Recall that a topological ring $A$ is Tate if there is an open subring $A\_0$ such that the induced topology on $A\_0$ is t-adic for some $t \in A\_0$ that becomes a unit in $A.$ One can, given a Tate ring define a notion of powerbounded elements, denote the ring of powerbounded elements by $A^\circ.$ One says that the Tate ring $A$ is uniform if $A^\circ$ is bounded. 


Is there an easy example of a Tate ring $A$ which is complete and reduced but not uniform? 



Reduced complete Tate ring which is not uniform?
While I do not have an idea as to how easy or difficult the following is compared to other examples, here is *an* example:


Let $A\_0$ be the subring of $\mathbb{Z}\_p[[T]]$ consisting of all the power series $\sum\_i a\_iT^i$ satisfying the condition
$$v\_p(a\_i)\geq \sqrt{i}$$
(here $v\_p$ denotes the $p$-adic valuation written additively). Note that this is a ring - to check that $A\_0$ is closed under multiplication, one uses the fact that the function $\sqrt{x}$ is concave, hence subadditive. It is easy to see that $A\_0$ is $p$-complete (as an additive group, it is $\prod\_ip^{\lceil{\sqrt{i}}\rceil}\mathbb{Z}\_p$). The Tate ring in question is then $A=A\_0[\frac{1}{p}]$, with the topology given by $(A\_0, p)$. Note that $A$ contains $\mathbb{Q}\_p[T]$ as a subring.


To demonstrate some power-bounded elements, note that $p\mathbb{Z}\_p[T] \subseteq A^{\circ}$. This follows from the fact that for a polynomial $f=\sum\_{i=0}^n p a\_iT^i$ with $a\_i \in \mathbb{Z}\_p$, we have $f^N \in A\_0$ for all large enough $N$. To see that, note that $$f^N=p^N\left(\sum\_{i=0}^na\_iT^i\right)^N,$$
and so for $f^N$ to be in $A\_0,$ it is enough to have $N \geq \sqrt{nN},$ which is achieved by taking any $N \geq n.$


However, the set $p\mathbb{Z}\_p[T]$ is not itself bounded, since it contains e.g. the unbounded set 
$$pT, pT^2, pT^3, \dots$$
Thus, $A$ is not uniform.


