Q : Find a quadratic extension of where has no nontrivial zero.
- Over , is a solution.
- Over , is a solution.
Sketch of how to go about it :
- Finding a zero of is equivalent to finding a zero of . In fact, more generally, finding zero of is equivalent to finding a zero of over any field . 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 which has generators : and relations : .
- A quaternion algebra is either a skew field (all elements have inverses) or it is , 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 is a matrix algebra over is to check that the form has a nontrivial zero.
- So from now on, we can assume our form is .
- 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 , all the completions of K.
- For example, is the field of -adic numbers, which is the completion of at prime . 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.
- So we try to find a and a completion such that our form has no solution over .
- In fact, let us try to imitate the proof of why this form has no solution over : If it does, it should have a nontrivial solution over (the 3 adics), so by clearing denominators, the solution will be in (the 3-adic integers).
- Let where are in and not all 0 simultaneously. Since is a DVR with parameter 3, this says that , so divide by 3, the whole equation to get . Continue this process till you can assume there is some such that and 3 does not divide one of v and w.
- is the field of 3 elements which is also modulo its maximal ideal , so the previous solution gives a nontrivial solution , which is a contradiction (you can check manually in )
- So if we find a and ring of integers and a prime of such that the residue field of (completion of K at P) is , and 3 is not ramified at P, the same explanation as above shows that has no solution over K.
- For where d is 2 or 3 mod 4, , and I think we can directly see how any prime p in splits in K [look at factorization of in ]
- For , where P and Q are distinct primes in (because In fact ). So 3 does not get ramified in . Also we see that residue field of
- In fact, there is a formula to check whether an odd prime is ramified, split or inert in depending on whether is a square modulo or not.