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

Question

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.

Answer 1

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.