Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be some finite group and $J_G$ be $ZG$ modulo the ideal $\{n\sum_{g} g | n\in\mathbb N\}$, or equivalently the dual of the augmentation ideal which is the kernel of $ZG\to Z$ sending $g$ to $1$. For a field $F$ we set $F[J_G]$ to be the group algebra of $J_G$ so that $F[J_G]=F[x_g]/\{\prod x_g = 1\}$. This algebra has an induced $G$ action defined by $h(x_g)=x_{hg}$ and fixing $F$. Finally, we set $F(J_G)$ to be the fraction field of $F[J_G]$ with the induced $G$ action.

The question that I'm interested in is whether the invariant field $(F(J_G))^G$ is a rational extension of $F$. The only case I know to be true is for cyclic groups and $F$ contains a primitive $|G|$-th root of unity (M. Hajja - On the rationality of monomial automorphism). Is this result true in other cases, and in particular for abelian groups?

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.