Division ring on a field

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.

• Presumably your field needs to have at least one quadratic extension? – Lubin Jun 14 '16 at 13:07
• What is the definition of quadratic extension? – MH.Fakharan Jun 14 '16 at 13:15
• 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. – Jason Starr Jun 14 '16 at 13:42
• Quadratic extension meaning a degree $2$ extension (a field extension that is $2$-dimensional over the given field). – Todd Trimble Jun 14 '16 at 13:45
• 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. – Glasby Jul 28 '16 at 2:51

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 $$a=\partial_x, b=\partial_x-x^{-1}.$$