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$f(x,y,z)=2x^2+3y^2-6z^2$.

Determine whether the form has a non-trivial zero. (You do not need to exhibit it).

I know that I need to use the Hasse-Minkowski Theorem somehow. From examples I've seen online, I can split finding roots into a couple of cases.

I know that if I can find a root in each case, then there is a non-trivial zero in $\mathbb Q$. I also know that if there is a case in which I can't find a non-trivial zero, then there is no non-trivial zero in $\mathbb Q$.

Case 1: $\mathbb Q_p,p\nmid 2,3$ I used Hensel's Lemma and found a solution here.

Case 2: $\mathbb Q_2$ I'm stuck here. I know basically to let $x_0=0$ but I can't figure out (guess?) what $z_0$ should be. I set $g(y)=2x_0^2+3y^2-6z_0^2$ and so $g'(y)=6y$. I want to have $|g(y_0)|_p<|g'(y_0)|^2_p$, where $y_0$ is a root, so that I can use Hensel's Lemma again.

Case 3: $\mathbb Q_3$ (I haven't got round to this yet.)

Firstly, is this the correct method to approach this question? Secondly, is there some nice way to determine what $z_0$ should be or is it just a typical trial and error to see which $z_0$ works?

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  • $\begingroup$ Why do you "know" you can take $x_0 = 0$? There are no non-trivial $2$-adic solutions to $3y^2 - 6z^2 = 0$, since $2$ is not a square in $\mathbb Q_2$. $\endgroup$ – Erick Wong Apr 7 '13 at 0:06
  • $\begingroup$ I just realised that too. So I'm supposed to "trial-and-error" guess what values $x_0$ and $z_0$ should be? Or is there some nice method to get them out such that my argument can carry on? $\endgroup$ – Haikal Yeo Apr 7 '13 at 8:46
  • $\begingroup$ There are systematic ways to compute Hilbert symbols, but there really aren't that many cases to check. WLOG $x_0$ is either $0$ or $1$, since you can divide through by any unit. $\endgroup$ – Erick Wong Apr 7 '13 at 14:45
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OR: if there is a root in $\mathbb Q$ there is a root in $\mathbb Z.$ In particular, there is a root in $\mathbb Z$ with $\gcd(x,y,z) = 1.$ ASSUME that.

As $2x^2 = -3 y^2 + 6 z^2,$ we know $3 | 2 x^2,$ so $3 | x$ and $9 | 2 x^2.$

Since $9 | 3 y^2 - 6 z^2,$ next $3 | y^2 - 2 z^2. $ But Legendre symbol $(2|3)=-1.$ So, actually $3|y,z,$ and $3 | \gcd(x,y,z).$

Detail: Let $y^2 - 2 z^2 \equiv 0 \pmod 3.$ Then $y^2 \equiv 2 z^2 \pmod 3.$ Assume $z \not\equiv 0 \pmod 3,$ then it has a multiplicative inverse $\pmod 3.$ And $$ \left( \frac{y}{z} \right)^2 \equiv 2 \pmod 3, $$ which contradicts the Legendre symbol. So, actually $z$ is divisible by $3,$ and so is $y.$

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  • $\begingroup$ Thanks for the solution. Did you use Hasse-Minkowski Theorem implicitly or is that too much to solve this problem? I copied my idea of page 19,20 and 21 of math.umass.edu/~hatley/Capstone.pdf $\endgroup$ – Haikal Yeo Apr 7 '13 at 7:23
  • $\begingroup$ No Hasse-Minkowski here, just working modulo $3$. $\endgroup$ – Gerry Myerson Apr 7 '13 at 9:24
  • $\begingroup$ I'm also trying to do the same for $g(x,y,z)=2x^2+3y^2-10z^2$ and $h(x,y,z)=x^2+y^2-64z^2$. Will the method provided in this solution suffice for these or even in general for that matter? $\endgroup$ – Haikal Yeo Apr 7 '13 at 11:28
  • $\begingroup$ Why do you say that if there is a root in $\mathbb Q$, then there is a root in $\mathbb Z$? Surely that's not true in general? Are you referring to a specific case related to this question? $\endgroup$ – Haikal Yeo Apr 7 '13 at 11:50
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    $\begingroup$ @HaikalYeo Of course $\mathbb Q \to \mathbb Z$ is true for general homogeneous equations. $\endgroup$ – Erick Wong Apr 7 '13 at 14:37

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