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My question concerns the following:

For a given polynomial $F(x) = a_n x^n + \cdots + a_0$ with a root $F(x) = 0 , x \in \Omega_p$, (that is the completion of the p-adic numbers) and $a_j \in Z_p$, we can find the valuation of the root through Newton Polygons. We also know that an arbitrary non-zero element $x \in \Omega_p$ can be written as $p^r ab$, where '$a$' is a $(p^l -1)$th root of unity (i.e. a Teichmuller lift) and '$b$' is an element in the open unit disk about 1 (i.e. $b = 1 + M, |M|_p < 1$) (page 100 here).

The question is, for a given root of the polynomial with non-zero valuation, can we find out information concerning which kind of root of unity it is a product of, i.e. the order of '$a$' or terms of its algebraic expansion (such as discussed here), from information we have regarding the polynomial?

Thank you.

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You didn’t say anything about the nature of the number $r$ in the expression $p^r$. In fact, $r$ may be any rational number. This requires giving an unambiguous definition of $p^r$ for any rational number $r$. This may be done, but only with an appeal to the Axiom of Choice, and amounts to a choice of a homomorphism $\xi:\Bbb Q^+\to\Omega^\times$ satisfying $\xi(1)=p$, where $\Omega$ is your complete algebraically closed field.

Once this is done, your task is easy. Take your $p$-adic quantity $x$, say $v_p(x)=m/n\in\Bbb Q$, not necessarily an integer since $x$ is not necessarily in $\Bbb Q_p$. Then $x/\xi(m/n)$ is a unit in the field $\Bbb Q_p(x,\xi(1/n))$, reducing modulo the maximal ideal to $\tilde a$ in some finite field $\Bbb F$. This lifts to a (unique) root of unity $a$, so we have $x/\xi(m/n)=ab$, where $b$ is a “principal unit”, i.e. congruent to $1$ modulo the maximal ideal. And there you are.

But notice that everything depends on your uniform choice of roots of $p$.

I think the utility of the decomposition $x=p^rab$ is in defining a logarithm on all of $\Omega^\times$; you define $\log(p)=1$, and use the analytic definition of $\log$ on $1+\mathfrak M$; the logarithm of any root of unity is necessarily $0$ to preserve the homomorphism property (and comes for free when the root of unity is in $1+\mathfrak M$). Then the ambiguity of our definition of $p^r$ no longer matters.

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  • $\begingroup$ Thank you, this is precisely the sort of information I needed. The motivation for this question was trying to understand whether the p-adic value of the elementary symmetric polynomials appearing in Viete's relations could be bounded by understanding the form the roots take. $\endgroup$
    – Niklas
    May 16, 2017 at 10:53
  • $\begingroup$ That is given no precise knowledge of the polynomial's coefficients beyond some broad properties of the polynomial's Newton Polygon (e.g. a single line segment). $\endgroup$
    – Niklas
    May 16, 2017 at 10:59
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    $\begingroup$ I’m sure you know that each segment of the N-Polygon corresponds to a polynomial factor of the original. Even for a single-segment polygon, where all roots have the same $v_p$-value, I don’t think there’s much you can say about the coefficients from the “form” of the roots, especially if by form you mean the expression of a root as $p^rab$. $\endgroup$
    – Lubin
    May 16, 2017 at 12:40
  • $\begingroup$ True, there are many sets of coefficients which would lead to the same NP. $\endgroup$
    – Niklas
    May 16, 2017 at 13:09

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