I'm stuck with this thing:
Let $F$ be a field of characteristic $\neq 2$, let $D$ be a 4 dimensional noncommutative division algebra over $F$. For $x\in D\smallsetminus F, F[x]$ is a field of degree 2 over $F$, as $D$ is not commutative.
Why does this hold? I can show that $F[x]$ is a field only by knowing that $\dim_F F[x]=2$. We have the generalization for division rings of the multiplicative formula for the degree of a field extension (here). But in order to be able to use it (by saying that $n\mid 4, n\neq 1,4 \Rightarrow n=2$), I need to know in advance that $F[x]$ is a division ring (and I don't know that, since I haven't shown yet that $F[x]^{-1} \subseteq F[x]$.
It looks like a vicious circle. Thank you in advance!
EDIT: The question arose by reading the proof of Prop. 1.2 (bottom of the first page here).
The subring $F[x]$ of $D$ generated by $F$ and $x$ is commutative, because $F$ is contained in the center of $D$.
Since $D$ is finite-dimensional over $F$, also $F[x]$ is. Let $y\in F[x]$, $y\ne0$; in particular $y$ satisfies a polynomial of minimal degree over $F$, say $a_0+a_1X+\dots+a_nX^n$. Then $a_0\ne0$, otherwise the degree would not be minimal.
Since $D$ is a division ring, $y^{-1}\in D$; therefore $$ a_0y^{-1}+a_1+a_2y+\dots+a_ny^{n-1}=0 $$ and $a_0\ne0$ implies $y^{-1}\in F[x]$.
Hence $F[x]$ is a field.