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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?

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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).

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  • $\begingroup$ $q(1,1,2)=2\mod 8$, but I thought Hensel's lemma didn't apply for dyadic fields? $\endgroup$ – Nick Dec 9 '11 at 7:44
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    $\begingroup$ 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. $\endgroup$ – Ted Dec 9 '11 at 7:51

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