Isomorphic finite fields of a skew field

Question

Let $D$ be a skew field and $F$ and $E$ be isomorphic finite subfields of $D$, is it true that $F=E$?

Answer 1

No, this is not true. Suppose that $D$ is a skew field containing a finite field $F$ and $F$ is not contained in the center of $D$. Take $d \in D$ which does not commute with $F$. If $dFd^{-1}$ is distinct from $F$, then it is a subfield of $D$ isomorphic to $F$, and we're done. Otherwise, we have that $dFd^{-1} = F$, which means that conjugation by $d$ acts as a non-trivial automorphism of $F$.

Let $f \in F$ be a generator of the multiplicative group and then $dfd^{-1} = f^n$, where $f^n \neq f$. Now consider $(1+d)F(1+d)^{-1}$, which will also be a subfield of $D$. Assume for contradiction that $(1+d)F(1+d)^{-1} = F$. Then, that means $(1+d)f(1+d)^{-1} = f^m$ for some $m$. Rearranging, we get: $$f^m + f^m d = f + df = f + f^nd$$ and so: $$f-f^m = (f^m-f^n)d$$ If $f^n \neq f^m$, then $d$ is an element of $F$, which is a contradiction. On the other hand, if $f^n = f^m$, then $f = f^m$, and so $f = f^n$, which is also a contradiction. So we conclude that $(1+d)F(1+d)^{-1}$ is a subfield of $D$ isomorphic to $F$ but not equal to it.

It only remains to show that there actually is division ring $D$ with a finite subfield not contained in the center. The example I know of is to take the skew polynomial ring $\mathbb F_{p^k}[x; F]$ where $k \geq 2$ and $x$ acts by Frobenius, meaning that $x f = f^p x$ for $f \in \mathbb F_{p^k}$. Then, $\mathbb F_{p^k}[x; F]$ satisfies the Ore condition, so that it can be extended to a division ring. This gives example for $F$ isomorphic to any finite field other than a prime field, which is always unique.

Now you could ask whether any two finite subfields of a division ring which are isomorphic are always conjugate, and that I don't know.

Answer 2

The following answer is nearly identical to @DustinCartwright's, which I saw after posting had been posted a few minutes earlier. I only construct a specific division algebra with a non-central finite subfield differently.

No. Let $\theta$ be a non-square in $\mathbb F_3$, put $k = \mathbb F_3(X)$ (i.e., rational functions in one variable over $\mathbb F_3$), and consider the quaternion algebra $D \mathrel{:=} (\theta, X)_k$ (in the usual notation), so that there are elements $i, j \in D$ with $i^2 = \theta$, $j^2 = X$, and $i j = -j i$. Then $D$ is a division algebra because the quadratic form $(a, b, c, d) \mapsto a^2 - b^2\theta - c^2 X - d^2\theta X$ on $k^4$ does not non-trivially represent $0$.

We have that $D$ contains a finite field $F \mathrel{:=} \mathbb F_3(i)$ with 9 elements. One computes directly that the centralizer $C_D(F)$ equals $k[i] = F(X)$. Since $N_{D^\times}(F)$ contains $j \notin C_{D^\times}(F)$, and since $N_{D^\times}(F)/C_{D^\times}(F)$ injects into the order-$2$ group $\operatorname{Gal}(F/\mathbb F_3)$, we have that $N_{D^\times}(F)$ equals $\langle j\rangle\cdot k[i]^\times$, which is a proper subgroup of $D^\times$. For any element $g$ of $D^\times \setminus \langle j\rangle\cdot k[i]^\times$, we have that $E \mathrel{:=} g F g^{-1}$ is a subfield of $D$ that is isomorphic, but not equal, to $F$.