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In "On definable susbsets of p-adic fields", Macintyre 1976, he states the following fact that, let $x\in Q_p^*$ such that $v(x) = 0$ (where $v$ is the p-adic valuation) and $k\in N$ then there exists $n\in Z$ such that $x/n\in (Q_p)^k$ (ie the $k$-th powers).

This property is evident if $k$ is prime to $p$ by Hensel's lemma (the form I use is that let $f\in Z_p[X]$ and $a\in Z_p$ such that res$(f(a)) = 0$ and res$(f'(a))\neq 0$ then there exists $b\in Z_p$ such that $f(b) = 0$ and res$(a) =$ res$(b)$). But if not, then I don't know how to prove it.

If anyone has any ideas on the subject I'd be very grateful.

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The basic idea is that it is possible to extract $k^\mathrm{th}$-roots of elements of $\mathbb{Z}_p$ that are close enough to 1. This is because you can compute explicitely the radius of convergence of the series $(1+X)^{1/k}$ (because you know the $p$-adic valuation of the factorials, hence of the binomial coefficients). Let us say that the radius of convergence is greater that $p^{-r}$.

Now you can write $x= n + yp^r$, with $y\in \mathbb{Z}_p$ and $n$ invertible in $\mathbb{Z}_p$, since $v(x)=0$. Then $x/n = 1+ n^{-1}yp^r$ has a $k^\mathrm{th}$-root.

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You can notice that $(Z_p^\times)^k$ is open. It avoids the use of series. –  Auguste Hoang Duc Jul 6 '11 at 12:26
    
Writing x=n+yp^r makes absolute sense if x is in Q (it is a simple euclidian division) but how does it work when x is not? –  Silvain Rideau Jul 6 '11 at 12:37
    
@Auguste: Indeed! That's a nice argument. @Silvain: Write x as a series $\sum_{i\ge 0} a_i p^i$ and let y be $\sum_{i\ge r} a_i p^{i-r}$. –  Jérôme Poineau Jul 6 '11 at 13:20
    
@Auguste, how do you see the open-ness? Perhaps one wishes to argue based on the fact that the derivative is a surjection, but I suspect that a proof based on this observation will still involve series somewhere. –  L Spice Jul 6 '11 at 19:03
    
@Jérôme, or, again, one can avoid series and say that integers are dense, so that there exists one in $x + \wp^r$. –  L Spice Jul 6 '11 at 19:05
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There are many different ways of attacking this, and though I have nothing against series, I always prefer a polygon argument, when one exists. First, by dividing by a suitable $n_0$, you can assume that $v(x-1)\ge1$; second, since Silvain has already taken care of the question when $k$ is prime to $p$, you can assume that $k=p^s$ for some $s\ge1$.

Now consider the polynomial $f(t)=(1+t)^p-1$, its Newton polygon, and especially the polygon of $f(t)-\alpha$, for $\alpha\in\mathbb{Z}_p$. If $v(\alpha)\le1$, then there's a single segment of (negative) slope $1/p$ or zero, while if $v(\alpha)\ge2$, then there's a segment of slope $v(\alpha)-1$ and width $1$, and a segment of slope $1/(p-1)$ and width $p-1$. Consequence is that $f$ maps $p\mathbb{Z}_p$ onto $p^2\mathbb{Z}_p$ in such a way that $v(f(z))=v(z)+1$, and more, that $f^{\circ s}$ maps $p\mathbb{Z}_p$ onto $p^ {s+1}\mathbb{Z}_p$. (Argument needs slight modification when $p=2$) So for $x/n$ to be a $p^s$-th root of a $p$-adic integer, it's enough for the condition $v(x-1)\gt s$ to hold, which can be done because the reciprocals of integers are dense in the $p$-adic units.

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You can further reduce to the case $s=1$ and study the map $(\ )^p$ on $1+p{\mathbf Z}_p$ as in Chapter 15 of Hasse's {\it Number Theory\/} or in III.3 of arxiv.org/abs/0711.3878v2. This works for finite extensions of ${\bf Q}_p$ and avoids the use of the Newton polygon or the $p$-adic exponential. –  Chandan Singh Dalawat Jul 7 '11 at 2:54
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Another way to do this is to use the Hensel-Rychlick version of Hensel's Lemma. This says that if you have $f\in {\mathbb Z}_p[X]$ and $a\in{\mathbb Z}_p$ such that $v(f(a))>2v(f^\prime(a))$, then $f$ has a zero in ${\mathbb Z}_p$.

As in Jerome's answer we can find an integer $n$ such that $x=n+p^3\alpha$ where $\alpha$ is a unit. Let $f(X)=X^p-{x\over n}$. Then $v(f(1))=3$ while $v(f^\prime(1))=1$. Thus $x\over n$ has a $p$-th root.

Indeed, the Hensel-Rychlick Lemma says you can find a zero $c$ with $v(c-a)\ge v(f(a))-v(f^\prime(a))$.

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I guess that is probably the proof that Macintyre is hinting at in his paper, but I could not find what Hensel-Rychlik said... –  Silvain Rideau Jul 11 '11 at 13:56
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I think you just need to ensure that $\frac{x}{n} \equiv 1 (mod p^{1 + \nu(k)})$ if $p$ is odd, and $\frac{x}{n} \equiv 1 (mod p^{2 + \nu(k)})$ if $p = 2.$ Then you can start to use Hensel's Lemma, starting with $x_{0} = 1$ for the first iterate, obtaining a sequence $(x_m)$ whose limit $y$ has $y^k = \frac{x}{n}$.

Added later: This is much the same as the argument given in the post above, except that we prefer the iterative method, rather than the infinite series method using the binomial theorem for fractional exponents. We set $x_0 = 1$ and $x_{m+1} = x_{m} - \frac{x_m ^{k} - a}{kx_{m}^{k-1}}$, where $a = \frac{x}{n}$. Suppose that $p$ is odd and $\nu(x_{m} ^{k} - a) = d_m + \nu(k).$ Then $x_{m+1}^{k} - a = \frac{k(k-1)}{2} x_{m}^{k-2}\left(\frac{x_m^{k}-a}{kx_m^{k-1}}\right)^{2} + \ldots$, and we see then that $\nu(x_{m+1}^{k} - a) \geq 2d_m + \nu(k).$ Hence $(x_m)$ converges to $a$. A slightly more complicated, but similar, argument works for $p=2$.

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I am under the impression this generalises nicely if v(x) is not 0. Indeed, at some point I began to wonder if Qp*/(Qp*)^k was finite and the explicit bound here seems to point that way. –  Silvain Rideau Jul 11 '11 at 10:03
    
Thanks. Well, it seems to be fine if $\nu(x) >0$, since we can just choose replace $x$ by $\frac{x}{m}$ for some integer $m$ with $\nu(m) = \nu(x)$. Since the conclusion is just about $\frac{x}{n}$ for some integer $n$, this doesn't change anything. But if $\nu(x) <0$, I don't see what to do, if you're only allowed to divide by inetgers. –  Geoff Robinson Jul 11 '11 at 11:04
    
Well I guess then you have to allow n to be rationnal (or at least of the form 1/n where n is an interger). –  Silvain Rideau Jul 11 '11 at 14:03
    
Sure, it's OK if you allow rational $n$. –  Geoff Robinson Jul 11 '11 at 15:17
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