November 5th
Today I learned the law of quadratic reciprocity in function fields. This is really the sort of thing I should have tried on my own in advance, but I run out of time. Let $q$ be an odd prime power. We define the Legendre-analog as expected, for irreducible $\pi\in\FF_q[x]$ and $\varphi\in\FF_q[x]/(\pi)\cong\FF_{q^{\deg\pi}}$ by\[\left(\frac\varphi\pi\right)=\begin{cases} \phantom-1 & \varphi\text{ is a nonzero square}\pmod\pi, \\ \phantom-0 & \pi\mid\varphi, \\ -1 & \varphi\text{ is non-square}\pmod\pi.\end{cases}\]Because (for example) $\FF_{q^{\deg\pi}}$ is cyclic, we pick a generator and conclude that nonzero $\varphi$ are squares if and only if their multiplicative order divides $\left(q^{\deg\pi}-1\right)/2.$ Namely, we get "Euler's Criterion'' by writing\[\left(\frac\varphi\pi\right)\equiv\varphi^{\left(q^{\deg\pi}-1\right)/2}\pmod\pi.\]Before dealing with the actual quadratic reciprocity law, we note that the elementary quadratic reciprocity requires us to choose a particular "associate'' of the primes (namely, the positive one), and we have to do similar here. Our reciprocity law will involve monic irreducibles, but this means that we would like a way to deal with constants before factoring them out. Well, we can just say, for $c\in\FF_q,$\[\left(\frac c\pi\right)\equiv c^{\left(q^{\deg\pi}-1\right)/2}\pmod\pi\]from "Euler's Criterion.'' Both sides are elements of $\FF_q,$ so we may restrict the modular arithmetic to\[\left(\frac c\pi\right)=c^{\left(q^{\deg\pi}-1\right)/2},\]where arithmetic is done in $\FF_q.$ This is a bit nicer.
The proof of the quadratic reciprocity law is actually quite simple. For a monic irreducible $\pi,$ we have that its splitting field is $\FF_q[x]/(\pi)\cong\FF_{q^{\deg\pi}}.$ However, the neat trick we apply is that the automorphisms $\FF_{q^{\deg\pi}}/\FF_q$ must also permute the roots of $\pi,$ and we understand those automorphisms very well: they're generated from the Frobenius $\sigma:\alpha\mapsto\alpha^q.$ So we'd like to say, in $\FF_{q^{\deg\pi}},$ that for a root $\alpha$ of $\pi,$\[\pi(x)=\prod_{k=0}^{\deg\pi-1}\left(x-\sigma^k(\alpha)\right)=\prod_{k=0}^{\deg\pi-1}\left(x-\alpha^{q^k}\right).\]Recall $\pi$ is monic. Indeed, our automorphisms permute the roots, we get $\deg\pi$ total roots, and all these roots are distinct because $\sigma^k(\alpha)\equiv\sigma^\ell(\alpha)$ would imply $\sigma^{k-\ell}(\alpha)=1.$ Well, $\FF_q[\alpha]\cong\FF_q[x]/(\pi)$ by construction, so if $\sigma^{k-\ell}(\alpha)=1,$ then this projects everyone in $\FF_q[\alpha]$ to $\FF_q,$ so this isn't actually an automorphism, and $\deg\pi\mid k-\ell.$ So we indeed get $k=\ell.$
We can stretch this trick to get to quadratic reciprocity very quickly. Fix monic irreducibles $\pi_1$ and $\pi_2,$ where $\alpha_1$ is a root of $\pi_1$ and $\alpha_2$ a root of $\pi_2.$ We work in the splitting field $\FF_q[\alpha_1,\alpha_2],$ partially for convenience. Euler's Criterion says that\[\left(\frac{\pi_1}{\pi_2}\right)\equiv\pi_1^{\left(q^{\deg\pi_2}-1\right)/2}\pmod{\pi_2}.\]But $\pi_1$ is a polynomial here, so we may plug in $\alpha_2$ to little detriment. Do realize that we cannot plug in $\alpha_1$ because $\alpha_1\pmod{\pi_2}$ does not necessarily make sense, but $\alpha_2\pmod{\pi_2}$ does. We get\[\left(\frac{\pi_1}{\pi_2}\right)\equiv\pi_1(\alpha_2)^{\left(q^{\deg\pi_2}-1\right)/2}\pmod{\pi_2},\]and this is in fact an equality in $\FF_q[x]/(\pi_2)$ and so expands to an equality in our splitting field. Our goal is to get as symmetric expression as possible. We start by using our splitting field to say\[\left(\frac{\pi_1}{\pi_2}\right)=\pi_1(\alpha_2)^{\left(q^{\deg\pi_2}-1\right)/2}=\prod_{k=0}^{\deg\pi_1-1}\left(\alpha_2-\alpha_1^{q^k}\right)^{\left(q^{\deg\pi_2}-1\right)/2}.\]To make symmetry appear, we write $q^{\deg\pi_2}-1=(q-1)\left(1+q+\cdots+q^{\deg\pi_2-1}\right).$ So we get\[\left(\frac{\pi_1}{\pi_2}\right)=\prod_{k=0}^{\deg\pi_1-1}~\prod_{\ell=0}^{\deg\pi_2-1}\left(\alpha_2-\alpha_1^{q^k}\right)^{q^\ell(q-1)/2}.\]The Frobenius automorphism lets us bring those powers of $q$ inside, which looks like\[\left(\frac{\pi_1}{\pi_2}\right)=\prod_{k=0}^{\deg\pi_1-1}~\prod_{\ell=0}^{\deg\pi_2-1}\left(\alpha_2^{q^\ell}-\alpha_1^{q^{k+\ell}}\right)^{(q-1)/2}.\]The orbit $\alpha_1^{q^\bullet}$ cycles with length $\deg\pi_1$ as established, so we may shift the $k+\ell$ back to indexing by $k.$ This tells us that\[\left(\frac{\pi_1}{\pi_2}\right)=\prod_{k=0}^{\deg\pi_1-1}~\prod_{\ell=0}^{\deg\pi_2-1}\left(\alpha_2^{q^\ell}-\alpha_1^{q^k}\right)^{(q-1)/2}.\]Now this is completely symmetrical. In particular, we could run the entire argument again to say that\[\left(\frac{\pi_2}{\pi_1}\right)=\prod_{\ell=0}^{\deg\pi_1-1}~\prod_{k=0}^{\deg\pi_2-1}\left(\alpha_1^{q^k}-\alpha_2^{q^\ell}\right)^{(q-1)/2},\]which is the same as our $\left(\frac{\pi_1}{\pi_2}\right),$ with an added $(\deg\pi_1)(\deg\pi_2)(q-1)/2$ signs. It follows\[\left(\frac{\pi_1}{\pi_2}\right)\left(\frac{\pi_2}{\pi_1}\right)=(-1)^{(\deg\pi_1)(\deg\pi_2)(q-1)/2},\]which is our law of quadratic reciprocity.