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Let $f$ be a formal power series with coefficients in the ring of integers of a finite extension of ${\mathbb Q}_p$. Is there a simple algorithm to compute a positive lower bound for $|\alpha - \beta|$ for the distinct roots $\alpha, \beta$ of $f$ in ${\mathbb C}_p$? Here, ${\mathbb C}_p$ denotes the field of complex $p$-adic numbers and $|\cdot|$ denotes the absolute value on ${\mathbb C}_p$ extending the usual $p$-adic absolute value on ${\mathbb Q}_p$.

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That sounds like a tough problem. If you just consider power series of the form $\prod_i(1+a_iX)$ with the $a_i$ tending to zero then the problem is that even after you've computed 10^8 of the $a_i$ to $10^8$ places of precision, the reciprocals of the next two $a_i$ might happen to be very very close together (and the size of the $a_i$ might be much smaller than the precision you've been working at). I guess what I'm saying is that the problem is "unstable" in some sense. – Kevin Buzzard Aug 23 2011 at 8:28
What if you restrict to roots of bounded norm, say, $|\alpha|, |\beta| < 1$? (I suppose then using the Weierstrass preparation theorem you could assume that $f$ is a polynomial.) – Jesse Elliott Aug 23 2011 at 8:44
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 If you restrict to roots of bounded norm then here's a simple algorithm: just use the theory of the Newton Polygon to compute the valuations of all the roots in the disc, and then use Hensel's Lemma to compute them all one by one. – Kevin Buzzard Aug 23 2011 at 9:02
As Kevin suggests, this may be a harder problem than may appear at first glance. Do you have such a power series in hand? How infinite is it, that is, does it have infinitely many segments to the Newton polygon, like $\sum_n p^{n^3} x^{n^2}$? In the finite (polynomial) case, there's certainly no problem if the thing is irreducible, and maybe factoring into irreducibles might be the strategy to take. – Lubin Aug 23 2011 at 20:00
The only simplification I can make is to restrict to roots of absolute value less than 1. The motivation was the need for an algorithm to compute a positive lower bound for the absolute value of the discriminant $\Delta$ of the Weierstrass polynomial of $f$ (defined if $f$ isn't in the maximal ideal of the power series ring) in the case where it has no repeated roots. Since $\Delta=\Prod_{i<j}(\alpha_i-\alpha_j)^2$, where the $\alpha_i$ are the roots of $f$ in ${\mathbb C}_p$ of absolute value less than 1, if I can bound the $|\alpha_i-\alpha_j|$ from below then I can bound $|\Delta|$. – Jesse Elliott Aug 23 2011 at 20:37
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