If $X$ is a smooth rigid analytic space over a $p$-adic field $K$ (of characteristic zero), then every coherent $\mathcal{O}_X$-module with integrable connection is locally free. In his paper "Finiteness theorems for the cohomology of an overconvergent isocrystal on a curve" Crew gives a proof of this result in the particular case of a 1-dimensional annulus over $K$ as follows.

Let $I$ be a closed interval of $[0,\infty)$, and $A_I$ the ring of Laurent series in 1 variable over $K$ which are convergent for $\vert x \vert\in I$. Let $M$ be a finitely generated $A_I$-module with connection, we wish to prove that $M$ is a projective $A_I$-module. Since $A_I$ is a PID it suffices to show that $M$ is torsion free. Let $f$ generate the annihilator of $M_{\mathrm{tors}}$, one can easily show using the axioms of a connection that $f'\in (f)$. (Here $f'$ means differentiation with respect to the variable).

Crew now argues as follows: "Since $A_I$ is a PID and contains the rational numbers, $(f)$ must be the unit ideal, and hence $M_{\mathrm{tors}}=0$".

**Why does this implication hold? That is, why does the fact that $A_I$ is a PID and a $\mathbb{Q}$-algebra, together with the fact that $f'\in (f)$, imply that $(f)=A_I$?**

For example, would this implication still hold if I worked instead over the ring $K\otimes_{\mathcal{O}_K} \mathcal{O}_K[[t]]$ of power series with bounded coefficients?