Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

First I define the level of field. The level of a field $\mathbb K$ is the least $n$ such that $−1$ is a sum of $n$ squares in field, and is denoted by $S(\mathbb K)$. I know that the level of $\mathbb Q_2$ (the field of $2$-adic numbers) is $4$. It is an easy calculation:

You can see that every $2$-adic integer which is congruent to $1 \bmod 8$ is a square. So $-7$ is a square and hence $S(\mathbb K) \leqslant 4$. On other hand any integral square in $\mathbb{Q}_{2}$ is congruent to $0$, $1$, or $4 \bmod 8$ so $S(\mathbb K) = 4$.

I am looking for the level of $p$-adic number field. Any suggestions? Thanks.

share|cite|improve this question
I made some $\TeX$ edits and fixed (I hope) some minor English errors. Feel free to revert! – Dylan Moreland May 25 '12 at 6:33
up vote 4 down vote accepted

For odd $p$, $x^2 + y^2 + 1 \equiv 0 \mod{p}$ always has a solution because the sets $\{x^2 : x \in \mathbb{F}_p\}$ and $\{-y^2-1 : y \in \mathbb{F}_p\}$ both have cardinality $(p+1)/2$ and hence have nonempty intersection. By Hensel's Lemma (since $p$ is odd), $x^2 + y^2 + 1 = 0$ has a solution in $\mathbb{Q}_p$. Therefore, $S(\mathbb{Q}_p) \le 2$.

$S(\mathbb{Q}_p) = 1$ iff -1 is a square mod $p$ iff $p \equiv 1 \mod{4}$ (again by Hensel's Lemma).

So $S(\mathbb{Q}_p) = 2$ iff $p \equiv 3 \mod{4}$.

share|cite|improve this answer

Your Answer


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.