I am reading chapter 1 of Gille and Szamuely's book Central Simple Algebras and Galois Cohomology, on quaternion algebras. In it they prove (remark 1.3.1 on p. 18) that the conic associated to a quaternion algebra is an intrinsic invariant of the algebra, i.e. does not depend (at least up to isomorphism of the conic) on the basis chosen for the algebra. The argument almost convinces me, but not quite (see below). I am wondering if the ingredient that I need to convince me is to cast the argument in scheme-theoretic language. Thus, my first question is, can you help me formulate the argument in a way that speaks to my discomfort with it? My second question is, can we use schemes to do this?
Background: Let $k$ be a field of characteristic $\neq 2$. The quaternion algebra
$$A = \binom{a,b}{k}$$
for $a,b\in k^\times$ is the four-dimensional associative $k$-algebra with basis $1,i,j,ij$ and (noncommutative) multiplication defined by $i^2=a, j^2=b, ij = -ji$; $1$ is identified with the unit of $k$. (Thus the usual Hamilton quaternions are the case $a=b=-1, k=\mathbb{R}$.) A change of basis is elements $1,i',j',i'j'$ in $A$ satisfying $i'^2=a', j'^2=b', i'j'=-j'i'$ for some (possibly different) $a',b'\in k^\times$. The subspace spanned by $i,j,ij$ does not depend on the basis since a calculation shows that this is exactly the set of elements of $A\setminus k$ whose square is $\in k$. This is called the subspace of pure quaternions.
The conic associated to $A$ is the conic in $\mathbb{P}^2_k$ defined by the form $ax^2+by^2-z^2$. Up to a simple substitution, this is the same as the conic defined by $ax^2+by^2-abz^2$.
$A$ is said to be split if it is isomorphic to $M_2(k)$. For example, if $a=1$, $A$ is split, because it can be mapped isomorphically onto $M_2(k)$ by
$$i\mapsto \begin{pmatrix}1& \\ &-1\end{pmatrix},\; j\mapsto \begin{pmatrix} &b\\ 1& \end{pmatrix}$$
A basic theorem says that $A$ is split if and only if the associated conic has a $k$-rational point.
The argument: Gille and Szamuely argue that changing basis in a quaternion algebra, which may change $a,b$, nonetheless does not change the isomorphism class of the associated conic. Their argument is essentially that the form $ax^2 + by^2 - abz^2$ which defines the conic is just given by squaring the pure quaternion $xi + yj + zij$. Since the space of pure quaternions is basis independent, so is this form, and thus so is the conic.
My discomfort with it: I like this argument, and I buy it in essence, but a detail troubles me. Namely, when I try to actually write down a basis-independent description of the conic, I come up empty (literally):
Okay, so the conic "is" the vanishing set in $\mathbb{P}_k^2$ of the form $ax^2+by^2 - abz^2$. Great, I can conceive of $\mathbb{P}_k^2$ as being the quotient of the space of pure quaternions by the scalar $k^\times$ action. Then the conic is the vanishing set of the form $\alpha^2$, where $\alpha$ is a pure quaternion.
The problem is this. In all interesting cases, i.e. whenever $A$ is not split, the conic I've just described is, as a set, empty. (Because the conic does not have $k$-rational points when $A$ is not split.) This empty set has lost the ability to distinguish between nonsplit nonisomorphic quaternion algebras, so it is not doing the job I came to it for.
This objection does not feel very substantive. The conic is not really the set of its $k$-rational points. (This seems like just the type of distinction that scheme theory is good at, hence the title of the question.)
My question: Clearly I need a more robust basis-independent description of the conic. I'd like it to be the $\operatorname{Proj}$ of $k[x,y,z]/(ax^2+by^2-abz^2)$, but this description is not basis-independent. Your thoughts?
Thanks in advance.