# Does $x, y\in \mathcal{R}$, $z\in (\mathcal{E}^\dagger)^*$ with $x\cdot y= z$ imply $x,y\in (\mathcal{E}^\dagger)^*$

Let $\mathcal{R}$ be the Robba ring and $\mathcal{E}^{\dagger}$ the elements of $\mathcal{R}$ that are bounded at 0 (so the coefficients of the powerseries are bounded. Is it true that $x, y\in \mathcal{R}$, $z\in (\mathcal{E}^\dagger)^*$ with $x\cdot y= z$ implies $x,y\in (\mathcal{E}^\dagger)^*$?

If not, is it then true that $x\in \mathcal{R}$, $y, z\in (\mathcal{E}^\dagger)^*$ with $x\cdot y= z$ implies $x\in (\mathcal{E}^\dagger)^*$?

This kind of seems to be implied in some of the papers of Colmez about $(\varphi, \Gamma)$-modules, for example here https://webusers.imj-prg.fr/~pierre.colmez/triangulines in Proposition 4.2, but I might be mistaken and some other argument is implicitly used.

• I think it is you who have written a thesis about the robba ring over a $p$-adic field. I have read it and I have learned many things from it. – Ang Nov 18 '19 at 16:04
• Thanks, that is nice to hear. I am glad you found it helpful :-) – Quentchen Nov 20 '19 at 10:11
• Have you known how to prove your lemma 5.26? In your remark, you said that Colmez gave an alternative proof with the use of an equivance of categories, but when I see Colmez's paper which you refered to, I find it is short and not detailed... – Ang Nov 20 '19 at 10:16

Yes, both statements are true. Note also that $\mathcal{E}^\dagger$ is a field. The simplest way to see this would be to use the fact that if you work over a fixed annulus $A = \{ z, r \leq |z| < 1\}$, then an analytic function on $A$ is bounded if and only if it has finitely many zeroes. You can prove this using the theory of Newton polygons. This now implies your first claim, as well as the fact that $\mathcal{E}^\dagger$ is a field.
• How to use the boundedness of the analytic function $f$ if we consider the Newton polygons associated to $f$? Thanks! – Ang Nov 21 '19 at 13:29