6
$\begingroup$

Let $f \in \mathbb{Z}[x_1, \ldots, x_n]$ and $p$ be a prime. Let $\nu_t(p)$ denote the number of solutions $\mathbf{x} \in ((\mathbb{Z}/p^t \mathbb{Z}))^*)^n$ to the congruence $$ f( \mathbf{x} ) \equiv 0 \ (\text{mod }{p^t}). $$ We define (something similar to the $p$-adic density) $$ \mu(p) = \lim_{t \rightarrow \infty} \frac{ p^t \nu_t(p) }{ \phi(p^t)^n }. $$ Could someone please explain how to show $$ \mu(p) > 0 $$ provided the equation $f( \mathbf{x} ) = 0$ has a non-singular solution in $\mathbb{Z}_p^{\times}$, the units of $p$-adic integers? Thank you very much!

PS I asked this question in mathstack exchange https://math.stackexchange.com/questions/1579876/showing-the-positivity-of-p-adic-density-of-zeroes-of-a-polynomial. I wasn't able to get an answer there so I thought I would try it on overflow.

PPS This came up when I was reading an article "Diophantine equations in the primes" http://link.springer.com/article/10.1007%2Fs00222-014-0508-1 I was interested in seeing the details but I couldn't quite figure it out.

$\endgroup$
  • $\begingroup$ I guess the strategy is to use Hensel's lemma to show that, given a p-adic solution, if you move all but one of the variables a little bit (thus breaking the solution), then you can also move the last one to fix it. That's geometrically what a non-singular solution gives you. How little a little bit is depends on the equation, but there will be some concrete epsilon depending on eveything for which it will all be OK. Once you have this, you're then capable of generating p^{n-1} solutions mod p^{M+1} for every solution mod p^M (for all suff large M) and this will be enough. $\endgroup$ – eric Dec 30 '15 at 22:27
  • $\begingroup$ It think this involves a standard argument using Hensel's lemma $\endgroup$ – Stanley Yao Xiao Dec 30 '15 at 22:27
  • $\begingroup$ It looks that the problem with applying lifting argument like Hensel lemma occurs when gradient at a p-adic root is p-adic 0. $\endgroup$ – Fedor Petrov Dec 30 '15 at 22:36
  • $\begingroup$ Thank you very much for all the comments. Would it be possible that someone give me a detailed solution for a simple case say when $n=1$ or something, because I am not quite seeing how it works still... Thank you very much. $\endgroup$ – Johnny T. Dec 30 '15 at 23:58
  • 2
    $\begingroup$ Did you know that Hensel's lemma can be formulated in a way that does not require nonsingular roots mod p? Instead of requiring an $a \in \mathbf Z_p$ such that $f(a) \equiv 0 \bmod p$ and $f'(a) \not\equiv 0 \bmod p$, it's enough to have an $a$ such that $|f(a)|_p < |f'(a)|_p^2$. And this inequality is satisfied by all $a$ that are close to a nonsingular root in $\mathbf Z_p$. See Theorem 7.1 of math.uconn.edu/~kconrad/blurbs/gradnumthy/hensel.pdf. Your multivariable situation can be reduced to the one-variable situation by fixing all but one coordinate of a nonsingular solution. $\endgroup$ – KConrad Jan 5 '16 at 0:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.