In the book Algebra IX: Finite Groups of Lie type and Finite Dimensional Algebra, the authors Kostrikin-Shafarevich mention (p. 159) that

If $A$ is a central simple algebra over $F$ of finite dimension and if $K$ is a maximal subfield of $A$, then $K$ may not be Galois extension of $F$.

On the other hand, in a paper on Central Simple Algebras, Rowen mentiones

Every division algebra of degree $2, 3, 4, 6$, or $12$ is a crossed product (i.e. maximal subfields are Galois extensions; am I right? Here degree of $D$ over $F$ is $n$ means $\dim_F(D)=n^2$.)

Q. What is example of a non-commutative simple (or division) algebra of finite dimension over $F$ in which a maximal subfield is not Galois extension of $F$?

(I have no idea of how difficult is this problem in the literature, and I have never seen simple algebras other than quaternions and matrix algebras over quaternions or fields.)

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    $\begingroup$ In anothe question you asked for a 9-dimensional division algebra, central over $\Bbb{Q}$ containing a copy of the non-Galois extension field $\Bbb{Q}(\root3\of2)$. Is that not an answer to this question as well :-) $\endgroup$ – Jyrki Lahtonen Jan 22 at 20:11
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    $\begingroup$ Agree with Jyrki; also note that this gives a counterexample to your claim in the second highlighted paragraph. In a crossed product, there exists a maximal subfield which is Galois; but that does not mean that every maximal subfield is Galois. $\endgroup$ – Torsten Schoeneberg Jan 23 at 18:33
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    $\begingroup$ Further: while maximal subfields of a central division algebra over a given field $K$ may still be somewhat restricted, I claim that in the central simple $K$-algebra $M_n(K)$, you will find a copy of any (!) field extension $L\vert K$ with degree $[L:K] \le n$ (this should be a standard exercise in linear algebra); further, I am quite sure that those with dimension $n$ are maximal, although I don't see an elementary argument for that right now. $\endgroup$ – Torsten Schoeneberg Jan 23 at 19:00
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    $\begingroup$ @JyrkiLahtonen: When I wrote the comment, I was actually keeping Grothendieck's Theorem about conjugacy of maximal tori (here, in $GL_n$) up my sleeve, hoping I would not be put on the spot to even formulate it. See, they use it for a more general question here: mathoverflow.net/q/251953/27465. Compared to that, the double centraliser theorem is indeed elementary, and I doubt one can get more elementary than that! -- Beginner: About the (weak but complicated) restrictions in the division algebra case I mentioned, compare math.stackexchange.com/q/178999/96384. $\endgroup$ – Torsten Schoeneberg Jan 30 at 17:55
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    $\begingroup$ @Torsten I am still hoping that there is something even simpler than the double centralizer. Like using the fact that a commuting set of diagonalizable matrices is simultaneously diagonalizable. But, to make the matrices representing the field diagonalizable I need to go to an algebraic closure and... $\endgroup$ – Jyrki Lahtonen Jan 30 at 19:43

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