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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 dimension $n$) 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 $dim (D)\geq 3$, 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 $K$.

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 $K$.

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.

Suppose $D$ is a skew field, and is finite dimensional (of dimension $n$) 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 $K$). Then if $dim (D)\geq 3$, then $(xy-yx)^2$ 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 $K$.

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 dimension $n$) 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 $dim (D)\geq 3$, 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 $K$.

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.

Suppose $D$ is a skew field, and is finite dimensional (of dimension $n$) 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 $K$). Then if $dim (D)\geq 3$, then $(xy-yx)^2$ 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 $1,a,b,ab,ba$ 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 $K$.

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.

Suppose $D$ is a skew field, and is finite dimensional (of dimension $n$) 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 $K$). Then if $dim (D)\geq 3$, then $(xy-yx)^2$ 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 $K$.