Finding zeros (or not) of quadratic forms over number fields

Q : Find a quadratic extension K of \mathbb{Q} where u^2-3v^2-3w^2+9z^2 has no nontrivial zero.

Non-solutions :

  1. Over \mathbb{Q}(\sqrt{2}), (0,\sqrt{2},1,1) is a solution.
  2. Over \mathbb{Q}(\sqrt{3}), (0,\sqrt{3},0,1) is a solution.

Solution : \mathbb{Q}(\sqrt{7})

Sketch of how to go about it :

  • Finding a zero of <1,-3,-3,9> is equivalent to finding a zero of <1,-3,-3> . In fact, more generally, finding zero of <1,-a,-b,ab> is equivalent to finding a zero of <1,-a,-b> over any field K. A nice explanation of this fact is via the theory of quaternions. Our form is a really nice form, it is the norm form of the quaternions called (a,b) which has generators : i, j and relations : i^2 = 3 =a , j^2 = 3 = b , ij=-ji= k.
  • A quaternion algebra is either a skew field (all elements have inverses) or it is M_2(K), the 2×2 matrices.
  • The norm is multiplicative, so it is a good test to check whether our quaternion is a skew field or not. The norm form has a nontrivial zero if and only if the quaternion is a matrix algebra.
  • Another if and only if test for whether the quaternion (a,b) is a matrix algebra over K is to check that the form <1,-a,-b> has a nontrivial zero.
  • So from now on, we can assume our form is <1,-3,-3>.
  • Hasse-Minkowski’s theorem : If q is a quadratic form over K, a number field, then q has a nontrivial zero if and only if it has a nontrivial zero over K_v, all the completions of K.
  • For example, \mathbb{Q}_p is the field of p-adic numbers, which is the completion of Q at prime p. In fact, here we only need the trivial direction of this theorem, namely the fact that if there is a rational solution, there is a local solution.
  1.  So we try to find a K=\mathbb{Q}(\sqrt{d}) and a completion K_v such that our form q=<1,-3,-3> has no solution over K_v.
  2. In fact, let us try to imitate the proof of why this form has no solution over \mathbb{Q} : If it does, it should have a nontrivial solution over \mathbb{Q}_3 (the 3 adics), so by clearing denominators, the solution will be in \mathbb{Z}_3 (the 3-adic integers).
  3. Let a^2 - 3b^2 - 3c^2 =0 where a,b,c are in \mathbb{Z}_3 and not all 0 simultaneously. Since \mathbb{Z}_3 is a DVR with parameter 3, this says that 3|a, so divide by 3, the whole equation to get 3x^2 - b^2 - c^2 =0 . Continue this process till you can assume there is some u,v,w\in \mathbb{Z}_3 such that 3u^2 - v^2 - w^2 =0 \in \mathbb{Z}_3and 3 does not divide one of v and w.
  4. F_3 is the field of 3 elements which is also \mathbb{Z}_3 modulo its maximal ideal (3), so the previous solution gives a nontrivial solution v^2+w^2=0 \in F_3, which is a contradiction (you can check manually in F_3)
  5. So if we find a K= Q(\sqrt{d}) and ring of integers O_K and a prime P of O_K such that the residue field of K_P (completion of K at P) is F_3, and 3 is not ramified at P, the same explanation as above shows that u^2 - 3v^2 - 3 w^2 has no solution over K.
  6. For K = Q(\sqrt{d}) where d is 2 or 3 mod 4, O_K = Z[\sqrt{d}], and I think we can directly see how any prime p in \mathbb{Z} splits in K [look at factorization of x^2-d in F_p[x]]
  7. For Q(\sqrt{7}) , 3 = PQ where P and Q are distinct primes in O_K (because x^2-7 = x^2 - 1 = (x+1)(x-1) \in F_3[x] In fact P=(3,\sqrt{7}+1), Q=(3,\sqrt{7}-1)). So 3 does not get ramified in K_P. Also we see that residue field of K_P = \mathbb{Z}[\sqrt{d}]/P = F_3
  8. In fact, there is a formula to check whether an odd prime p is ramified, split or inert in \mathbb{Q}(\sqrt{d}) depending on whether d is a square modulo p or not.
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