For any field of $p$-adic numbers $\mathbb{Q}_p$ there is a unique, up to an isomorphism, extension $\mathbb{C}_p$ that is both algebraically and metrically complete, the field of $p$-adic complex numbers. Further, there exists a unique spherically complete extension $\Omega_p\equiv \mathbb{C}_p^{sph}$ of $\mathbb{C}_p$, whose elements are easily characterized:

The elements are exactly those power series $\sum_{r \in \mathbb{Q}} a_r p^r$ with coefficients given by Teichmüller representatives of $\overline{\mathbb{F}_p}$, such that the set of exponents with nonzero coefficients forms a well-ordered subset of the rationals.

(see the post https://sbseminar.wordpress.com/2007/08/21/p-adic-fields-for-beginners/, and Poonen's article http://www-math.mit.edu/~poonen/papers/amsval.pdf)

The question:

Is there any alghorithm known that produces a power $p$ series, like above, for the roots of a given polynomial $f\in\mathbb{Q}_p[x]$ ?

The situation is more or less clear when equation $f(x)=0$ is solvable in radicals, e.g. $f(x)=x^2-3 x +1 \in\mathbb{Q}_5[x]$ has a root $$x_1=\frac{3+\sqrt{5}}{2}= 4 + 3\cdot 5^{1/2} + 2\cdot 5^1 + 2\cdot 5^{3/2} + 2\cdot 5^{2} + \ldots\; \in \mathbb{C}_5 .$$ But what about general case?

UPD. Elements $x\in\mathbb{C}_p$ inside $\mathbb{C}_p^{sph}$ can be quite easily characterized: for any rational $t\in\mathbb{Q}$ there are only finitely many $a_r\neq 0$ with $r<t$. To see this one should approximate $x$ by $y\in\mathbb{Q}_p^{alg}$ such that $|x-y|_p < p^{-t}$. Since $y$ belongs to some finite extension of $\mathbb{Q}_p$ with valuation group of the form $\{ p^ {-k/n} : k\in \mathbb{Z}\}\subset(0,+\infty)$, $n$ fixed, the expansion $y=\sum_{r\in\mathbb{Q}}{b_r p^r}$ can have nonzero $b_r$ only for $r=k/n$. Since $a_r=b_r$ for $r<t$, the statement follows.

UPD2. Numerically finding roots from coefficients is an ill-defined problem for polynomials over $\mathbb{C}$, as an example of Wilkinson's polynomial shows (see https://en.wikipedia.org/wiki/Wilkinson%27s_polynomial). I guess, the situation is as bad over $\mathbb{C}_p$. Or not?

  • $\begingroup$ You can use Hensel's lemma to find (at least some of) the roots in $\Bbb{Q}_p$. I don't know about the others, though. $\endgroup$ – A.P. Jun 6 '15 at 11:50
  • $\begingroup$ The more I look at this question, the more interesting it is. But your example works only over $\Bbb Q_5$. Have you expressed the same irrational quantity as an element of $\Omega_p$ for other values of $p$? No matter what, you should do lots of examples. By hand! $\endgroup$ – Lubin Jun 15 '15 at 1:31
  • $\begingroup$ @Lubin: Yes, the example is over $\mathbb{Q}_5$ (I've updated the post accordingly). The root can be expressed for other values of $p$. The resulting series belong either to $\mathbb{Q}_p$ itself, or to its unramified quadratic extension, depending on the value of Legendre's symbol $\left(\frac{5}{p}\right)$. The whole problem seems to boil down to finding roots of polynomial equations in finite fields (and to writing them down in some appropriate form). $\endgroup$ – Yauhen Radyna Jun 15 '15 at 20:25
  • $\begingroup$ @YauhenRadyna: I see a potential flaw in the reasoning of your UPD. For $y\in\overline{\mathbb{Q}_p}$, we have that $|y|\in R=\{p^{-k/n}:k\in\mathbb{Z}\}$ for some fixed $n\in\mathbb{N}$, but I don't see why the support of $y$ is contained in the valuation group $R$ of $\mathbb{Q}_p(y)$. $\endgroup$ – Chilote Dec 7 '16 at 0:58
  • $\begingroup$ I was right, an element in $\mathbb{Q}^{alg}_p$ may have accumulation points in its support i.e. there exists $y=\sum a_rp^r$ such that for some $t\in\mathbb{N}$, the set $\{r\in\mathbb{Q}:r<t, a_r\neq0 \}$ is infinite. In fact, in the article "Algebraic p-adic expansions." Journal of Number Theory 23.3 (1986): 279-284, by David Lampert, it is proved that for $p>2$, the exponents of the p-adic expansion of the $p^2$-th root of $1$ are $r_i=(p^i-p+1)/(p^{i+1}-p^i)$ for $i\in\mathbb{N}$. Hence $\{r\in\mathbb{Q}:r<1, a_r\neq0 \}$ is infinite. Therefore your characterization of $C_p$ is wrong. $\endgroup$ – Chilote Dec 7 '16 at 15:40

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