2. Seminorm and integral extension [026B]

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2. Seminorm and integral extension

Let $\mathscr{A}$ be a finitely generated $\mathfrak{o}_{k}$ -algebra, which contains $\mathfrak{o}_{k}$ as a subring. We set $A:=\mathscr{A}\otimes_{\mathfrak{o}_{k}}k$ . Note that $A$ coincides with the localization of $\mathscr{A}$ with respect to ${S}:=\mathfrak{o}_{k}\setminus\{0\}$ . Let $\operatorname{Spec}(A)^{\mathrm{an}}$ be the analytification of $\operatorname{Spec}(A)$ , that is, the set of all seminorms of $A$ over the absolute value of $k$ . For $x\in\operatorname{Spec}(A)^{\mathrm{an}}$ , let $\mathfrak{o}_{x}$ and $\mathfrak{m}_{x}$ be the valuation ring of $(\hat{\kappa}(x),|\raisebox{1.72218pt}{.}|_{x})$ and the maximal ideal of $\mathfrak{o}_{x}$ , respectively (see §1.1.3 for the definition of $\hat{\kappa}(x)$ ). We denote the natural homomorphism $A\to\hat{\kappa}(x)$ by $\varphi_{x}$ . It is easy to see that the following are equivalent:

(1)

$\operatorname{Spec}(\hat{\kappa}(x))\to\operatorname{Spec}(A)$ extends to $\operatorname{Spec}(\mathfrak{o}_{x})\to\operatorname{Spec}(\mathscr{A})$ , that is, there is a ring homomorphism $\tilde{\varphi}_{x}:\mathscr{A}\to\mathfrak{o}_{x}$ such that the following diagram is commutative:

$\begin{CD}\mathscr{A}@>{\tilde{\varphi}_{x}}>{}>\mathfrak{o}_{x}\\ @V{}V{}V@V{}V{}V\\ A@>{\varphi_{x}}>{}>\hat{\kappa}(x)\end{CD}$
(2)

$|a|_{x}\leq 1$ for all $a\in\mathscr{A}$ .

Moreover, under the above conditions, the image of $\mathfrak{m}_{x}$ of $\operatorname{Spec}(\mathfrak{o}_{x})$ is given by $\tilde{\varphi}_{x}^{-1}(\mathfrak{m}_{x})=(\mathscr{A},|\raisebox{1.72218pt}{.}|_{x})_{<1}$ , and $(\mathscr{A},|\raisebox{1.72218pt}{.}|_{x})_{<1}\in\operatorname{Spec}(\mathscr{A})_{\circ}$ , where

\begin{cases}(\mathscr{A},|\raisebox{1.72218pt}{.}|_{x})_{<1}:=\{a\in\mathscr{A}\mid|a|_{x}<1\},\\ \operatorname{Spec}(\mathscr{A})_{\circ}:=\{P\in\operatorname{Spec}(\mathscr{A})\mid P\cap\mathfrak{o}_{k}=\mathfrak{m}_{k}\}.\end{cases}

Let $\operatorname{Spec}(A)^{\mathrm{an}}_{\mathscr{A}}$ be the set of all $x\in\operatorname{Spec}(A)^{\mathrm{an}}$ such that the above condition (2) is satisfied. The map $r_{\mathscr{A}}:\operatorname{Spec}(A)^{\mathrm{an}}_{\mathscr{A}}\to\operatorname{Spec}(\mathscr{A})_{\circ}$ given by

x\mapsto(\mathscr{A},|\raisebox{1.72218pt}{.}|_{x})_{<1}

is called the reduction map (cf. §1.1.6). Note that the reduction map is surjective (cf. [1, Proposition 2.4.4] or [5, 4.13 and Proposition 4.14]).

Theorem 2.1.

If we set $\mathscr{B}:=\{\alpha\in A\mid\text{$\alpha$ is integral over $\mathscr{A}$}\}$ , then

\mathscr{B}=\bigcap_{x\in\operatorname{Spec}(A)^{\mathrm{an}}_{\mathscr{A}}}(A,|\raisebox{1.72218pt}{.}|_{x})_{\leq 1},

where $(A,|\raisebox{1.72218pt}{.}|_{x})_{\leq 1}:=\{\alpha\in A\mid|\alpha|_{x}\leq 1\}$ .

Proof.

First let us see that $\mathscr{B}\subseteq(A,|\raisebox{1.72218pt}{.}|_{x})_{\leq 1}$ for all $x\in\operatorname{Spec}(A)^{\mathrm{an}}_{\mathscr{A}}$ . If $a\in\mathscr{B}$ , then there are $a_{1},\ldots,a_{n}\in\mathscr{A}$ such that $a^{n}+a_{1}a^{n-1}+\cdots+a_{n}=0$ . We assume that $|a|_{x}>1$ . Then

	$\displaystyle\|a\|_{x}^{n}$	$\displaystyle=\|a^{n}\|_{x}=\|a_{1}a^{n-1}+\cdots+a_{n}\|_{x}\leq\max_{i=1,\ldots,n}\{\|a_{i}\|_{x}\|a\|_{x}^{n-i}\}$
		$\displaystyle\leq\max_{i=1,\ldots,n}\{\|a\|_{x}^{n-i}\}=\|a\|_{x}^{n-1},$

so that $|a|_{x}\leq 1$ , which is a contradiction.

Let $a\in A$ such that $a$ is not integral over $\mathscr{A}$ . We show that there exists a prime ideal $\mathfrak{q}$ of $\mathscr{A}$ such that the canonical image of $a$ in $A/S^{-1}\mathfrak{q}$ is not integral over $\mathscr{A}/\mathfrak{q}$ . In fact, since $A$ is a $k$ -algebra of finite type, it is a noetherian ring. In particular, it admits only finitely many minimal prime ideals $S^{-1}\mathfrak{p}_{1},\ldots,S^{-1}\mathfrak{p}_{n}$ , where $\mathfrak{p}_{1},\ldots,\mathfrak{p}_{n}$ are prime ideals of $\mathscr{A}$ which do not intersect $S=\mathfrak{o}_{k}\setminus\{0\}$ . Assume that, for any $i\in\{1,\ldots,n\}$ , $f_{i}$ is a monic polynomial in $(\mathscr{A}/\mathfrak{p}_{i})[T]$ such that $f_{i}(\lambda_{i})=0$ , where $\lambda_{i}$ is the class of $a$ in $A/S^{-1}(\mathfrak{p}_{i})$ . Let $F_{i}$ be a monic polynomial in $\mathscr{A}[T]$ whose reduction modulo $\mathfrak{p}_{i}[T]$ identifies with $f_{i}$ . One has $F_{i}({a})\in S^{-1}\mathfrak{p}_{i}$ for any $i\in\{1,\ldots,n\}$ . Let $F$ be the product of the polynomials $F_{1},\ldots,F_{n}$ . Then $F({a})$ belongs to the intersection $\bigcap_{i=1}^{n}S^{-1}\mathfrak{p}_{i}$ , hence is nilpotent, which implies that $a$ is integral over $\mathscr{A}$ . To show that there exists $x\in\operatorname{Spec}(A)^{\mathrm{an}}_{\mathscr{A}}$ such that $|a|_{x}>1$ we may replace $\mathscr{A}$ (resp. $A$ ) by $\mathscr{A}/\mathfrak{q}$ (resp. $A/S^{-1}\mathfrak{q}$ ) and hence assume that $\mathscr{A}$ is an integral domain without loss of generality.

We set $b=a^{-1}$ . Let us see that

b\mathscr{A}[b]\cap\mathfrak{o}_{k}\not=\{0\}\quad\text{and}\quad 1\not\in b\mathscr{A}[b].

We set $a=a^{\prime}/s$ for some $a^{\prime}\in\mathscr{A}$ and $s\in{S}$ . Then $s=ba^{\prime}\in b\mathscr{A}[b]\cap\mathfrak{o}_{k}$ , so that $b\mathscr{A}[b]\cap\mathfrak{o}_{k}\not=\{0\}$ . Next we assume that $1\in b\mathscr{A}[b]$ . Then

1=a^{\prime}_{1}b+a^{\prime}_{2}b^{2}+\cdots+a^{\prime}_{n^{\prime}}b^{n^{\prime}}

for some $a^{\prime}_{1},\ldots,a^{\prime}_{n^{\prime}}\in\mathscr{A}$ , so that $a^{n^{\prime}}=a^{\prime}_{1}a^{n^{\prime}-1}+\cdots+a^{\prime}_{n^{\prime}}$ , which is a contradiction.

Let $\mathfrak{p}$ be the maximal ideal of $\mathscr{A}[b]$ such that $b\mathscr{A}[b]\subseteq\mathfrak{p}$ . As $\mathfrak{p}\cap\mathfrak{o}_{k}\not=\{0\}$ and $\mathfrak{p}\cap\mathfrak{o}_{k}\subseteq\mathfrak{m}_{k}$ , we have $\mathfrak{p}\cap\mathfrak{o}_{k}=\mathfrak{m}_{k}$ , and hence $\mathfrak{p}\in\operatorname{Spec}(\mathscr{A}[b])_{\circ}$ . Note that $\mathscr{A}[b]$ is finitely generated over $\mathfrak{o}_{k}$ and $\mathscr{A}[b]\otimes_{\mathfrak{o}_{k}}k=A[b]$ . Thus, since the reduction map

r_{\mathscr{A}[b]}:\operatorname{Spec}(A[b])^{\mathrm{an}}_{\mathscr{A}[b]}\to\operatorname{Spec}(\mathscr{A}[b])_{\circ}

is surjective, there is $x\in\operatorname{Spec}(A[b])^{\mathrm{an}}_{\mathscr{A}[b]}$ such that $r_{\mathscr{A}[b]}(x)=\mathfrak{p}$ . Clearly $x\in\operatorname{Spec}(A)^{\mathrm{an}}_{\mathscr{A}}$ . As $b\in\mathfrak{p}$ , we have $|b|_{x}<1$ , so that $|a|_{x}>1$ because $ab=1$ . Therefore,

a\not\in\bigcap_{x\in\operatorname{Spec}(A)^{\mathrm{an}}_{\mathscr{A}}}(A,|\raisebox{1.72218pt}{.}|_{x})_{\leq 1},

as required. ∎

We assume that $X$ is projective. Let $\mathscr{X}\to\operatorname{Spec}(\mathfrak{o}_{k})$ be a flat and projective scheme over $\operatorname{Spec}\mathfrak{o}_{k}$ such that the generic fiber of $\mathscr{X}\to\operatorname{Spec}(\mathfrak{o}_{k})$ is $X$ . Let $\mathscr{L}$ be an invertible sheaf on $\mathscr{X}$ such that $\left.{\mathscr{L}}\right|_{{X}}=L$ . We set $h:=\{|\raisebox{1.72218pt}{.}|_{\mathscr{L}}(x)\}_{x\in X^{\mathrm{an}}}$ . For the definition of the metric $|\raisebox{1.72218pt}{.}|_{\mathscr{L}}(x)$ at $x$ , see §1.1.6.

Corollary 2.2.

Fix $l\in H^{0}(X,L)$ . If $|l|_{\mathscr{L}}(x)\leq 1$ for all $x\in X^{\mathrm{an}}$ , then there is $s\in\mathfrak{o}_{k}\setminus\{0\}$ such that $sl^{\otimes n}\in H^{0}(\mathscr{X},\mathscr{L}^{\otimes n})$ for all $n\geq 0$ .

Proof.

Let $\mathscr{X}=\bigcup_{i=1}^{N}\mathscr{\operatorname{Spec}}(\mathscr{A}_{i})$ be an affine open covering of $\mathscr{X}$ with the following properties:

(1)

$\mathscr{A}_{i}$ is a finitely generated over $\mathfrak{o}_{k}$ for every $i$ .
(2)

$\operatorname{Spec}(\mathscr{A}_{i})_{\circ}\not=\emptyset$ for all $i$ .
(3)

There is a basis $\omega_{i}$ of $\mathscr{L}$ over $\operatorname{Spec}(\mathscr{A}_{i})$ for every $i$ .

We set $l=a_{i}\omega_{i}$ for some $a_{i}\in A_{i}:=\mathscr{A}_{i}\otimes_{\mathfrak{o}_{k}}k$ . By our assumption, $|a_{i}|_{x}\leq 1$ for all $x\in\operatorname{Spec}(A_{i})^{\mathrm{an}}_{\mathscr{A}_{i}}$ . Therefore, by Theorem 2.1, $a_{i}$ is integral over $\mathscr{A}_{i}$ , so that, by the following Lemma 2.3, we can find $s_{i}\in{S}$ such that $s_{i}a_{i}^{n}\in\mathscr{A}_{i}$ for all $n\geq 0$ . We set $s=s_{1}\cdots s_{N}$ . Then, as $sa_{i}^{n}\in\mathscr{A}_{i}$ for all $n\geq 0$ and $i=1,\ldots,N$ , we have the assertion. ∎

Lemma 2.3.

Let $A$ be a commutative ring and $S$ a multiplicatively closed subset of $A$ , which consists of regular elements of $A$ . If $t\in{S^{-1}A}$ and $t$ is integral over $A$ , then there is $s\in S$ such that $st^{n}\in A$ for all $n\geq 0$ .

Proof.

As $t$ is integral over $A$ , there are $a_{1},\ldots,a_{r-1}\in A$ such that

t^{r}=a_{1}t^{r-1}+\cdots+a_{r-1}t+a_{r}.

We choose $s\in S$ such that $st^{i}\in A$ for $i=0,\ldots,r-1$ . By induction on $n$ , we prove that $st^{n}\in A$ for all $n\geq 0$ . Note that

t^{n}=a_{1}t^{n-1}+\cdots+a_{r-1}t^{n-r+1}+a_{r}t^{n-r}.

Thus, if $st^{i}\in A$ for $i=0,\ldots,n-1$ , then $st^{n}\in A$ because

st^{n}=a_{1}(st^{n-1})+\cdots+a_{r-1}(st^{n-r+1})+a_{r}(st^{n-r}).

∎