An example about a non-commutative division ring with finite characteristic

Question

After reading the proof of the theorem

“For every central division $F$-algebra $D$ with $D$ $\neq$ $F$, $D$ contains a separable extension $K \supsetneqq F$“,

I have a question: dose there exist a non-commutative division ring $D$ with finite characteristic, which contains an $x\in D$ such that $x^n \not\in Z(D)$ for all positive integers n?

First I tried quaternion algebra over field $F_p(X)$.

Let $i^2= x$, $j^2= y$, $ij=k$, $ji=-k$ where $x, y$ are any non-zero elements in $F_p(X)$, the quaternion algebra $H(F_p(X))$ is a vector space over $F_p(X)$ with a basis $\{1,i ,j, k\}$.

Let $α =a+ib+jc+kd$, $\bar{α}=a-ib-jc-kd$, $α\bar{α}= \bar{α}α= a^2-xb^2-yc^2+xyd^2$, then $H(F_p(X))$ is a division ring if and only if $α\bar{α}= 0$ implies $α= 0$. After trying some examples, I find that the consequent processes are hard. Does there exist some good example that satisfies the property in the question?

Answer 1

All you need to do is to produce an Ore domain which isn't commutative and has finite characteristic.

In fact there's an easy way to produce left principal ideal domains (which are necessarily left Ore). The construction is that of the skew-polynomial ring $k[X;\sigma]$, where $\sigma$ is an automorphism of $k$ that isn't the identity. Addition is plain addition of polynomials, but multiplication is governed by the rule $Xa=\sigma(a)X$. When the automorphism isn't the identity, the ring is not commutative.

To achieve this we'll select $k$ to be $F_4$, the field of four elements, and use the Frobenius automorphism $a\mapsto a^2$ (which isn't the identity on $F_4$.)

So now you have $F_4\langle X;\sigma\rangle$, a not-commutative left Ore domain. By Ore's theorem, the classical left ring of fractions exists, and is a division ring that is not commutative, and it's easy to see that it has characteristic $2$.

With regards to the way you were going: there are quaternion-like constructions that do produce division rings from finite fields. You have elements $i,j,k$ such that $i^2=a$, $j^2=b$ and $ij=-ji=k$.

It is possible for this to be a division ring. When the characteristic of $F$ is not $2$, and $-1$ is not the sum of two squares in $F$, then the quaternion construction I described is a division ring (See Corollary 4.23 of K. Conrad's notes.

But also note Corollary 4.24: every quaternion algebra over $F_p$ is isomorphic to $M_2(F_p)$ (hence not a division ring.) So the only hope would be to use more complex fields of finite characteristic.