Isotropy over $p$-adic numbers

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

Solution 1:

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

http://www.amazon.com/Rational-Quadratic-Forms-Dover-Mathematics/dp/0486466701

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.

Solution 2:

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 $\mathbb Q_5.$