How to construct a $p$-adic power series $g(x)$ over $O$ or over $\text{Frac}(O)$ from a given series $f(x) $?

Question

Let $K \supseteq \mathbb{Q}_p$ be the $p$-adic field with ring of integer $O$ and maximal ideal $M$. Let $\bar M$ be the closure of $M$.

Let $f(x)$ be a noninvertible power series in $ x \cdot O[[x]]$ having only simple roots.

Consider an infinite subset $S \subset \bar M$ such that $f(\alpha)$ takes values in $S$ for all $\alpha \in S$.

I want to construct a power series $g(x) \in x \cdot O[[x]]$ or over $\text{Frac}(O)$ which interpolate $f(x)$ on $S$ and has at least one multiple root that is:

(i) $g(\alpha)=f(\alpha)$ $\forall \alpha \in S$,

(ii) $g(x)$ has at least one root with multiplicity $\geq 2$.

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The note on $p$-adic interploation says a function $f : \mathbb{N} \to \mathbb{Q}_p$ extend to a continuous function $\mathbb{Z}_p \to \mathbb{Q}_p$. I am not sure whether this extends to a finite extension $K/\mathbb{Q}_p$.

Any comment here, please.

But I am thinking the following possible way:

Let $g(x)=x^m \circ f(x)=(f(x))^m$ for $m \geq 2$.

Then clearly $g(x)$ has multiple roots.

To conclude this infinite case, we have to show $f(\alpha)=g(\alpha)$ for all $\alpha \in S$ i.e., we have to show $$f(\alpha)=(f(\alpha))^m.$$

Assume $f(\alpha)$ is idempotent for all $\alpha \in S$ (which is not always true though). In that case $f(\alpha)=(f(\alpha))^2$ implies $f(\alpha)=(f(\alpha))^m$ for $m \geq 2$.

I have doubt the approach is correct.

Can you please suggest me any authentic way to anwer my question ?

Thanks

Answer 1

If $f\in p^{-N}O[[x]]$ (well-defined as a function $p^\epsilon O_{\overline{\Bbb{Q}}_p}\to O_{\overline{\Bbb{Q}}_p}$) then for any $r>0$, $f$ has finitely many roots of valuation $\ge r$.

If $S$ contains infinitely many elements of valuation $\ge r$ and $f|_S=g|_S$ for another $g\in p^{-N}O[[x]]$ then $f=g$.

The case $S$ finite is already treated by Merosity.

Answer 2

Here's an approach that will work in the finite case, but not sure if/when it will work in the infinite case, since it depends on what $S$ looks like. Since we know $g(\alpha)-f(\alpha)=0$ we can define it as,

$$g(x) - f(x) = xh(x)\prod_{\alpha \in S} (1-\frac{x}{\alpha})$$

Since we want $g$ to not have a constant term, the $x$ was placed on the product. Furthermore since we want $g$ to have a double root, we'll specify $h$ with two free variables which we can use to solve for a condition on a root. We'll put the root at $x=r$, which must lie in the region of convergence of $f$.

$$h(x)=a+b(x-r)$$

This specifically gives us $h(r)=a$ and $h'(r)=b$ which will give us a system of equations to solve for $g(r)=g'(r)=0$.

Specifically,

$$a = -\frac{f(r)}{r}\prod_{\alpha \in S} (1-\frac{r}{\alpha})^{-1}$$

$$b = \frac{f'(r)}{f(r)} - 1 + \sum_{\alpha \in S} \frac{1}{\alpha-r}$$