The quadratic equation $x^2=c$ in a division ring Let $D$ be a division ring. We denote $D'$ by the derived subgroup of the multiplicative group $D\setminus\{0\}$, that is, the subgroup generated by all the commutators of $D\setminus\{0\}$. For $c\in D',$ does there exist an element $a$ of $D$ such that $a^2=c$? In other words, is the quadratic equation $x^2=c$ solvable for each $c\in D'$?
It is true for $D=\mathbb{H},$ which is the real quaternion division ring. Thanks for all your support.
 A: Let us analyze the problem in a particular case.
Assume that $D$ has prime degree $p$ over its center $F$. Then it is known that $D'$ is the set of elements of reduced norm $1$.
Assume that $c$ has reduced norm $1$, and $c\notin F$. Since $F[c]$ is a commutative subfield of $D$, its degree over $F$ divides $p$. It cannot be $1$ since $c\notin F$. Hence $F[c]/F$ has degree $p$, and $[F[c]:F]=p$. Assume that there exists $a\in D$ such that $a^2=c$. Once again, $F[a]/F$ has degree $p$ (otherwise $c=a^2\in F$ !). Since $F[c]\subset F[a]$, we get $F[a]=F[c]$. Hence $a\in F[c]$.
So we have to solve the equation $a^2=c$ in $F[c]$. Since $F[c]/F$ is a MAXIMAL commutative subfield of $D$, $Nrd_D(x)=N_{F[c]/F}(x)$ for all $x\in F[c]$.
Hence, we need to solve the equation $a^2=c$ in the field $F[c]$, where $N_{F[c]/F}(c)=1$. However, this equation does not have always solutions.
We can now construct explicit counter examples. Take $D=(d,-1)_\mathbb{Q}$, where $d$ is not a sum of two squares (for example $d$ is a prime congruent to $3$ mod $4$, or negative), so $D$ is division. Let $j\in D$ the basis element such that $j^2=-1$.
Then $Nrd_D(j)=-j^2=1$, so $j\in D'$. Now the above analysis shows that, if $a^2=j$, then $a\in \mathbb{Q}[j]$.
Write $a=x+yj,x,y\in \mathbb{Q}$. Then $x^2-y^2+2xyj=j$, so $y=\pm x$ and $2xy=1$. We then get $\pm 2x^2=1\in\mathbb{Q}$, which has no solution.
Of course, you can avoid the general analysis if you only work with this particular example, and I am pretty sure you can prove by hand that $j\in D'$, but I think it is still interesting to know how counterexamples are found...
