Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I understand the concepts used in then Poincare series, but I don't know how to compute the Poincare serie of an specific polinomyal, for example $x^2-a$, what must I do?

share|improve this question
    
It's my understanding that the Poincare series of a polynomial $f$ is $\sum a_nx^n$, where $a_n$ is the number of solutions of $f(x)\equiv0\pmod{p^n}$ --- but you have to be given a prime $p$ to make this definition go. Anyway, let's start with this question: given a prime $p$, what can you say about the number of solutions of $x^2-a\equiv0\pmod p$? –  Gerry Myerson Nov 15 '12 at 4:42
add comment

1 Answer 1

up vote 1 down vote accepted

Let's work out your example for $f(x)=x^2-a \in \mathbb{Z}[x]$. For a prime $p>2$, suppose first that $x^2 \equiv a \mod p$ has no solutions (i.e. $a$ is not a residue mod $p$). Then $x^2 \equiv a \mod p^n$ has no solutions for any $n \geq 1$, as these would project down to roots of $x^2-a \mod p$. In this case, we find that the Poincare series is $0$.

If instead $x^2 \equiv a \mod p$ admits a root $r \not\equiv 0 \mod p$, the fact that $f'(r) =2r \not\equiv 0 \mod p$ implies by Hensel's lemma that there is a unique solution $s$ to $x^2 \equiv a \mod p^2$ such that $s \equiv r \mod p$. Thus our two solutions mod $p$ lift to two solutions mod $p^2$, and so on (again by Hensel's lemma). So we always obtain two solutions, and our series is $$\sum_{n=1}^\infty 2 x^n = \frac{2x}{1-x}.$$

Else $a\equiv 0 \mod p$, a situation that requires more work. Write $a=p^k m$ with $(p,m)=1$. Then $x^2 \equiv p^km \mod p^n$ (with $n$ sufficiently large) forces $x$ to be divisible by $p$, with multiplicity $\lceil k/2 \rceil$. Dividing through yields $$(x/p^{\lceil k/2 \rceil})^2 \equiv m \mod p^{n-\lceil k/2\rceil},$$ (so in fact $k$ must be even to obtain solutions mod $p^n$ with $n$ large). For $y:=x/p^{\lceil k/2 \rceil}$, we see that this problem can be solved with one of the cases above.

Note that some of the above work falls apart for $p=2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.