Let $k$ be a field, $D$ be a crossed product division algebra over $k$, namely $D$ has a maximal subfield which is Galois over $k$. Is it possible $D$ contain some other maximal subfield which is non-Galois over $k$



Crossed product division algebra
Take a prime number $p>2$ and a field $k$ of characteristic zero which does not contain a $p$-th root of unity. Assume that $K/k$ is a Galois extension of order $p$ with a Galois group $\langle g\rangle$ and that $D = K\*\_{\alpha}\langle g \rangle$ where $[\alpha]\in H^2(\langle g\rangle,K^{\times})$ is non-trivial. Then by dimension consideration $D$ is a division algebra. The element $U\_g$ generates a subfield of dimension $p$, which is not a Galois extension of $k$, because $k$ does not contain a $p$-th root of unity.


As a concrete example, take $p=3$, $k=\mathbb{Q}[x\_1,x\_2,x\_3]^{\tau}$ where $\tau=(123)$ permutes the elements $x\_1,x\_2,x\_3$. Take $K=\mathbb{Q}[x\_1,x\_2,x\_3]$ and take the crossed product in which $U\_g^3 = x\_1x\_2^2 + x\_2x\_3^2+x\_3x\_1^2$.


Another example in case $p=3$, where third roots of unity are present arises from this paper by Haile:
<http://www.ams.org/journals/proc/1989-106-02/S0002-9939-1989-0972232-0/S0002-9939-1989-0972232-0.pdf>


Assume that $D$ is any division algebra of dimension $3^2$ containing a non-Galois extension.
Then by the above paper you can find an element $x\in D$ such that $x^3=a\in k$. 
Then $k(x)/k$ is a Galois extension, and you can describe $D$ as a crossed product of $k(x)$. But then $D$ contains a non-Galois extension. 


