I was wondering if there is a way to prove that the multiplicative of a finite field is cyclic by looking at the character table of such a group. In particular, I was wondering if there is a way to avoid direct mention of classification of finite abelian groups (I am aware of the standard proofs from root counting of polynomials etc.).

My initial thoughts were as follows:

Suppose we somehow manage to find the character table of $\mathbb{F}_q^{\times}$. We know this group is abelian so compare it to the character tables of other abelian groups.

Over $\mathbb{C}$ the character table of a finite abelian group uniquely distinguishes the group, so if we were to find the character table for $\mathbb{F}_q^{\times}$ we would be done.

However the finding of the character table seems like the hard thing to do- we must have to use the field structure somehow. So instead maybe we should look for representations over characteristic $p$ rather than $\mathbb{C}$, but I do not know anything about such representations.

I know somewhere we may be indirectly using the classification of abelian groups in these results, but I think it would be an instructive thing to see it all link up still.

Any thoughts would be appreciated.

  • $\begingroup$ If you use an algebraic closure $K$ of $\Bbb{F}_q$ as the range of the characters, then the group of characters $Hom(\Bbb{F}_q^* ,K^*)$ is generated by the identity mapping. In light of this I don't think you will make any headway looking at the $K$-valued characters. $\endgroup$ – Jyrki Lahtonen Feb 27 '18 at 17:14
  • $\begingroup$ You might be able to locate a suitable prime ideal $\mathfrak{p}$ of $R=\Bbb{Z}[\zeta]$, $\zeta=e^{2\pi i/(q-1)}$, and show that $R/\mathfrak{p}$ is a field of $q$ elements and all the powers of $\zeta$ are in distinct cosets. But, that doesn't look very natural to me. $\endgroup$ – Jyrki Lahtonen Feb 27 '18 at 17:20
  • $\begingroup$ I guess that you are looking for a way to get around the fact that the proof of the isomorphism $\hat G\simeq G$, $G$ a finite abelian group, relies on the structure theorem. The problem is that there is no natural isomorphism here. For the characters of the additive group $(\Bbb{F}_q,+)$ we can take advantage of the fact that the non-degenerate bilinear trace form $(x,y)=tr(xy)$ gives a somewhat naturally available isomorphism between the additive group and its dual. $\endgroup$ – Jyrki Lahtonen Feb 27 '18 at 17:23

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