1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

This isn't a complete solution but it may be possible to finish it.

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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