All maximal subfields of a Division Algebra are isomorphic.

Question

How can I show that two maximal subfields of a division-algebra have the same dimension over k. I found a simple proof here on page 4:http://www.math.northwestern.edu/~len/d70/chap17.pdf. Is there a simpler way to show this with even less theory needed?

And is there a very simple way to prove it only for finite division-algebras.

I would like to prove Wedderburns Theorem with this Lemma and the Skolem-Noether Theorem, that I already proved .

There is no page $4$ in the text you have linked. On page $160$ it is shown that $[D:k]=[L:k]^2$. So where is this result you are asking about? — Dietrich Burde, Jan 19 '18 at 15:39
Does your definition of a maximal subfield mean that it has ot be proper? — rschwieb, Jan 19 '18 at 15:40
No it not necesarilly proper. Sorry I mean page 160. Sorry I made a mistake they are only isomorphic if D is finite. If D is finite, theory of finite fields shows every two maximal subfields are isomorphic. — Maximilian Böhmermann, Jan 19 '18 at 15:55
Re: the new version. The double centralizer theorem may be a good starting point. Or may be it is more complicated than the proof you linked to? — Jyrki Lahtonen, Jan 19 '18 at 16:26
I think its more complicated. But im looking for something simpler then the proove I linked. — Maximilian Böhmermann, Jan 20 '18 at 11:07

Jyrki Lahtonen · Answer 1 · 2018-07-10T06:34:18.277

The original version of the question asked to prove that maximal subfields are isomorphic. This is wrong, see below.

Many counterexamples can be found inside the division algebra of quaternions with rational coefficients.

Let $$ D=\{a+bi+cj+dk\mid a,b,c,d\in\Bbb{Q}\} $$ with the usual relations $i^2=j^2=k^2=-1=ijk$. Because $\dim_{\Bbb{Q}}D=4$ we know that all the maximal subfields are quadratic extensions of $\Bbb{Q}$. Conversely, any field $L, \Bbb{Q}\subset L\subset D$, such that $[L:\Bbb{Q}]=2$ is maximal.

But then we have several non-isomorphic maximal subfields.

Because $i^2=-1$ we see that $D$ has a copy of $\Bbb{Q}(\sqrt{-1})$ as a subfield.
Because $u=i+j+k$ satisfies $u^2=-3$ we also get a copy $\Bbb{Q}(u)$ of $\Bbb{Q}(\sqrt{-3})$.
In fact, if $m=n_1^2+n_2^2+n_3^2$ we have $(n_1i+n_2j+n_3k)^2=-m$, and consequently a maximal subfield isomorphic to $\Bbb{Q}(\sqrt{-m})$. By the celebrated three squares theorem any squarefree number $m\not\equiv7\pmod8$ can be written as a sum of three squares.
But $\Bbb{Q}(\sqrt{-n_1})$ and $\Bbb{Q}(\sqrt{-n_2})$ are non-isomorphic unless $n_1/n_2$ is a square of a rational number.

There is no shortage of non-isomorphic maximal subfields of $D$.

A "standard" proof for the fact that all the maximal subfields have the samed dimensiongoes by applying the double centralizer theorem. More precisely, the extension degree of any maximal subfield equals the square root of the dimension of the division algebra over its center. After all, any subfield is contained in its centralizer. And if the containment is proper then the field can be extended by adjoining any element from its centralizer.

Yes you`re right. Im interested in the result that there dimension over k to the power of two of the Dimension of D over k. — Maximilian Böhmermann, Jan 19 '18 at 16:00
Sorry for that mistake and thank you for the counterexample. I changed the question. — Maximilian Böhmermann, Jan 19 '18 at 16:01
@Jyrki, so do these maximal subfields have some property in common? Say, do they have isomorphic galois group over the base field? — , Jul 08 '18 at 14:51
@Qixiao: Some people know quite a bit about this in the case where the center is a number field. Class field theory gives results constraining the possibilities. — Jyrki Lahtonen, Jul 08 '18 at 15:12
@Qixiao For example the algebra of rational quaternions has its non-trivial Hasse invariants at $\infty$ and $2$. Meaning that a maximal subfield has to A) be quadratic (double centralizer), B) imaginary (behavior at $\infty$) and C) the prime $2$ must either ramify or be inert. — Jyrki Lahtonen, Jul 08 '18 at 15:16

All maximal subfields of a Division Algebra are isomorphic.

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