4
$\begingroup$

I am currently reading some introductory material on Brauer groups ("Noncommutative Algebra", by Farb and Dennis) and the following two questions came to my mind:

1) Are all crossed products algebras, division algebras?

2) Are all division algebras, crossed product algebras?

In chapter 4 the authors state that the answer to question 2 is affirmative in the case of central division algebras over algebraic number fields. I would appreciate an answer to those two questions or maybe a counterexample (if available that is) or even a reference. I should mention that crossed products in the context of Dennis and Farb are associative algebras. Thanks!

$\endgroup$
  • $\begingroup$ Not all crossed product algebras are division algebras. For example a cyclic algebra $(E/F,\gamma)$ with $E/F$ cyclic Galois of order $n$, is a division algebra iff the element $\gamma$ giving the normalized factor set has the property that $n$ is the smallest positive integer $\ell$ with the property that $\gamma^\ell$ is in the image of the norm map $N_{E/F}$. Therefore with a suitable choice of $\gamma$ such a crossed product algebra (here a cyclic one) will not be a division algebra. $\endgroup$ – Jyrki Lahtonen Feb 2 '16 at 20:38
  • $\begingroup$ If your division algebra has a maximal subfield that is a Galois extension of the center, then the algebra is a crossed product (using the Galois group), see for example Jacobson's Basic Algebra II, chapter 8. I recall having read somewhere that when the center is a number field, then the existence of a maximal subfield that is also a cyclic extension of the center is roughly equivalent to the main results of global class field theory. I'm afraid I'm not at all conversant with what goes on there. $\endgroup$ – Jyrki Lahtonen Feb 2 '16 at 20:45
  • $\begingroup$ Thank you for your comments. Jacobson's algebra II contains as an exercise (in chapter 8) the "if" part of the result you mention in your first comment. I was wondering if the "only if" part in your comment is a typo or if indeed is correct and if it is how one could go about proving it. $\endgroup$ – Econm Feb 5 '16 at 11:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.