# non-division quaternion algebra is isomorphic to $2\times 2$ matrices

Let $k$ be a field of characteristic $\neq2$. Let $a,b\in k$ be nonzero elements. Let $A:=\left(\frac{a,b}{k}\right)$ be the quaternion algebra over $k$ with parameters $a,b$. Suppose $A$ is not a division algebra. Prove that $A$ is isomorphic to the matrix algebra $M_2(k)$ over $k$.

How to get a proof without using a lot of theorems?

Let's use $F$ for the field instead of $k$ so that I can use $k$ for an element of the quaternion algebra.
The quaternion algebra $A$ is spanned over $F$ by $\{1,i,j,k\}$ with relations $i^2 = a$, $j^2 = b$, $k=ij=-ji$. We know that $A$ has a conjugation operator $x \mapsto \overline{x}$ which negates the $i,j,k$ components, and that $N(x) = x\overline{x}$ is an element of $F$. Therefore, if $N(x) \ne 0$, $x$ is invertible: The inverse of $x$ is $\overline{x}/N(x)$. So if $A$ is not a division algebra, then there must exist an $x \ne 0$ such that $N(x) = 0$. Writing $x = x_0 + x_1 i + x_2 j + x_3 k$, this implies that the following equation has a nonzero solution in $F$: $$x_0^2 - a x_1^2 - b x_2^2 + ab x_3^2 = 0.\,\,\,\,\,\,\,(1)$$
Now let's try to figure what the isomorphism between $A$ and $M_2(F)$ would look like. Since $i^2 = a$, the matrix corresponding to $i$ has minimal polynomial $X^2 - a$. (Because $a \ne 0$ and char $F \ne 2$.) So the trace of this matrix is 0 and determinant is $-a$, and without loss of generality, we may assume that the isomorphism sends $$i \mapsto \pmatrix{0&1\\a&0}.$$ Similarly, $j^2 = b$ implies that the image of $j$ has trace 0 and determinant $-b$, so $$j \mapsto \pmatrix{x&y\\z&-x} \text{where x^2 + yz = b.}$$ We also have the relation $ij + ji = 0$. By calculating the image of $ij+ji$ using the matrices above, we conclude that $ay + z = 0$ and hence the equation we must solve is $$x^2 - ay^2 = b.\,\,\,\,\,\,\,(2)$$ So we have a solution to (1), and we are looking to solve (2). This is where I get stuck. The equations are similar, but not quite similar enough to produce a solution to (2) from a solution to (1). If we could find a solution to (1) where $x_2 = 1$ and $x_3 = 0$, for example, we would be done. (We still have to check that the map is bijective, but this should be the easy part.)