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).

  • $\begingroup$ What is the meaning of the notation $F[x]$? $\endgroup$ – Lorenzo Jun 7 '17 at 15:26
  • $\begingroup$ It is the smallest subring of $D$ containing both $F$ and $x$, i.e. $F[x]=\{f(x)\mid f \in F[X]\}$, where $F[X]$ is the polynomial ring in one variable. $\endgroup$ – nbayo Jun 7 '17 at 15:31
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    $\begingroup$ So in other words, there is a surjection from $F[X]$ to $F[x]$. $F[X]$ is still a PID, so the kernel is of the form $(g)$. You would need to check that $g$ is irreducible, so then $F[X]/g(x) \cong F[x]$ is a field. Maybe the proof is the same as in the case when $D$ is a field? $\endgroup$ – Lorenzo Jun 7 '17 at 15:43
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    $\begingroup$ (What happens if $g$ is not irreducible?) $\endgroup$ – Lorenzo Jun 7 '17 at 15:48
  • $\begingroup$ I think I got it! $F[x]$ is an integral domain, this is equivalent to "$(g)$ is a prime ideal", which is equivalent (since $(g)\neq (0)$ and $F[X]$ is a PID) to "$(g)$ is a maximal ideal", which is equivalent to "$F[x]$ is a field". Thank you! $\endgroup$ – nbayo Jun 7 '17 at 16:01

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.

  • $\begingroup$ Thank you! This makes much sense. I think the argumentation in the comments of the question holds too, doesn't it? $\endgroup$ – nbayo Jun 7 '17 at 16:13
  • $\begingroup$ @nbayo Yes, the argument is sound $\endgroup$ – egreg Jun 7 '17 at 16:48
  • $\begingroup$ Yup. This argument shows up often enough. $x$ commutes with its powers due to associativity and with the center by definition. $\endgroup$ – Jyrki Lahtonen Jun 12 '17 at 15:16

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