ and numerators holomorphic on some connected neighborhood $U$ of $0$; they are then all defined at the points of $U\setminus \{ h=0\} $, a dense open subset of $U$. We shall use the following boundedness criterion: if the values of an element $a/h$ of $K$ at the points of $V\setminus \{ h=0\} $ are bounded, for some neighborhood $V\subset U$ of $0$, then $a/h\in \mathcal O_ d$. Indeed, $|a|\leq C|h|$ on $V\setminus \{ h=0\} $, hence on $V$, since both sides are continuous functions on $V$ and $V\setminus \{ h=0\} $ is dense in $V$; so $a/h\in \mathcal O_ d$ by lemma 9.2.9. (For $d=0$, the space $\mathbb C^0$ is a point and $K=\mathcal O_0=\mathbb C$.)

Proof of (1). Let $x\in L$ be integral over $\mathcal O_ d$, say $f(x)=0$ with $f(T)=T^ m+f_1T^{m-1}+\cdots +f_ m$ and $f_ j\in \mathcal O_ d$. Then $f([x])=[f(x)]=0$. Since $\chi _ x(x)=0$, we can write $\chi _ x=\mu _ x\rho $ with $\rho \in K[T]$. Write the entries of $[x]$ and the coefficients of $\mu _ x$ and $\rho $ with a common denominator $h$, so that they are all defined at the points of $U\setminus \{ h=0\} $ for a connected neighborhood $U$ of $0$; the coefficients of $\chi _ x$, polynomials in the entries of $[x]$, can then be written with the denominator $h^ q$.

At the points $z'\in U\setminus \{ h=0\} $ near $0$ the complex matrix $[x](z')$ satisfies $f(z',[x](z'))=0$. Applying this to an eigenvector, we see that every eigenvalue $\lambda $ of $[x](z')$ satisfies $f(z',\lambda )=0$, so that $|\lambda |\leq C_ f$ by (9.2.1), where $C_ f$ is an upper bound of $2\max _ j|f_ j|^{1/j}$ near $0$. The coefficients of $\chi _ x(z',T)$, the characteristic polynomial of $[x](z')$, are up to sign the elementary symmetric functions of these eigenvalues, so they are bounded at the points of $V\setminus \{ h=0\} $ for a neighborhood $V\subset U$ of $0$, and $\chi _ x\in \mathcal O_ d[T]$ by the boundedness criterion; in particular $\operatorname {Tr}(x)\in \mathcal O_ d$. The roots of $\mu _ x(z',\cdot )$ are roots of $\chi _ x(z',\cdot )=\mu _ x(z',\cdot )\rho (z',\cdot )$, so they are also bounded by $C_ f$; hence the coefficients of $\mu _ x(z',\cdot )$, which are up to sign the elementary symmetric functions of these roots, are bounded at the points of $V\setminus \{ h=0\} $. Therefore $\mu _ x\in \mathcal O_ d[T]$, again by the boundedness criterion, which proves (1).

Proof of (2). We evaluate the multiplication matrices at a point $z'_0$, where they span, in a suitable basis, the algebra of all diagonal matrices; then a generic combination of the matrices $[b_ k](z'_0)$ has distinct eigenvalues. We first show that the trace form is nondegenerate. Let $\Theta =\det (\operatorname {Tr}(e_ ie_ j))_{i,j}\in K$. If $\Theta =0$, there are $x_1,\ldots ,x_ q\in K$, not all zero, with $\sum _ ix_ i\operatorname {Tr}(e_ ie_ j)=0$ for all $j$; then $x=\sum _ ix_ ie_ i\neq 0$ satisfies $\operatorname {Tr}(xy)=0$ for all $y\in L$, by linearity, and $y=x^{-1}$ gives $\operatorname {Tr}(1)=q=0$, which is absurd. Write $e_ ie_ j=\sum _ l\gamma _{ijl}e_ l$ and $b_ k=\sum _ l\beta _{kl}e_ l$ with $\gamma _{ijl},\beta _{kl}\in K$, and, since $B$ is generated by the $b_ k$ as an $\mathcal O_ d$-algebra, $e_ l=G_ l(b_1,\ldots ,b_ r)$ for polynomials $G_ l$ with coefficients in $\mathcal O_ d$. Choose a connected neighborhood $U$ of $0$ and a point $z'_0\in U$ at which the finitely many elements $\Theta $, $\gamma _{ijl}$, $\beta _{kl}$, the entries of the matrices $[e_ l]$ and $[b_ k]$, and the coefficients of the $G_ l$ are defined, such that $\Theta (z'_0)\neq 0$.

Let $\mathcal N$ be the complex linear span of the matrices $N_ l=[e_ l](z'_0)$, $1\leq l\leq q$. Since $[e_ i][e_ j]=[e_ ie_ j]=\sum _ l\gamma _{ijl}[e_ l]$, the set $\mathcal N$ is a commutative algebra of complex $q\times q$ matrices, and it contains $N_1=I$. The matrix $(\operatorname {tr}(N_ iN_ j))_{i,j}$ is the value of $(\operatorname {Tr}(e_ ie_ j))_{i,j}$ at $z'_0$, which is invertible; hence the $N_ l$ are linearly independent, $\dim \mathcal N=q$, and the bilinear form $(N,N')\mapsto \operatorname {tr}(NN')$ on $\mathcal N$ is nondegenerate. If $N\in \mathcal N$ is nilpotent, then $NN'$ is nilpotent for every $N'\in \mathcal N$, because $N$ and $N'$ commute; so $\operatorname {tr}(NN')=0$ for all $N'$, and $N=0$. Consequently every $N\in \mathcal N$ is diagonalizable. Indeed, if $\lambda _1,\ldots ,\lambda _ p$ are the distinct eigenvalues of $N$, then the minimal polynomial of $N$ divides a power of $(T-\lambda _1)\cdots (T-\lambda _ p)$, so the element $(N-\lambda _1I)\cdots (N-\lambda _ pI)$ of $\mathcal N$ is nilpotent, hence zero, and the minimal polynomial of $N$ has simple roots. Commuting diagonalizable matrices are simultaneously diagonalizable; so in a suitable basis of $\mathbb C^ q$ all elements of $\mathcal N$ are diagonal, and since $\dim \mathcal N=q$, the algebra $\mathcal N$ consists of all diagonal matrices.

The matrices $[b_ k](z'_0)=\sum _ l\beta _{kl}(z'_0)N_ l$ belong to $\mathcal N$; let $\lambda ^{(i)}_ k$ be the $i$th diagonal entry of $[b_ k](z'_0)$ in this basis, and $\lambda ^{(i)}=(\lambda ^{(i)}_1,\ldots ,\lambda ^{(i)}_ r)$. For $i\neq i'$ we have $\lambda ^{(i)}\neq \lambda ^{(i')}$: otherwise every matrix $N_ l=G_ l(z'_0)([b_1](z'_0),\ldots ,[b_ r](z'_0))$ would have equal $i$th and $i'$th diagonal entries, and so would every element of $\mathcal N$. Therefore, for $c$ outside the finitely many proper linear subspaces $\{ c:\sum _ kc_ k(\lambda ^{(i)}_ k-\lambda ^{(i')}_ k)=0\} $, $i<i'$, the matrix $[u](z'_0)=\sum _ kc_ k[b_ k](z'_0)$ is diagonal with $q$ distinct diagonal entries $\tau _1,\ldots ,\tau _ q$.

Fix such a $c$, and let $\Lambda \in K^{q\times q}$ be the matrix whose columns are the coordinate vectors of $1,u,\ldots ,u^{q-1}$; its entries are polynomials in the entries of the $[b_ k]$, so they are defined at $z'_0$. The linear map $\mathcal N\to \mathbb C^ q$ that takes a matrix to its first column, its product with the coordinate vector $(1,0,\ldots ,0)^{\mathsf T}$ of $e_1=1$, maps $N_ l$ to the $l$th unit vector, the coordinate vector of $e_ le_1=e_ l$; so it is bijective. It maps $[u](z'_0)^ j=[u^ j](z'_0)$ to the $j$th column of $\Lambda (z'_0)$, counting from $j=0$. In the diagonal basis, $[u](z'_0)^ j=\operatorname {diag}(\tau _1^ j,\ldots ,\tau _ q^ j)$, and these matrices are linearly independent for $0\leq j<q$, by the Vandermonde determinant, since the $\tau _ i$ are distinct. Hence $\det \Lambda (z'_0)\neq 0$, so $\det \Lambda \neq 0$, and $1,u,\ldots ,u^{q-1}$ is a basis of $L$. Consequently $\mu _ u$ has degree $q$, and since $u\in B$ is integral over $\mathcal O_ d$, its coefficients lie in $\mathcal O_ d$ by (1). This proves (2).

Proof of (3). In the basis $1,u,\ldots ,u^{q-1}$, multiplication by $u$ has the companion matrix $C_ W$, because $u\cdot u^{q-1}=-\sum _ jw_ ju^{q-j}$. Hence $[u]=\Lambda C_ W\Lambda ^{-1}$, and $\operatorname {Tr}(u^ k)=\operatorname {tr}(C_ W^ k)=S_ k$ for $k\geq 0$, with $S_ k$ as in the definition of $\operatorname {Disc}(W)$. The $w_ j$, being the coordinates of $-u^ q$ in the basis $1,\ldots ,u^{q-1}$, are defined at $z'_0$, because $\det \Lambda (z'_0)\neq 0$; and $C_ W(z'_0)=\Lambda (z'_0)^{-1}[u](z'_0)\Lambda (z'_0)$ has the $q$ distinct eigenvalues $\tau _1,\ldots ,\tau _ q$. Since the characteristic polynomial of $C_ W(z'_0)$ is $W(z'_0,\cdot )$, these are its roots, and (9.3.2) shows that the value of $\delta $ at $z'_0$ is not zero; so $\delta \neq 0$. Now let $g\in L$ be integral over $\mathcal O_ d$, and write $g=\sum _{j<q}a_ ju^ j$ with $a_ j\in K$. Multiplying by $u^ l$ and taking traces, we obtain

The elements $gu^ l$ are integral over $\mathcal O_ d$, as products of integral elements, so $\operatorname {Tr}(gu^ l)\in \mathcal O_ d$ by the proof of (1). The matrix $(S_{j+l})$ has entries in $\mathcal O_ d$ and determinant $\delta $; multiplying the system by its adjugate gives $\delta a_ j\in \mathcal O_ d$, which is (3) with $d_ j=\delta a_ j$. $\square$