# 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 $3u^2 - v^2 - w^2 =0 \in \mathbb{Z}_3$and 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.