# Isotropy over $p$-adic numbers

Over what $p$-adic fields $\mathbb{Q}_p$ is the form $\langle3, 7, -15\rangle$ isotropic?

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Is this a binary form or a diagonal ternary? – Will Jagy Dec 7 '11 at 4:41
It's ternary instead of binary. – Eric Dec 7 '11 at 4:42
From your questions, you really ought to buy Cassels. It is an inexpensive Dover reprint, store.doverpublications.com/0486466701.html – Will Jagy Dec 7 '11 at 5:23

Everything is in Cassels, Rational Quadratic Forms, pages 41-44, then pages 55-59. In particular Lemma 2.5 on page 59. Your forms is isotropic at the primes $\{2,5,\infty\}.$ It is anisotropic at primes $\{3,7\}.$ And isotropic at everything else.

To give the simplest looking version, what happens if $$3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}?$$ Well, $$3 x^2 - 15 z^2 \equiv 0 \pmod {7},$$ $$x^2 - 5 z^2 \equiv 0 \pmod {7},$$ $$x^2 \equiv 5 z^2 \pmod {7}.$$ If we assume $z$ is not divisible by 7, it has a multiplicative inverse $\pmod 7,$ so $$\frac{ x^2}{z^2} \equiv 5 \pmod {7}$$ and $$\left(\frac{ x}{z}\right)^2 \equiv 5 \pmod {7}.$$ However, 5 is a quadratic nonresidue $\pmod 7,$ written $(5|7)= -1.$ So the equation is impossible, contradicting the assumption, and, in fact, $z$ is divisible by 7. From $x^2 \equiv 5 z^2 \pmod {7}$ we find that $x$ is also divisible by 7. So, actually, $$x^2 - 5 z^2 \equiv 0 \pmod {49}.$$ It follows from $3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}$ that $7 y^2 \equiv 0 \pmod {49}$ which tells us that $y^2 \equiv 0 \pmod {7}$ and $y \equiv 0 \pmod {7}.$ In sum, $3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}$ implies that $(x,y,z)$ are all divisible by 7. It follows that we can divide out 7 from each of the variables and 49 from whatever the result was. Or, we cannot express a number divisible by an arbitrarily high power of 7 unless $(x,y,z)$ are all divisible by 7 to roughly half that exponent.

Or, put another way, the only solution to $3 x^2 + 7 y^2 - 15 z^2 = 0$ in $\mathbb Q_7$ is $(x=0, y=0, z=0).$

It is a worthwhile exercise for you to confirm things about the prime 3, as above. Also plow through Lemma 2.5 for each prime.

Finally, from Lemma 2.4 on page 58, $7 y^2 - 15 z^2$ is isotropic in $\mathbb Q_2,$ and $3 x^2 + 7 y^2$ is isotropic in $\mathbb Q_5.$

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Great, thanks. I was wondeirng if you could explain a little bit how you got that though. I don't have access to Cassels, I'm using Lam's Introduction to Quadratic Forms Over Fields. I know that it's isotropic over primes not equal to 2 3 5 7 (and infinity). I'm just not sure how you check 2 3 5 and 7 like you did. – Eric Dec 7 '11 at 5:23

Given your questions, you absolutely must purchase the Cassels book, and quickly, otherwise you are completely screwed.

We find tables of the Hilbert Norm Residue Symbol $(a,b)_p$ on pages 43 and 44, one for $\mathbb Q_p$ with odd prime $p,$ a bigger table for $\mathbb Q_2,$ then a smaller one for $\mathbb Q_\infty.$
On page 55, for a quadratic form $$f(x_1, \ldots, x_n) = a_1 x_1^2 + \ldots + a_n x_n^2, \; \; a_j \in \mathbb Q_p^\ast,$$ he gives the definition $$c_p(f) = \prod_{i < j} (a_i, a_j)_p$$ which is his version of the Hasse-Minkowski Invariant.
On page 59, we have Lemma 2.5, which says a (nondegenerate) ternary is isotropic in $\mathbb Q_p$ if and only if $$c_p(f) = (-1, - \det(f))_p$$
Then we have the calculations. This is intricate and prone to error, but not conceptually difficult at this stage. $$\begin{array}{ccccccc} p & (3,7)_p & (3,-15)_p & (7,-15)_p & c_p & (-1,315)_p & \mbox{comment} \\ 2 & -1 & 1 & 1 & -1 & -1 & 315 \equiv 3 \pmod 8 \\ 3 & 1 & -1 & 1 & -1 & 1 & 35 = \frac{315}{9} \equiv -1 \pmod 3 \\ 5 & 1 & -1 & -1 & 1 & 1 & 63 = \frac{315}{5} \equiv 3 \pmod 5 \\ 7 & -1 & 1 & -1 & 1 & -1 & 45 = \frac{315}{7} \equiv 3 \pmod 7 \\ \infty & 1 & 1 & 1 & 1 & 1 & \mbox{indefinite} \end{array}$$
Some authors define their version of the Hasse-Minkowski Invariant as the quantity that Cassels would call $$c_p(f) \; \cdot \; (-1, - \det(f))_p,$$ in which case you get isotropy if and only if the number is 1.