# 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' $$8$$ proof 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 $$\pm 3$$. 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.

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