The following lemma is in [Bosch](http://www.math.purdue.edu/~tongliu/seminar/rigid/Bosch.pdf)'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?



A translation between formal and rigid geometry
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 $R$). 


