Crossed product division algebra

Question

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 $k$

Answer 1

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.

Answer 2

Perhaps the simplest example is a quaternion division algebra over a field of characteristic two. Let $F$ be a field of characteristic $2$. For $a\in F$ and $b\in F\backslash\{0\}$, let $Q=[a,b)$ be the quaternion algebra generated by a basis $\{1,i,j,k\}$ over $F$ with the relations $$i^2+i=a\in F, \ j^2=b\in F^\times, \ ij+ji=j,\ ij=k.$$ Then $Q$ has a Galois maximal subfield, namely $L=F(a)$ and a purely inseparable maximal subfield namely $L=F(b)$.