Aareyan Manzoor

I am a 3rd year math UG at ASU. My interest lies in the interplay of different field but I am leaning towards operator theory for my PhD. I am a former bangladeshi but current arizonian, and I am a member of the LGBTQ+ community

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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