# How many solutions for $x^2 = 1$?

Let $F$ be an non-archimedean local field, let $o$ be its ring of integers, and let $p$ be the maximal ideal

Is there a closed form for the cardinality

$$| \{ x \in o / p^N: x^2 = -1 \bmod p^N\} | =A(o, p^N)?$$

I am actually interested in $\chi(p^N) = (-1)^{\frac{p^{N} - p^{N-1}+A(o,p^N)}{2}}$ or $\chi(p^N) / \chi(p^{N-1})$. Are these functions characters? Do they have a name and are they well understood?

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Doesn't Hensel apply? – Gerry Myerson Mar 3 '12 at 12:01
I would guess that it only applies for odd residue characteristic, since the derivative vanishes in even characteristic. – plusepsilon.de Mar 3 '12 at 12:07
Interesting question! I checked that taking $F$ to be the unramified degree $k$ extension of $\mathbf{Q}_2$, the answer seems to be $A(\mathbf{Z}_{2^k}, N) = \begin{cases} 1 & \text{if$N = 1$} \\\ 2^{k} & \text{if$N = 2$} \\\ 2^{k+1} & \text{if$N \ge 3$.}\end{cases}$ based on computations for small $k$ and $N$. – David Loeffler Mar 3 '12 at 14:09
Did you want square roots of 1, or square roots of -1? – Hurkyl Mar 3 '12 at 15:22
-1,...................... – plusepsilon.de Mar 3 '12 at 17:43

Here's a solution for the case when $F$ is an unramified extension of $\mathbf{Q}_2$ (of degree $k$, say).

If $N = 1$, then the only solution is $x = 1$ (obviously, since the Frobenius on $\mathbf{F}_{2^k}$ is injective).

I claim that for $N \ge 2$, we have $x^2 = 1$ in $\mathcal{O} / 2^N$ if and only if $x = \pm 1$ in $\mathcal{O} / 2^{N - 1}$. Indeed, we must have $x = 1 \mod 2$, so suppose $x = 1 + 2y$ for $y \in \mathcal{O}$. Then $x^2 = 1 + 4y(y+1)$, and $y$ and $y+1$ cannot both be divisible by 2, and the result follows by checking cases. This gives the formula I stated in a comment above,

$A(\mathcal{O}, 2^N) = \begin{cases} 1 & \text{if$N = 1$} \\ 2^{k} & \text{if$N = 2$} \\ 2^{k+1} & \text{if$N \ge 3$.}\end{cases}$

In the ramified 2-adic case, things are going to be messier, but I expect the picture will be similar, with $A(\mathcal{O}, \mathfrak{p}^N)$ stabilizing for $N$ sufficiently large. For instance, when $F = \mathbf{Q}_2(\sqrt{-1})$ we have

$A(\mathcal{O}, 2^N) = \begin{cases} 1 & \text{if$N = 1$} \\ 2 & \text{if$N = 2, 3$} \\ 4 & \text{if$N = 4$} \\ 8 & \text{if$N \ge 5$}\end{cases}$

(I checked as far up as $N = 10$).

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Very good. But OP has changed the problem to $x^2=-1$. – Gerry Myerson Mar 4 '12 at 0:13
Right, in which case the answer is either 0 or the number I gave, since the ratio of any two square roots of -1 is a square root of +1. – David Loeffler Mar 4 '12 at 9:17