What are in units of an affinoid algebra? Suppose that $K$ is a complete local field and $A$ is an affinoid $K$-algebra. Is there a known way to produce an explicit description of the units of $A$?
Here is what I already know: write $A^\circ$ for the subring consisting of elements of $A$ of norm at most $1$ and $A^{\circ\circ}$ for the ideal of $A^\circ$ consisting of all elements of norm strictly less that $1$. Certainly every element of $1+A^{\circ\circ}$ is a unit in $A$. So is every non-zero element of $K$. Let's call the group generated by these units the group of standard units of $A$. I believe that if the ring $A^\circ/A^{\circ\circ}$ is prime then all units in $A$ are standard. 
I also know that in general there can be non-standard units. Perhaps an easier question than the one above is `must the group of units of $A$ modulo the group of standard units be finitely generated?'.   
I have a particular application in mind for a solution to this but I feel that the question is sufficiently interesting in its own right and the application sufficiently distant from the problem that it is not worth explaining it now.
Edit: given some of the comments/answers below I probably want to modify my definition of standard units to include any non-zero element of a finite field extension of $K$ inside $A$.
Edit 2: thanks for the help so far... I'm actually happy to consider as 'standard' anything in $A^\circ$ that is a unit as an element of $A^\circ$ if that makes things easier.   
 A: Not an answer, but some perhaps pertinent examples.
One catch is that any unit in $A$ can be scaled until it's in $A^0$, but I don't think you can also guarantee that it's not in $A^{00}$. Here's a funny affinoid: $\mathbf{Q}_p\langle X,Y\rangle/(Y^2-pX)$. This is the unit disc of radius $1/\sqrt{p}$ with parameter $Y$, but it's defined over $\mathbf{Q}_p$! So $|Y|=1/\sqrt{p}$ has norm not in $|K|$. It's not a unit, but if you remove an open disc containing 0 (e.g. by throwing in another variable $Z$ and adding the relation $YZ=1$) it will be. So there's an example where non-standard units will exist, I guess.
Even more silly example: let $A$ be a finite field extension $L$ of $K$ with the induced norm. Then $A^0$ will be the integers of $L$, the standard units will be $K^\times$ times the 1-units of $L$, and if $L$ is ramified then again you have non-standard units. However in this case $A^0/A^{00}$ might just be the field with $p$ elements, which is surely "prime" whatever your definition of "prime ring" is, so I don't understand the thing you said you believe in the question.
