Today I Learned

April 13th

Today I learned about characters of finite fields, from here . The surprise here is that the characters of finite fields behave a lot like the characters of local fields. We begin by exhibiting a "standard'' character for each finite field $\FF_q,$ analogously to local fields.

For our base fields, we fix a prime $p$ and let\[\psi_p(x):=e^{2\pi ix/p}\]be our "standard'' character $\FF_p\to S^1.$ Indeed, this is a character because $x\mapsto e^{2\pi ix/p}$ is actually a character of $\ZZ$ (note $e^{2\pi i(x+y)/p}=e^{2\pi ix/p}e^{2\pi iy/p}$) with kernel when $e^{2\pi ix/p}=1,$ which is equivalent to $x/p\in\ZZ$ or $x\in p\ZZ.$ So $\psi$ induces a (nontrivial) character of $\ZZ/p\ZZ\cong\FF_p.$

To continue as with local fields, we would like to build our characters of finite fields $\FF_q/\FF_p$ by defining a trace mapping $\op{Trace}:\FF_q\to\FF_p$ and then fixing\[\psi_q:=\psi_p\circ\op{Trace}.\]So we have to think about what the $\op{Trace}$ map should be. Well, in number fields, Galois extensions $L/K$ have\[\op{Trace}(\alpha)=\sum_{\sigma\in\op{Gal}(L/K)}\sigma(\alpha).\]We can imitate this with finite fields because $\FF_q/\FF_p$ is automatically Galois and in fact generated by $\op{Frob}_p:x\mapsto x^p.$ Fixing $q=p^n,$ we thus take\[\op{Trace}(\alpha):=\sum_{k=0}^{n-1}\op{Frob}_p^k(\alpha)=\sum_{k=0}^{n-1}\alpha^{p^k}.\]We can check that $\op{Trace}$ does indeed have image in $\FF_p$ because applying $\op{Frob}_p$ merely permutes the sum, so the image of $\op{Trace}$ lies in the fixed field of $\op{Frob}_p,$ which is $\FF_p.$

So we take the following definition.

I am under the impression that it is surprisingly annoying to prove that $\op{Trace}$ doesn't send everything to $0$; a proof in a more general case can be found in Milne's notes around Proposition 2.26. I will just say that, indeed, $\op{Trace}$ is non-degenerate, it can be proven, and it means that our $\psi_q$ is a nontrivial character.

We will need there to exist a nontrivial character of $\FF_q,$ so we remark that we can exhibit one without appealing to non-degeneracy of the trace. Indeed, $\FF_q$ is an $\FF_p$-vector space, so\[\FF_q\cong\FF_p^n\]as additive groups. Thus, we can exhibit a character $\FF_q\to S^1$ by using $\psi_p$ on the first coordinate and ignoring the remaining coordinates.

Now here is the big parallel between local fields and finite fields.

We imitate the argument with local fields. Fix $\Psi_\bullet:\FF_q\to\widehat{\FF_q}$ by $\Psi_a(x):=\psi(ax).$ This is well-defined because\[\psi(a(x+y))=\psi(ax+ay)=\psi(ax)\psi(ay).\]In fact, we claim that $\Psi_\bullet$ exhibits an isomorphism $\FF_q\cong\widehat{\FF_q},$ which will complete the proof. We split this into parts.

• We see $\Psi_\bullet$ is a homomorphism because \[\Psi_{a+b}=\big(x\mapsto\psi((a+b)x)\big)=\big(x\mapsto\psi(ax)\psi(bx)\big)=\Psi_a\Psi_b.\]

• To show that $\Psi_\bullet$ is injective, we show it has trivial kernel, which follows from $\psi$ being nontrivial. In particular, we may fix some $x$ for which $\psi(x)\ne1.$ Then for any $a\in\FF_q\setminus\{0\},$ we actually have $a\in\FF_q^\times,$ so \[\Psi_a\left(a^{-1}x\right)=\psi\left(aa^{-1}x\right)=\psi(x)\ne1.\] Thus, $\Psi_a$ is also a nontrivial character.

• Showing that $\Psi_\bullet$ is surjective can probably be done in the same way we did with local fields, but it's easier to abuse the finiteness of these objects. Because $\Psi_\bullet$ is already injective, it suffices to show \[\#\widehat{\FF_q}=\#\FF_q=q.\] Well, $\FF_q$ is being viewed as a finite group, so this holds automatically from some character theory. Alternatively, we could let $q=p^n$ with $p$ prime so that \[\FF_q\cong\FF_p^n=(\ZZ/p\ZZ)^n.\] We can now directly count the number of characters of $(\ZZ/p\ZZ)^n$ by noting each character is determined by where it sends each of the $n$ generators for the $\ZZ/p\ZZ,$ and the generators has $p$ options among the $p$th root of unity. So indeed $\#\widehat{\FF_q}=p^n=q.$

Thus, we have finished checking that $\Psi_\bullet$ is an isomorphism, which competes the proof. $\blacksquare$