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Problem of the Week #197 - March 8, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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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.
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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
No one answered this week's problem. You can read my solution below.


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, $D = R$.
 
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