# The rigid-analytic open disk

Let $K$ be a local field and $D_K$ the open unit disk, considered as a rigid space or adic space over $K$. What is the algebra of analytic functions on $D_K$? Proposition 1.1 of this article describes functions on the punctured open disk as certain "Laurent series" (possibly with infinitely many terms of negative degree). Can we write functions on $D_K$ as power series satisfying some convergence condition?

A related question: suppose $K = k((t))$ for some field $k$. If I am not mistaken, $D_K = \text{Spa } k[[t]] \times \text{Spa } k((t))$ with the product taken in the category of adic spaces over $k$. So it seems to me that $D_K$ will have a "diagonal" $K$-point. Does this make sense?

• The second half becomes much easier if you don't use $t$ so much: if $F$ is any non-archimedean field containing $k$ then ${\rm{Spa}}(k[\![t]\!],k[\![t]\!])\times_{{\rm{Spa}}(k,k)}{\rm{Spa}}(F,F^0)$ exists and is the (adic space associated to the) open unit disc over $F$, verified by universal property (exercise) using its "standard coordinate" (pulled back from the first factor) denoted as $t$ by abuse of notation. For $a \in F$ with $|a|<1$ there is a "classical point" cut out by the condition $t=a$. Now set $F=k(\!(t)\!)$ and $a=t\in F$ to get the diagonal point with abuse of notation. – user74230 Jan 22 '15 at 17:52
• I should have said that $F$ contains $k$ as a discrete subfield (equivalently, $k$ lies in $F^0$). – user74230 Jan 23 '15 at 2:50

If I translate "closed unit disc" by "maximal spectrum of the Tate algebra" (but they are not the same), then the open unit disc is the union of the closed discs with radius smaller than one. Let $A_\varepsilon$ denote the algebra of analytic functions on $B_\varepsilon (0)$. In our case, $A_\varepsilon$ equals the rescaled Tate algebra $K \langle \varepsilon ^{-1} T \rangle$. For $\alpha < \beta$ we get an inclusion of function algebras $A_{\beta} \subset A_{\alpha}$ and the algebra of analytic functions on $D_K$ is the (projective) limit of those.