# Skew fields inside quaternion division algebras

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...

• Just to be sure about your notation, do you use "division algebra" and "skew field" interchangeably? So, the $2\times 2$ matrices are neither a skew field nor a division algebra, right? – Filippo Alberto Edoardo Feb 4 '13 at 16:19
• @Filippo: I use "division algebra" for "skew field which is finite-dimensional over its center". – Tom De Medts Feb 5 '13 at 9:02

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.

 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 $K$.

• What do you mean by "Zariski dense"? Are you speaking of the Zariski density of $D$ or of $\mathrm{Spec}(D)$, which is a bit weird as $D$ is not commutative? – Filippo Alberto Edoardo Feb 5 '13 at 0:30
• the space $V=D\otimes {\overline K}$ is a vector space and hence an affine space. If $K$ is infinite, then $D$ is Zariski dense in the affine space $V$. Let $x,y,z\in V$ be arbitrary and let $w=(xy-yx)^2$. Then the condition $wz=zw$ for all $x,y,z \in V$ is a collection of polynomial equations on $V$; if this does not hold on $V$, then by Zariski density, it does not hold on $D$ either. – Venkataramana Feb 5 '13 at 1:22
• @Aakumadula: Thanks for your answer! The fact that $(xy - yx)^2$ is a central element is a nice observation. Two questions/comments though: (1) I assume you mean $\deg(D) \geq 3$ instead of $\dim(D) \geq 3$ (where the degree is the square root of the dimension)? (2) I am not assuming $D$ to be finite-dimensional over its center, and the Zariski density argument seems to fail in the infinite-dimensional case. – Tom De Medts Feb 5 '13 at 9:15
• @Tom De Medts: you are right. I should have said $Deg (D)\geq 3$. As to the other observation, I think (I have not thought it through), that you can find $x,y \in D$ of trace zero such that they do not commute, and since $(xy-yx)^2$ is central, the algebra generated by the centre $K$ of $D$ and $x,y$ is finite dimensional over $K$ because if you see $x^2$ in a (multiplicative) word in $x,y$ you may replace it a scalar; if you see yxyx in the word, it can be replaced by $xyxy$ and a sum of lower degree terms... . You can then argue that this algebra is 4 dimensional (i.e. quaternionic). – Venkataramana Feb 5 '13 at 10:26
• I am a bit lost. The OP seemed to ask for very general $D\hookrightarrow Q$ and you assume that $D$ be finite dimensional over its center $K$ (which, as the OP said in his answer to a comment, is his definition of division algebra...) – Filippo Alberto Edoardo Feb 5 '13 at 11:59

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.

• Your $GL$ should be $M_2$, the matrix ring. – Mariano Suárez-Álvarez Feb 5 '13 at 0:19
• Your tensor product seems to be over $k$, but the OP did insist that $D$ "has nothing to do with $k$", so it might not be a $k$-algebra... – Filippo Alberto Edoardo Feb 5 '13 at 0:31
• @Dmitry and Filippo: Indeed, I am not assuming that $D$ is a $k$-algebra (otherwise the result is rather obvious for dimension reasons). – Tom De Medts Feb 5 '13 at 9:00