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 would like to know whether $q=\langle 3,3,11\rangle$ (a diagonal ternary form) represents $2$ over $\mathbb{Q}_2$ (i.e. whether there exist $x,y,z\in\mathbb{Q}_2^\times$ such that $q(x,y,z)=2$). I have computed the Hasse invariant for $q$; it's $-1$, and I have computed the Hilbert symbol $(-1,-\mathrm{disc}\;q)_2=1$, so $q$ is anisotropic; no help there.

I'm now out of ideas. Anyone know what to do?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

First solve $q(x,y,z) = 2$ mod 8, then use Hensel's Lemma to lift to a solution in $\mathbb{Z}_2$.

When working with quadratic forms in characteristic 2, usually it helps to work mod 8. A good reason for this is because a unit in $\mathbb{Z}_2^{\times}$ is a square iff it is $\equiv 1$ (mod 8).

share|improve this answer
    
$q(1,1,2)=2\mod 8$, but I thought Hensel's lemma didn't apply for dyadic fields? –  Nick Dec 9 '11 at 7:44
2  
The "easy" form of Hensel's Lemma doesn't apply, but in a more general form, it still applies (which is why you need to work mod 8 and not just mod 2). For any prime $p$, if $f(x) \in \mathbb{Z}_p[x]$ is a monic polynomial and $a \in \mathbb{Z}_p$ is such that $v(f(a)) > 2 v(f'(a))$ (here $v$ denotes $p$-adic valuation), then $a$ lifts to a root of $f$ in $\mathbb{Z}_p$. By the "easy" form I mean the case $v(f'(a)) = 0$. But when $p=2$ and you are working with quadratic forms, you have $v(f'(a)) = 1$, so you need $v(f(a)) \ge 3$ so you need to work mod 2^3=8. –  Ted Dec 9 '11 at 7:51

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.