Subfield generated by traces of some elements in finite fields

Question

I'm doing exercises on Bonnafe's beautiful book Representations of $SL_2(\mathbb{F}_q)$. The first exercise is:

1.1. Let $k$ be the subfield of $\mathbb{F}_{q}$ generated by $\{ \operatorname{Tr}_{2}(\xi):\xi\in \mathbb{F}_{q^2} ~,\xi^{q+1}=1\}$. Show that $k=\mathbb{F}_{q}$. (Hint: Set $q^{\prime}=|k|$ and show that, if $\xi \in \mu_{q+1}:=\{a\in\mathbb{F}_{q^2} ~:a^{q+1}=1\}$, then $1+\xi^{2}+\xi^{ q^{\prime}}+\xi ^{q^{\prime}+1}=0$.)

I met some difficulty, as follows.

If $\xi\in\mathbb{F}_{q^2}$, then $\operatorname{Tr}(\xi):=\xi+\xi^q$. Then by the hint we have $$(\xi+\xi^q)^{q'}=\xi+\xi^q.$$ Expanding this, and remember that $\xi^q=\xi^{-1}$, we obtain $\xi+\xi^{-1}=\xi^{q'}+\xi^{-q'}$, and this is $1+\xi^{2}=\xi^{q'+1}+\xi^{-q'+1}$. Still I couldn't see how this is equivalent to the hint.

By the way, even assuming the claim of the hint. I still had no idea how to deduce the result of the exercise. Can anybody give me some help? Any suggestion or hint will be welcome. Thanks a lot in advance!

Answer 1

I don't understand the hint.

• The point is that $$\Bbb{F}_{q^2}=\Bbb{F}_p(\zeta_{q+1})$$

  (proof: write $q=p^n$, then $\Bbb{F}_p(\zeta_{q+1})=\Bbb{F}_{p^m}$ where $m$ is the least integer such that $p^n+1|p^m-1$. As $\gcd(p^{2n}-1,p^m-1)=p^{\gcd(2n,m)}-1$ we get that $\gcd(p^n+1,p^m-1)$ divides $p^{\gcd(2n,m)}-1$. So $p^n+1|p^m-1$ gives that $\gcd(2n,m)> n$ ie. $m\ge 2n$)

• Whence $$Tr_{\Bbb{F}_{q^2}/\Bbb{F}_q}(\Bbb{F}_p(\zeta_{q+1}))=\Bbb{F}_q$$

Conclude by noting that $\Bbb{F}_p(\zeta_{q+1})$ is the subgroup of $\Bbb{F}_{q^2}$ generated by the roots of $x^{q+1}-1$ so that the LHS is the subring of $\Bbb{F}_{q^2}$ generated by the traces of roots of $x^{q+1}-1$.

Answer 2

I might do this as follows, related to the hint in a way. Turning it into a counting argument.

Consider the set $$S=\{z\in\Bbb{F}_{q^2}\mid z^{q+1}=1\}.$$ As $\Bbb{F}_{q^2}^*$ is cyclic of order $q^2-1=(q-1)(q+1)$ it follows that $S$ is a cyclic subgroup of order $q+1$. As $\gcd(q+1,q-1)$ is $1$ or $2$ according to whether $q$ is even or odd, we see that $S\cap \Bbb{F}_q=\{\pm1\}$.

If $z\in S\setminus\{\pm1\}$ then the minimal polynomial of $z$ over $\Bbb{F}_q$ is $$m_z(x)=(x-z)(x-z^q)=x^2-Tr(z) x+z^{q+1}=x^2-Tr(z)x+1.$$ This is shared by exactly two elements of $S$, namely $z$ and $z^q=z^{-1}$. It follows that the trace function takes at least $(q-1)/2$ distinct values in $\Bbb{F}_q$. That's quite a few for them to fit into a proper subfield, don't you think?

The largest proper subfield of $\Bbb{F}_q$ has $\sqrt q$ elements, but $(q-1)/2>\sqrt q$ whenever $q\ge 6$. Meaning that the claim is in doubt only when $q\in\{2,3,4,5\}$. The prime fields are obviously out of the reckoning, so that leaves $q=4$. But with $q=4$ zero is not a trace of an element of $S\setminus\{1\}$. For if $z+z^{-1}=0$ then $0=z^2+1=(z-1)^2$. Hence the trace takes two non-zero values on $S$ ruling out proper subfields.

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The end game with $q=4$ became a bit kludgier than I anticipated. Sorry about that. Anyway, the subgroup $S$ appears in many a trick. It is a lot like the unit circle of the complex plane. Here $\Bbb{F}_{q^2}$ is the "plane" over $\Bbb{F}_q$.