Gauss Sums And Kronecker Weber

07 Aug 2023 -

Recently I was a counselor at the Ross math program where quadratic reciprocity was one of the main things we built up to. (as a side note, I seriously recommend this job, some of the most fun I have had.) Quadratic reciprocity has always been mysterious to me. It is a simple enough statement: knowing when $p$ is a QR mod $q$ can be determined if we know when $q$ is a QR mod $p$. However this has led itself to 246 published proofs somehow, which always suggested to me something deep should be embedded into the statement. However before last month I could not really convince myself of this being a deep statement. A lot of proofs of it is often just e.g. counting or doing a computation and it just comes out.

We will look at in this blog, one of Gauss' $8$ proofs of this. The proof by so called Gauss sums. This is a common proof presented in many elementary number theory class. but usually without any intuition: its just computation. We will assume some Field theory and Galois theory, basically just know the Galois correspondence. Also for our intents and purposes, $2$ doesn't exist!

When is -3 a quadratic residue?

We start of with a simple question, when is $-3$ a quadratic residue? So we are asking, let $p$ be prime, does there exist $x$ such that $$x^2 = -3 \pmod{p}?$$ To do this, we will do a trick. Often you can figure out something for $\mathbb{F}_p$ by thinking about it in $\mathbb{C}$ first, or vice versa. An example of this is proving $\mathbb{F}_p$ has cyclic multiplicative group by counting elements of order $p$ in $\mathbb{C}^\times$. Consider $\omega$ to be a third root of unity, say $$\omega = \dfrac{-1+\sqrt{-3}}{2}.$$ Notice that you can hence write $\sqrt{-3}$ interms of third roots of unity, lets say $\sqrt{-3} = \omega-\omega^2$. So if I have the analogue of a third root of unity in $\mathbb{F}_p$, then i expect $-3$ to be a square root. Well there is an element of order $3$ if and only if $3|p-1$, i.e if $p\equiv 1 \pmod{3}$.

Let $p\equiv 1 \pmod{3}$, and $z$ an element of order $3$ in $\mathbb{F}_p$. Then we claim, motivated by the complex case, that $(z-z^2)^2 = -3$. Indeed, this is a simple computation: $$(z-z^2)^2 = z^2-2z^3+z^4 = z^2-2+z = (z^2+z+1)-3 = -3.$$ So actually we just proved that if $p\equiv 1\pmod{3}$, then $-3$ is a quadratic residue mod $p$.

What about when $p\equiv 2 \pmod{3}$? We do not have an element of order $3$ so we cannot do this nice trick. But what we can do is adjoint an element of order 3, i.e look at $\mathbb{F}_p[z]$ where $z$ is a root of $z^2+z+1=0$. In this case, we also have $(z-z^2)^2 = -3$. Since this is a field, the equation $x^2=-3$ can only have two solutions, i.e $z-z^2$ and its negative. So the question now: is $z-z^2$ actually in the basefield $\mathbb{F}_p$? Here we will use some light Galois theory: notice $z\mapsto z^2$ is an automorphism but doesnt fix $z-z^2$. This means $z-z^2$ isnt in the base field, i.e $-3$ has no square root.

The trick we did for $p\equiv 2\pmod{3}$ will not generalize unfortunately. The reason is that we do not easily know what the minimum polynomial of $z$, an element of order $q$, is in $\mathbb{F}_p$. So we do not know which $z\mapsto z^a$ are the correct automorphisms to consider! However, once we look at the general case, actually an even nicer trick will help us.

Gauss sums

We managed to completely classify when $-3$ is a quadratic residue mod $p$ by writing its square root interms of roots of unity. We will try to ask a more general question, when is $\sqrt{n}\in \mathbb{Q}(\zeta_p)$ where $\zeta_p$ is a primitive $p$th root of unity. Is there even such an element in the field? Actually the answer is yes. To see this all we need is some Galois theory. Note $$\Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q}) = \{\sigma_j| j \text{ coprime to } p\} \cong \left(\mathbb{Z}/p\mathbb{Z}\right)^\ast \cong \mathbb{Z}/(p-1)\mathbb{Z}, \quad \sigma^j(\zeta_p) = \zeta_p^j.$$ And since $2$ isnt a prime, this is an even cyclic group and hence has a subgroup of index $2$. The fixed field of this is hence of degree 2, i.e a quadratic extension $\mathbb{Q}(\sqrt{n})$ so in particular, some square root is in the cyclotomic field. Note the subgroup of index $2$ will be all things of the form $\{\sigma_{j^2}\}$.

But lets not get too lost in the Galois theory, and lets say simply that we have $$\tau = \sum_{i=0}^{p-1} a_i \zeta_p^i = \sqrt{n}$$ for some $n$. Then since an automorphism sends $\sqrt{n}$ to one of its conjugates, i.e itself or its negation, we would have $\sigma_j(\tau) =\pm \tau$. But in either case, we could just apply $\sigma_j$ again to have $\sigma_{j^2} (\tau) = \tau$. If we expand this out, it says $$\sigma_{j^2}(\tau) = \sum_{i=0}^{p-1} a_i \zeta_p^{j^2i} = \tau = \sum_{i=0}^{p-1} a_i \zeta_p^i.$$ Comparing the coefficient of $\zeta_p^{j^2}$, we have $a_1 = a_{j^2}$. So this tells us that all quadratic residues have the same coefficient. Since $\tau$ isnt fixed by everything, for non-residues we should have $\sigma_{j}(\tau) = -\tau$, i.e $a_j=-a_1$ and $a_0=0$. We can just divide by $a_1$ and hence we can assume $a_1=1$. So the coefficient of $\zeta^j$ is +1 or -1 depending on wether it is a quadratic residue or not. I.e $$\tau = \sum_{i=1}^{p-1}\left(\dfrac{i}{p}\right) \zeta_p^i.$$ Now we compute $$\begin{aligned} \tau^2 &= \sum_{(i,i')\in \mathbb{F}_p^2} \left(\dfrac{ii'}{p}\right) \zeta_p^{i+i'} \\ &= \sum_{(i,k)\in \mathbb{F}_p^2} \left(\dfrac{i(ki)}{p}\right) \zeta_p^{i+ki}\\ &= \sum_{(i,k)\in \mathbb{F}_p^2} \left(\dfrac{k}{p}\right) \zeta_p^{i(k+1))}\\ &= \sum_{k\in \mathbb{F}_p} \left(\dfrac{k}{p}\right) \sum_{i \in \mathbb{F}_p} \zeta_p^{i(k+1)}\\ &= (p-1)\left(\dfrac{-1}{p}\right) + \sum_{k\in \mathbb{F}_p-\{-1\}} \left(\dfrac{k}{p}\right) \sum_{s \in \mathbb{F}_p} \zeta_p^{s}\\ &= (p-1)\left(\dfrac{-1}{p}\right) + \sum_{k\in \mathbb{F}_p-\{-1\}} \left(\dfrac{k}{p}\right)(-1)\\ &= p\left(\dfrac{-1}{p}\right) -\sum_{k\in \mathbb{F}_p} \left(\dfrac{k}{p}\right)\\ &= p\left(\dfrac{-1}{p}\right). \end{aligned}$$ We used several time that whenever $i\neq 0$, we have $k\mapsto ki$ is a bijetion of $\mathbb{F}_p^\times$. We already knew $\tau^2$ was going to be a rational, so its great to see it is actually just $\pm p$. Ofcourse when $p=3$, this is the case we already worked out. Also recall that $\tau$ is fixed by $\sigma_q$ only when $q$ is a quadratic residue mod $p$. So $$\sigma_q(\tau) = \sum_{i=1}^{p-1}\left(\dfrac{i}{p}\right) \zeta_p^{qi} = \left(\dfrac{q}{p}\right) \tau.$$ Unfortunately there is no nice way to finish this off in $\mathbb{C}$, so we move down to finite fields as we did for $p=3$.

Now consider $\mathbb{F}_q$, and the question is wether $p$ is a quadratic residue mod $q$. Again, consider an extension $\mathbb{F}_q(z)$, so we have an element of order $p$. We know $$\tau = \sum_{i=1}^{p-1}\left(\dfrac{i}{p}\right) z^i\implies \tau^2 = p\left(\dfrac{-1}{p}\right).$$ The computation would be the same but also one might argue using polynomials, the min poly of $\zeta_p$ in $\mathbb{C}$ is $\Phi_p$ and it must divide $\tau^2-p$, and the min poly of $z$ mod $q$ divides $\Phi_p$. Note that if we define $\sigma_q$ the same way as before, sending $z\mapsto z^q$, its not necessarily an automorphism but it does still satisfy $\sigma_q(z) = \left(\dfrac{q}{p}\right) \tau.$. Since $(a+b)^q = a^q+b^q$ in this field, we have $$\tau^q = \sigma_q(\tau) = \left(\dfrac{q}{p}\right) \tau.$$ As we know $\tau \in \mathbb{F}_q$ if and only if $\tau^q = \tau$. So actually we already know the answer, $p\left(\dfrac{-1}{p}\right)$ is a quadratic residue mod $q$ iff $q$ is a quadratic residue mod $p$. We have already accomplished our goal, we know how to relate $p$ being a quadratic residue mod $q$ to the other way around. But lets write an equation down with this and say $$\tau^q = \tau (\tau^2)^{(q-1)/2} = \tau \left(\dfrac{-1}{p}\right)^{(q-1)/2} p^{(q-1)/2} = \left(\dfrac{-1}{p}\right)\left(\dfrac{-1}{q}\right)\left(\dfrac{p}{q}\right)\tau.$$ Equating we'd have $$\left(\dfrac{-1}{p}\right)\left(\dfrac{-1}{q}\right)\left(\dfrac{p}{q}\right) = \left(\dfrac{q}{p}\right),$$ which we can write nicely as $$\left(\dfrac{-1}{p}\right)\left(\dfrac{-1}{q}\right)=\left(\dfrac{p}{q}\right) \left(\dfrac{q}{p}\right).$$

conclusion

Hopefully this injected some intuition into the idea of the Gauss sums. Basically, Gauss sums are a way to capture quadratic Kronecker-Weber, that we can write every $\sqrt{n}$ interms of roots of unities. And this was enough to solve quadratic reciprocity. This idea can be generalized to higher reciprocity laws also. Infact the full statement of Kronecker-Weber says that every abelian extension(i.e Galois group over $\mathbb{Q}$ is abelian) is a subfield of a cyclotomic extension. And basically this allows one to write down all such higher reciprocity laws. This is the content of Class field theory.