# A translation between formal and rigid geometry

The following lemma is in Bosch's book "Lectures on Formal and rigid geometry" p198.

Lemma Let $K$ be a non-archimedean field and $R$ its valuation ring. Let $X= \mathrm{Spf}A$ be an affine admissible formal $R$ scheme. Then there are canonical bijections between (1) the set of non-open prime ideals $\mathfrak{p}\subset A$ with $\dim A/\mathfrak p =1$ and (2) the set of maximal ideals in $A\otimes_RK$.

Now let's look at a easy case: $A=R\langle \zeta_1,\dots,\zeta_n\rangle$. Then $A\otimes_RK$ is simply the Tate algebra $T_n:=T_n(K)$. Then we know the set of maximal ideals in $T_n$ can be described by the points of the unit ball $\mathbb B^n(K)$(assume $K$ is algebraically closed): $\mathfrak{m}_x=\{f\in T_n\mid f(x)=0\}$ The question is that what the prime ideal does $\mathfrak m_x$ correspond by the above lemma?

Using your notation: it corresponds, like in the "classic" case of polynomials over a field, to the ideal generated by $\zeta_1-x_1,\dots,\zeta_n-x_n$, where $x=(x_1,\dots,x_n)\in \mathbb B^n(K)$. In another terms, it is as before the kernel of the evaluation map $A\to R$ given sending $f$ to $f(x)$ (which converges and it is an element of the valuation ring $R$).
• But I feel confused still. It doesn't make sense if $x_i\in K - R$, right? Maybe we should take a fraction$x_i=y_i/z_i$ and consider $z_i\zeta_i -x_i$. – Hang Jul 16 '18 at 18:00
• But $x_i\in R$ for all $i$. This is what it means to be in $\mathbb B^1(K)$. – Xarles Jul 17 '18 at 22:52