I am working through problems in Number Theory text by Borevich, Shafarevich and this problem has quite stumped me. Over what $\mathbb{Q_{p}}$, the quadratic form $f(x) = 2x^2-15y^2+14z^2 $ does not represent $0$? I have a few ideas on how to go on about this.. I will be using representation theorem from Serge's A course in Arithmetic which states that a non degenerate quadratic form in n variables over $\mathbb{Q_{p}}$ represents $0$ if and only if:.

  1. $n=1$ always,
  2. $n=2$ when discriminant is $-1$,
  3. $n=3$ when $(-1,-d) = \epsilon $.
    Here, $d$ denotes the discriminant of the quadratic form which is essentially the product of coefficients when it is in diagonal form and $\epsilon$ denotes it's hasse's invariant.

My strategy is to break it into three cases. Case 1 when $p$ is $\infty$ , the second case when $p=2$ and the third case when it is an odd prime. In the first case, by the theorem above f will represent $0$ and in the second case, since discriminant is $0$, it is degenerate quadratic form (now what??).
And I am lost on how to deal with the third csse since I have so many primes of different form to take care of in order to use the above theorem and calculate d and $\epsilon$. Should I further take two cases, when $p$ is 1 and 3 modulo 4?

It would be great if someone could suggest me how to proceed on this problem or suggest some other way to think about this. Any kind of help is appreciated. Thanks!

  • $\begingroup$ I apologise for any spelling mistakes, typing it on mobile is quite difficult! $\endgroup$ – Mojojojo Apr 21 '18 at 12:35
  • $\begingroup$ math.stackexchange.com/questions/738446/… $\endgroup$ – individ Apr 21 '18 at 13:01
  • $\begingroup$ I don't understand how that answers my question! Could you explain? $\endgroup$ – Mojojojo Apr 21 '18 at 16:43
  • $\begingroup$ Formula there is. Sets the solvability of the equation. Do you want to find out the solvability of the equation without solving it? $\endgroup$ – individ Apr 21 '18 at 17:04
  • 2
    $\begingroup$ $ax^2+by^2+cz^2$ always represents zero in $\mathbf{Q}_p$ when $p \nmid 2abc \infty$. $\endgroup$ – Infinity Apr 21 '18 at 20:30

I take up your calculation in case 3. For simplification, omit the index $p$ in the Hilbert symbol $(.,.)$. Modulo possible computational errors, your quadratic form $g=X^2 -\frac{15}{2}Y^2+7Z^2$ has dicriminant $d=-\frac{105}{2}$ and invariant $\epsilon=(-\frac{15}{2},7)$, and $g$ represents $0$ in $\mathbf Q_p$ iff $(-1,-d)=\epsilon$, or $1=(-1,-\frac{105}{2})(-\frac{15}{2},7)=(-\frac{15}{2},7)$(because the symbol is bilinear). NB: this means that $-\frac{15}{2}$ is a norm from $\mathbf Q_p (\sqrt 7)$, and could have been seen directly on the expression of $g$ (Cassels-Fröhlich, exercise 4.3).

Let us go on computing the Hilbert symbol. If $p$ is an odd prime, the symbol is tame : writing $a=p^{\alpha}a', b=p^{\beta}b'$, where $a', b'$ are $p$-adic units, we have the explicit formula $(a,b)=(-1)^{\frac{p-1}{2}\alpha\beta}(\frac{b'}{p})^{\beta}(\frac{a'}{p})^{\alpha}$ (Serre, "Local Fields", chap.XIV, §4). If $p>7$, obviously $(-\frac{15}{2},7)=1$, which simply means that in a non ramified extension, every unit is a norm. Taking for example $p=3$, we'll get $(-\frac{15}{2},7)=(\frac {-1}{3})\neq 1$, etc. I leave to you the other cases.

If $p=2$, there is an analogous but more complicated expression for the wild symbol (op. cit.).


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