Skew fields inside quaternion division algebras

Question

Suppose that $Q$ is a quaternion division algebra with center $k$, where $k$ is an arbitrary commutative field (let's say with $\operatorname{char}(k) \neq 2$ if necessary). Assume that $D$ is an arbitrary skew field (which a priori has nothing to do with $Q$ nor with the base field $k$), and assume that there is an injective ring homomorphism $\varphi \colon D \hookrightarrow Q$.

Is it true that $D$ is either a commutative field or a quaternion division algebra again?

My first guess was that this should be obviously true, but failing to see an obvious argument, I wonder whether it's true at all...

Answer 1

First note that if $A$ is the $2\times 2$ matrix algebra over a field, and $z\in A$ has trace zero, then by the Cayley Hamilton Theorem, $z^2=-det (z)$ is a scalar. Hence, if $x,y \in A$ then $(xy-yx)^2$ is a scalar matrix. Secondly, if $A$ is the $n\times n$ matrix algebra over a field, with $n\geq 3$, then it is easy to get two matrices $x,y\in A$ such that $(xy-yx)^2$ is $not$ a scalar.

Suppose $D$ is a skew field, and is finite dimensional (of degree $n$) over its centre $K$. We may assume that $D$ is not commutative. Then the centre cannot be a finite field and hence $D$ is Zariski dense in $D\otimes _K {\overline K}\quad $ (${\overline K}$ is the algebraic closure of $K$). If $degree (D)\geq 3$, then by the preceding paragraph, $(xy-yx)^2$ is not in $K$ for some $x,y\in D$.

Your condition says that for all $x,y\in Q$ we must have $(xy-yx)^2$ lies in the centre of $Q$. Hence the same is true for $D$. Hence $D$ is indeed quaternionic.

[Edit] Tom is right. You do not need to assume that $D$ is finite dimensional over its centre. This can be proved as follows. Let $K$ and $L$ be the centres of $D$ and $Q$ respectively. Since $Q$ is quaternionic, the $L$ vector space spanned by $D$ in $Q$ is (an algebra) and is therefore all of $Q$. Hence $K\subset L$. I will now prove that the trace of $b\in D$ and the norm of $b$ (all viewed in $Q$) lie in $K$ itself. If $b\in K$ this is clear.

If $b\notin K$, then there exists $a\in D$ which does not commute with $b$. The equation $$ab^2-b^2a= trace (b)(ab-ba)$$ follows from Cayley Hamilton in $Q$ and shows that $trace (b)$ is an element of $D$. Hence the norm also: $det (b)=trace (b)b-b^2$.

If $a,b\in D$ and don't commute, then it follows that the $K$ vector space spanned by $1,a,b,ab,ba, aba, bab$ is a sub-algebra $R$ and is hence quaternionic. This can be extended to any three generic elements elements $a,b,c$ as well. Hence every generic element $c\in D$ already lies in the subalgebra $R$. That is: $D=R$ is finite dimensional over $K$.

Answer 2

Yes, it is. This follows from the theory of central simple algebras. Here's another proof, possibly similar to Aakumadula's: let $D'=D\otimes \bar{k}\cong GL_2(\bar{k})$. If $D'$ has basis $1,a$ then it's commutative; otherwise, it has basis $1,a,b$. I claim that $a,b$ have a common eigenvalue in $\bar{k}^2$. Indeed, suppose not. Then $a,b$ are diagonalizable. Pick a basis in which $a$ is diagonal. Then $b$ won't be either upper- or lower-triangular, and you can see combinatorially that $1,a,b,ab$ are linearly independent, so $D'=Q'$ and $D=Q$. Thus $D'$ is the space of upper-triangular matrices in some basis. But this is impossible, e.g. since then $D'$ has an ideal fixed by the Galois action.