MHB Can Division Rings Be Generated by Products of Perfect Squares?

[Euge]

Here is this week's POTW:

-----
Let $D$ be a division ring. Show that if $D$ is not simultaneously of characteristic two and commutative, then $D$ is generated by products of perfect squares.
-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

Let $R$ be the subring of $D$ generated by products of squares of $D$. If $D$ has odd characteristic, then for every $x\in D$, $x = \left(\frac{x+1}{2}\right)^2 - \left(\frac{x-1}{2}\right)^2 \in R$. Hence $D = R$ if $D$ has odd characteristic.

If, on the other hand, $D$ is noncommutative with characteristic 2, then there exists $d\in D$ such that $d^2$ is non-central. So there exists $x\in D$ such that $d^2x \neq xd^2$. The element $a := d^2x + xd^2$ is nonzero, hence invertible in $D$. Now given $y\in D$,

$$ay = d^2xy + xd^2y = d^2xy + d^2yx + d^2yx + xd^2y = d^2(xy + yx) + (d^2y)x + x(d^2y) = d^2[(x+y)^2 - x^2 - y^2] + (d^2y + x)^2 - (d^2y)^2 - x^2\in R$$

Therefore $y = a^{-1}(ay)\in R$. Since $y$ was arbitary, $D = R$.