Power bounded elements of $\Bbb Q_p \langle T \rangle /(T^2)$

Question

I am rather confused by this example in 9, page 63.

Context. Let $A= \Bbb Q_p \langle T \rangle /(T^2)$.

$\Bbb Q_p\langle T \rangle$, which as pointed out, this is the completion of $\Bbb Q_p[T]$ wrt to the sup norm $\sup |a_k|$.

I would like to compute $A^\circ$. its set of power bounded elements.

The text claims that the answer is isomorphic to$$\Bbb Z_l \oplus \Bbb Q_l T$$

I would appreciate if someone explains why. My argument below shows after that it should be $$ \Bbb Z_l \oplus \Bbb Z_l T$$

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1. $\Bbb Q_l\langle T \rangle$ with the Gauss norm, $|\quad|= \sup |a_k|_p$ for $f= \sum_0^\infty a_k T^k \in \Bbb Q_l\langle T \rangle $. See 8,pg63,

2. we endow $A$ with the residue norm, $|\quad|_r$. For $\bar{f} \in A$, $$|\bar{f}|_r = \inf_{h \in \Bbb Q_l\langle T \rangle, \bar h=\bar{f}} |h|$$

3. An element $\bar{f} \in A$ is power bdd. iff $|\bar{f}|_r \le 1$.

4. By this formula, we see that for an arbitrary element $\bar{f} \in A$, represented $\sum a_kT^k$, we can really just take the lift of the first two elements. $a_0+a_1T$. But $$|\bar{f}|= |a_0+a_1T| = \sup \{ |a_0|_l, |a_1|_l\} \le 1 \Leftrightarrow a_0, a_1 \in \Bbb Z_l $$

What went wrong ? In retro spect, I think 3. is what went wrong. The residue norm doesn't seem to be power multiplicative in general.

Answer 1

Please avoid terminology.

$\Bbb Q_p \langle T \rangle $ is the completion of $\Bbb Q_p [T]$ for the norm $\sup |a_k|_p$ ?

You get a natural norm on the quotient by $T^2$, which is $|r+sT|_A = \max |r|_p,|s|_p$.

$A^\circ$ is the elements such that $\lim\sup_n |u^n|_A\ne \infty$.

$$|(r+sT)(v+wT)|_A=|r(v+wT)+v(r+sT)|_A\le \max |r|_p |v+wT|_A,|v|_p |r+sT|_A$$ and $|r+sT|_A^n \ge |r|_p^n$ so $$A^\circ=\Bbb{Z}_p+T \Bbb{Q}_p / [T^2=0]$$