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.

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Stupid question: what does "I believe that if the ring...is prime" mean? Does it mean "equal to a field with p elements"?? –  Kevin Buzzard Apr 30 '10 at 12:26
For a finite extension $K'/K$, the $K$-algebra $A = K'$ is affinoid and $A^0/A^{00}$ is the residue field, hence a "prime" ring (which I take to mean a domain), yet the group of standard units has index in the group of all units equal to $e := [|{K'}^{\times}|:|K^{\times}|] < \infty$ times $[{k'}^{\times}:k^{\times}]$ (with $k$ and $k'$ the respective residue fields of $K$ and $K'$). This is usually infinite when $k$ is infinite. So the answer to both questions appears to be negative. Maybe assuming "geometric reducedness" and "geometric irreducibility" things are better? It feels delicate. –  BCnrd Apr 30 '10 at 12:30
By prime ring I do mean domain. This is standard in non-commutative algebra. I suppose less so in commutative algebra. I am most interested in the case $K$ is a finite extension of $\mathbb{Q}_p$ however more general settings are also of interest. –  Simon Wadsley Apr 30 '10 at 12:54
Actually the second sentence of my comment above is not strictly accurate. It is standard to say a ring is prime if the zero ideal is prime but this is not equivalent to the ring being a domain in this case. –  Simon Wadsley Apr 30 '10 at 13:57
@Kevin, better: for $K$ local in classical sense and $A$ "geom. normal" and "geom. conn'd", $A^{\times}$ mod standard units is fin. gen'td! By geom. reducedness, Grauert-Remmert give finite sep'ble $K'/K$ so $(A_ {K'})^0$ (normal!) is top. fin. generated over val. ring $R'$ with geom. reduced special fiber. Any elt of $A^{\times}$ has $K^{\times}$-multiple in $A^0$; image in normal domain $(A_ {K'})^0$ has ord's at dvr generic pts of special fiber. By normality, two with same ords have unit ratio in $(A_ {K'})^0$. Reduced special fiber has fin. generated unit group (since $k$ finite). –  BCnrd Apr 30 '10 at 16:28
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.