Let $N \in \mathbb{Z}$ be a prime number, and let $B = \left( \dfrac{a, b}{\mathbb{Q}} \right)$ be the unique quaternion algebra over $\mathbb{Q}$ ramified at $N$ and at $\infty$. Then, in section 3 of his paper "Heights and the Special Values of L-series", Gross constructs a genus 0 curve $Y$ over $\mathbb{Q}$ associated to the quaternion algebra $B$ as follows.


For any $\mathbb{Q}$-algebra $E$, the points of $Y$ in $E$ are given by


$$
Y(E) = \{ \alpha \in B \otimes\_{\mathbb{Q}} E \mid \operatorname{Tr}{(\alpha)} = 0 = \operatorname{N}(\alpha) \} / E^{\times}
$$


where $\operatorname{Tr}(\cdot)$ and $\operatorname{N}(\cdot)$ are the reduced trace and the reduced norm in the quaternion algebra $B$, given as follows. For $h = x + iy + jz + ijw \in B$, with $x, y, z, w \in \mathbb{Q}$, 


\begin{align}
\operatorname{Tr}(h) &= h + \overline{h} = 2x\\
\operatorname{N}(h) &= h\overline{h} = x^2 - ay^2 - bz^2 + abw^2
\end{align}


Now, my confusion is that then I'm not sure about what it means to take the reduced trace and norm of an element $\alpha \in B \otimes\_{\mathbb{Q}} E$.


Finally, I have also seen in another paper describing the same construction, that then combining the two equations $\operatorname{Tr}{(\alpha)} = 0 = \operatorname{N}(\alpha)$, $Y$ is basically the conic $ay^2 + bz^2 = abw^2$.


**Questions**


1. How should I interpret taking the trace $\operatorname{Tr}{(\alpha)}$ and norm $\operatorname{N}{(\alpha)}$ for $\alpha \in B \otimes\_{\mathbb{Q}} E$ ?
2. According to the last observation about $Y$ being the conic $ay^2 + bz^2 = abw^2$, does that mean that $Y(E)$ can be thought of as $\{ (y, z, w) \in E^3 \mid ay^2 + bz^2 = abw^2 \}$?


Thank you very much for any help.



Question about the definition of the genus 0 curves in Gross' paper "Heights and the Special values of L-series"
$B\otimes E$ is an $E$-algebra, and an element $\alpha$ induces a linear map $B\otimes E\to B\otimes E$ given by $x \mapsto \alpha x$ and you can think of trace and norm of $\alpha$ as respectively the trace and determinant of this linear transformation. Since you are just extending scalars, you will get the same formulas you had before. That should answer 1.


For 2., you forgot to mod out by $E^{\times}$ so the actual result is the set of points in the projective plane satisfying the equation.


