number theoryquadratic-forms
Let $k$ be a global field and $q$ a ternary quadratic form over $k$.
Is it true that $q_v$ is isotropic for all but finitely many places $v$ of $k$?
I'm looking for a confirmation of this, and if possible a reference.
I know that this is true for $k = \mathbb Q$, because a ternary quadratic form is isotropic over $\mathbb Q_p$ iff its Hasse invariant equals the Hilbert symbol $(-1, -d)_p$, where $d$ is the discriminant of $q$.
Edit: I have a proof now, except when $char(k) = 2$.
Since this is only about $p$-adics, an elementary argument that comes to mind would be as follows (unless I totally missed something).
If $p\equiv 1\pmod4$, then $-1=a^2$ for some $a\in\mathbf{Z}_p^*$. Therefore $x^2$ and $(ax)^2=-x^2$ are equivalent.
If $p\equiv -1\pmod4$, then $-1=a^2+b^2$ for some integers $a,b\in\mathbf{Z}_p^*$. In this case $x^2+y^2$ is equivalent to $$ (ax+by)^2+(bx-ay)^2=(a^2+b^2)(x^2+y^2)=-(x^2+y^2). $$ Two or four such equivalences convert your $f$ to $g$.
And let's not forget the prime $p=2$. Taking norms of the product of quaternions $$ (2+i+j+k)(x_1+x_2i+x_3j+x_4k) $$ tells us that $$ \begin{aligned} 7(x_1^2+x_2^2+x_3^2+x_4^2)&=(2x_1-x_2-x_3-x_4)^2+(x_1+2x_2-x_3+x_4)^2\\ &+(x_1+x_2+2x_3-x_4)^2+(x_1-x_2+x_3+2x_4)^2. \end{aligned} $$ Combining this with the fact that $\sqrt{-7}\in\mathbf{Q}_2$ allows you to turn the sum of four squares to its negative by an equivalence transformation.
When $K$ is a field not of characteristic $2$, being able to solve $x^2 + y^2 + z^2 = 0$ nontrivially in $K$ is equivalent to being able to solve $x^2 + y^2 = -1$ in $K$, so I will think of the problem from that point of view.
I have checked explicitly that this can be done when $K$ is a quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$, as you say you did, and it can also be done when $K = \mathbf Q_2(\sqrt{2})$.
I'll list all seven quadratic extensions of $\mathbf Q_2$ in the order you did: $$ \mathbf Q_2(\sqrt{-1}), \ \mathbf Q_2(\sqrt{-3}), \ \mathbf Q_2(\sqrt{-5}), \ \mathbf Q_2(\sqrt{-2}), \ \mathbf Q_2(\sqrt{-6}), \ \mathbf Q_2(\sqrt{-10}), \ \mathbf Q_2(\sqrt{2}). $$ In the four fields $\mathbf Q_2(\sqrt{-1})$, $\mathbf Q_2(\sqrt{-5})$, $\mathbf Q_2(\sqrt{-2})$, and $\mathbf Q_2(\sqrt{-10}),$ it is very easy to see a solution to $x^2 + y^2 = -1$: $$ 0^2 + \sqrt{-1}^2, \ 2^2 + \sqrt{-5}^2, \ 1^2 + \sqrt{-2}^2, \ 3^2 + \sqrt{-10}^2. $$
In $\mathbf Q_2(\sqrt{-3})$ we have a cube root of unity $\zeta_3 = (-1 + \sqrt{-3})/2$, for which $1 + \zeta_3 + \zeta_3^2 = 0$, so $$ -1 = \zeta_3 + \zeta_3^2 = \zeta_3^4 + \zeta_3^2. $$
In $\mathbf Q_2(\sqrt{-6})$, after playing around a bit I got $$ \left(\frac{2+\sqrt{-6}}{2}\right)^2 + \left(\frac{2-\sqrt{-6}}{2}\right)^2 = -1. $$
That leaves us with $\mathbf Q_2(\sqrt{2})$, whose ring of integers is $\mathbf Z_2[\sqrt{2}]$. I am not going to record here all the numerical work I did (by hand), but just report the final result: we can write $$ -1 = x^2 + (\sqrt{2} + 2)^2 $$ for some $x \in \mathbf Z_2[\sqrt{2}]$ where $x \equiv 1 \bmod \sqrt{2}^3$. This follows from Hensel's lemma on $\mathbf Q_2(\sqrt{2})$ for the polynomial $f(X) = X^2 + 1 + (\sqrt{2}+2)^2 = X^2 + 7 + 4\sqrt{2}$: $$ f(1) = 8 + 4\sqrt{2} \Longrightarrow |f(1)|_2 = \frac{1}{4\sqrt{2}} $$ and $$ f'(1) = 2 \Longrightarrow |f'(1)|_2^2 = \frac{1}{4}. $$ Since $|f(1)|_2 < |f'(1)|_2^2$, Hensel's lemma tells us there is a unique solution to $f(x) = 0$ where $x \in \mathbf Z_2[\sqrt{2}]$ and $|x - 1|_2 < |f'(1)|_2 = 1/2$, or equivalently $x \equiv 1 \bmod (\sqrt{2}^3)$. A more precise form of Hensel's lemma tells us that $|x - 1|_2 = |f(1)/f'(1)|_2 = (1/4\sqrt{2})/(1/2) = 1/2\sqrt{2} = 1/\sqrt{2}^3$, so $x \not\equiv 1 \bmod \sqrt{2}^4$.
We can compute $x$ using the binomial series for the square root: $$ x = (-7 - 4\sqrt{2})^{1/2} = (1 - 4\sqrt{2} - 8)^{1/2} $$ so $$ x = \sum_{n \geq 0} \binom{1/2}{n}(-4\sqrt{2}-8)^n = 1 + \sum_{n \geq 1} \binom{1/2}{n}(-1)^n(4\sqrt{2}+8)^n $$ Since ${\rm ord}_2\binom{1/2}{n} = -2n + s_2(n)$, where $s_2(n)$ is the sum of the base $2$ digits of $n$, and ${\rm ord}_2((4\sqrt{2}+8)^n) = 5n/2$, the $n$th term of the series has $2$-adic valuation $5n/2 - 2n + s_2(n) = n/2 + s_2(n)$, so its $\pi$-adic valuation ($\pi = \sqrt{2}$ over the $2$-adics) is twice that: $n + 2s_2(n)$. Since $n+2s_2(n) \geq 7$ for all $n \geq 1$ other than $n = 1$, $2$, and $4$, $$ x \equiv 1 + \sum_{n = 1, 2, 4} \binom{1/2}{n}(-1)^n(4\sqrt{2}+8)^n \bmod \pi^7\mathbf Z_2[\pi]. $$ Working this out modulo $\pi^7$, $$ x \equiv 1 + \pi^3 + \pi^5 + \pi^6 \bmod \pi^7\mathbf Z_2[\pi]. $$ In "normal" notation, with $\pi = \sqrt{2}$, this approximate root of $f(X)$ is $9+6\sqrt{2} \bmod 8\sqrt{2}\mathbf Z_2[\sqrt{2}]$. As a check, $$ (9 + 6\sqrt{2})^2 + (7 + 4\sqrt{2}) = 160 + 112\sqrt{2} = 16\sqrt{2}(5\sqrt{2} + 7), $$ which is a multiple of $\pi^7 = 8\sqrt{2}$.
In what way did your work suggest that there might not be a solution to $x^2 + y^2 = -1$ in $\mathbf Q_2(\sqrt{2})$?
There is a high-brow explanation for why we found a solution in the particular quadratic extension $\mathbf Q_2(\sqrt{2})$. Being able to solve $x^2 + y^2 = -1$ in a local field $F$ (not of characteristic $2$) is the same as saying the Hilbert symbol $(-1,-1)_F$ is $1$ rather than $-1$. Letting $F = \mathbf Q(\sqrt{2})_v$ be a completion of $\mathbf Q(\sqrt{2})$, Hilbert reciprocity says the number of $v$ such that $(-1,-1)_{\mathbf Q_2(\sqrt{2})_v}$ is $-1$ is even. When $v$ is a place of odd residue field characteristic, $\mathbf Q(\sqrt{2})_v$ is a local field that is an extension of $\mathbf Q_p$ for an odd prime $p$, so that Hilbert symbol is $1$ since we can already solve $x^2 + y^2 = -1$ in $\mathbf Q_p$ (and actually in $\mathbf Z_p$). That leaves us with just three remaining places $v$ on $\mathbf Q(\sqrt{2})$: the two real places (corresponding to the two different embeddings of $\mathbf Q(\sqrt{2})$ into $\mathbf R$) and the unique place over $2$, corresponding to the prime $(\sqrt{2})$ in $\mathbf Z[\sqrt{2}]$. We obviously can't solve $x^2 + y^2 = -1$ in $\mathbf R$, so the Hilbert symbol $(-1,-1)_{\mathbf Q(\sqrt{2})_v}$ is $-1$ when $v$ is a real place. That leaves us with just one more place, over $2$, and since the Hilbert symbol is $-1$ at an even number of places it must be $1$ at the place over $2$.
By similar reasoning, if $K$ is an arbitrary real quadratic field, we can describe the non-archimedean places $v$ for which $x^2 + y^2 = -1$ has a solution in $K_v$:
(1) If $v$ lies over an odd prime then there is a solution in $K_v$ since there already is one in $\mathbf Q_p$ for all odd $p$, and $K_v$ contains some $\mathbf Q_p$.
(2) If $v$ lies over $2$ and $2$ does not split completely in $K$, then there is just one $2$-adic place on $K$. We can solve $x^2 + y^2 = -1$ in $K_v$ by Hilbert reciprocity because we can solve it in every other completion except the two real completions, and two is even but three is not.
(3) If $v$ lies over $2$ and $2$ splits completely in $K$ then we can't solve $x^2 + y^2 = -1$ in $K_v$ since $K_v \cong \mathbf Q_2$.
If $K$ is an arbitrary imaginary quadratic field, we can also determine the non-archimedean $v$ for which $x^2 + y^2 = -1$ has a solution in $K_v$:
(1) If $v$ lies over an odd prime then there is a solution in $K_v$ for the same reason as in the real quadratic case above.
(2) If $v$ lies over $2$ and $2$ does not split completely in $K$, then there is exactly one $2$-adic place on $K$. There is also exactly one archimedean place, with completion $\mathbf C$. We can solve $x^2 + y^2 = -1$ in $\mathbf C$, so Hilbert reciprocity implies we can solve $x^2 + y^2 = -1$ in $K_v$. In other words, we can solve $x^2 + y^2 = -1$ in every completion of $K$, so by Hasse-Minkowski we can solve $x^2 + y^2 = -1$ in $K$.
(3) If $v$ lies over $2$ and $2$ splits completely in $K$ then we can't solve $x^2 + y^2 = -1$ in $K_v$ since $K_v \cong \mathbf Q_2$.
The second item of each list for real and imaginary quadratic fields explains why we found solutions in every quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$ without needing to work in the completion itself (i.e., a solution in $\mathbf Q(i)$, $\mathbf Q(\sqrt{-3})$, and so on): you wrote every quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$ as a $2$-adic completion of an imaginary quadratic field, and in the imaginary quadratic fields that you chose, you can check that $2$ does not split completely (sometimes $2$ ramifies, sometimes $2$ is inert, but $2$ never splits completely). In the case of $\mathbf Q_2(\sqrt{2})$, we can write it as $\mathbf Q_2(\sqrt{-14})$, a completion of $\mathbf Q(\sqrt{-14})$ at its unique $2$-adic place. Then results above say we have to be to solve $x^2 + y^2 = -1$ in $\mathbf Q(\sqrt{-14})$, and in fact a solution is $x = (2+3\sqrt{-14})/10$ and $y = (6-\sqrt{-14})/10$.
More generally, if $\mathbf Q_2(\sqrt{2}) = \mathbf Q_2(\sqrt{d})$ for some integer $d$ then ${\rm ord}_2(d) = 1$, so $\mathbf Q(\sqrt{d})$ ramifies at $2$ and thus there is a unique $2$-adic place on $\mathbf Q(\sqrt{d})$, so for $d < 0$ we'd be in the same situation as $\mathbf Q(\sqrt{-14})$: there will be a solution to $x^2 + y^2 = -1$ in $\mathbf Q(\sqrt{d})$.
Best Answer
From the comments: For $k$ of characteristic not $2$, one can bring $q$ in the form $ax^2+by^2+cz^2$. When $v(2) = 0$, which is the case for almost all $v$, by Hensel's lemma any nonzero isotropic vector (automatically nonsingular) mod a prime element $\pi_v$ of the valuation ring $\mathcal O_{k_v}$ lifts to $k_v$.
For almost all $v$ one has $a, b, c \in O_{k_v}^\times$. Thus it suffices to show that a nondegenerate quadratic form $ax^2+by^2+cz^2$ over a finite field $F$ (of characteristic $\neq 2$) is isotropic, i.e. that $ax^2+by^2$ represents $-c$. Let $|F| = N$, so that $|F^2| = \frac{N+1}2$. The sets $aF^2$ and $-c-bF^2$ intersect, and we are done.
I don't know about $char(k) = 2$.