4
$\begingroup$

Suppose that $F$ is a field. Show that there exists a $F$-division algebra $D$ with two elements $a\neq b\in D$ such that $a^2-2ab+b^2=0$.

In the field extensions we know that $a^2-2ab+b^2=0$ if and only if $a=b$, because of $a^2-2ab+b^2=(a-b)^2$. But I know this is not true if the extension is division ring, but I can't prove it.

In a part of my research I want two elements like that in order to extend $F$ to $F(a)$ such that there exists $b\notin F(a)$ such that we know its minimal polynomial $p(x)=x^2-2ax+a^2$.

It is a little part of my research and I don't have any idea to prove or simplify this question. About 3 years ago one of my professors told me this is true but now, he isn't in touch.

$\endgroup$
  • $\begingroup$ Presumably your field needs to have at least one quadratic extension? $\endgroup$ – Lubin Jun 14 '16 at 13:07
  • $\begingroup$ What is the definition of quadratic extension? $\endgroup$ – MH.Fakharan Jun 14 '16 at 13:15
  • $\begingroup$ What, precisely, is the question you are asking in this post? For a finite field $F$, since the Brauer group of $F$ is trivial, every $F$-division algebra $D$ with $\text{dim}_F(D)$ finite has no such elements $a$ and $b$. Are you allowing us to consider divison algebras whose center is an infinite extension of $F$, e.g., $F(s,t)$ for transcendentals $s$ and $t$? In that case, the quaternion algebra generated by $s$ and $t$ might work. $\endgroup$ – Jason Starr Jun 14 '16 at 13:42
  • $\begingroup$ Quadratic extension meaning a degree $2$ extension (a field extension that is $2$-dimensional over the given field). $\endgroup$ – Todd Trimble Jun 14 '16 at 13:45
  • $\begingroup$ Polynomials such as $p(x)=x^2-2ax+a^2$ are called left polynomials. If $b$ is a root of $p(x)$, then $(x-b)$ is a factor, see Gordon and Motzkin, On the zeros of polynomials over division rings, Trans. Amer. Math. Soc 116 (1965), 218--226. $\endgroup$ – Glasby Jul 28 '16 at 2:51
12
$\begingroup$

I assume $\text{char}\,\mathbf F=0$.

Put $d:=b-a$. Because of $a^2-2ab+b^2=d^2-ad+da$ your equation is equivalent to $$ (*)\qquad d^{-1}a-ad^{-1}=1. $$ This precludes $\dim_{\mathbf F}D<\infty$ (take the reduced trace on both sides).

On the other side, $(*)$ is is the relation defining of the Weyl algebra $A_1(\mathbf F)=\mathbf F\langle x,\partial_x\rangle$ and it is well known that $A_1$ has a skew field of fractions $D$. Then $$ a=x,b=x+\partial_x^{-1}\in D $$ is a solution to your problem. Another would be $$ a=\partial_x, b=\partial_x-x^{-1}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.