# Relation in Brauer group coming from trace form

Let $$L/K$$ be a cubic (or, more generally, odd-order) extension of fields of characteristic $$0$$. To every element $$a \in L^\times$$ we can associate the quadratic form \begin{align*} q_a : L &\to K \\ x &\to \operatorname{tr}_{L/K}(ax^2) \end{align*} and then take its class in the $$2$$-Brauer group $$\operatorname{Br}(K)[2]$$ (using that Severi-Brauer varieties of order $$2$$ are conics). This class does not change if $$a$$ is scaled by an element of $$K$$ or a square in $$L$$, and so we get a map $$e : L^\times / ((L^\times)^2 \cdot K^\times) \to \operatorname{Br}(K)[2],$$ Thus $$e(a) = 0$$ exactly when $$\operatorname{tr}_{L/K}(ax^2) = 0$$ for some nonzero $$x$$. Alternatively, we can view $$e$$ as a map $$e : H^1(K, E[2]) \to H^2(K, \mu_2)$$ where $$E$$ is any elliptic curve over $$K$$ whose $$2$$-torsion has field of definition $$L$$.

The map $$e$$ is not in general a group homomorphism. Numerical evidence suggests that $$e$$ is a constant plus a quadratic form whose associated bilinear form is the Hilbert symbol: that is, for all $$a,b \in L^\times$$ of norm $$1$$, $$$$\tag{1}\label{eq:main} e(ab) - e(a) - e(b) + e(1) = \langle a, b\rangle,$$$$ where the Hilbert symbol $$\langle a, b\rangle$$ is the class in the $$2$$-Brauer group of the conic $$a x^2 + b y^2 = z^2.$$ When $$L/K$$ is Galois, I can prove \eqref{eq:main} by diagonalizing $$q_a$$ after a base change to $$L$$. But in the non-Galois case, I'm stuck with a nasty relation between the Hilbert symbols over $$L$$ and a quadratic extension. If it helps, I'm principally interested in the case when $$K \supseteq \mathbb{Q}_2$$ is a local field (and hence $$\operatorname{Br}(K)[2]$$ has just two elements).
