5
$\begingroup$

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 $1$, \begin{equation}\tag{1}\label{eq:main} e(ab) - e(a) - e(b) + e(1) = \langle a, b\rangle, \end{equation} 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).

$\endgroup$

1 Answer 1

1
$\begingroup$

$\newcommand{\tr}{\operatorname{tr}}\newcommand{\Ell}{\operatorname{Ell}}$That was quicker to solve than I expected. Consider the trace map $$ \tr : GW(L) \to GW(K) $$ between the Grothendieck-Witt rings of $L$ and $K$, where if $q : V \to L$ is a quadratic form, $\tr q : V \to K$ is given by postcomposition with $\tr_{L/K}$, viewing $V$ as a vector space over $K$ by restriction of scalars. The problem reduces to showing that $\tr(\Ell_L) = \tr(\Ell_K)$, where $\Ell_K$ is the unique class in $GW(K)$ of dimension $0$, determinant $1$, and Hasse symbol $-1$. This latter identity can be proved easily by computing the trace of a couple of well-chosen forms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.