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

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

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

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

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,

as required. ∎