# Subfield generated by traces of some elements in finite fields

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!

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 $$x^{q+1}-1$$.

• Oh you are right! The key point is that generating relation! Thanks! Commented Aug 11, 2022 at 9:31

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.

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 $$\Bbb{F}_q$$.

• Thank you! I'll try to digest your ideas. I fixed a latex syntex error in your typing. Commented Aug 11, 2022 at 9:29
• Thanks @tooweaktolearnmathematics. The relation between this solution and the hint is a bit hazy to say the least. Looking at the hint it somehow looked like a minimal polynomial to me :-) Commented Aug 11, 2022 at 12:51