# Is there a proof of quadratic reciprocity using $p$-adic numbers?

I asked same question on MSE before, but I didn't get any answer yet.

I know that the quadratic reciprocity can be regarded as a special case of Artin reciprocity (class field theory), and we can get it by considering the cyclotomic extension of $$\mathbb{Q}_{p}$$. However, I want to know if there's any proof of quadratic reciprocity that doesn't use any stronger result, but only uses some properties of $$p$$-adic numbers. Thanks in advance.

More precisely, we can do analysis on $$\mathbb{Q}_{p}$$. We have an exponential and logarithm function on $$\mathbb{Q}_{p}$$ (at least for $$p>2$$), and we understand unit group $$\mathbb{Z}_{p}^{\times}$$ well, etc. But I don't know how to prove it in a purely local and analytic way.

The quadratic reciprocity law is a special case of the product formula for Hilbert's symbol: for all $$(a,b)\in\mathbb{Q}^{\times}$$ $$$$\underset{v}{\prod}(a,b)_{v}=1$$$$ where $$v$$ ranges in the product other all prime numbers and $$\infty$$. The definition of the symbol $$(a,b)_{p}$$ can be given in purely $$p$$-adic terms, but of the course the proof of the product formula must involve global arguments, so is global in nature (as any proof of the quadratic reciprocity law must be at least in the sense that the statement of this theorem involves two primes). When done through the structure of the Brauer group and cyclic algebras, the proof of the product formula is however minimally global in some sense.

Is that a satisfying answer for you?

• One can find a nice proof of the product formula in Milnor "Introduction to algebraic K-theory", Chapter 11. According to Milnor, this proof of quadratic reciprocity is due to Tate and is very similar to Gauss' first proof of quadratic reciprocity. – François Brunault Apr 4 '19 at 8:09
• Salut François "and is very similar to Gauss' first proof of quadratic reciprocity" really? Isn't Gauss's first proof though Gauss's lemma, so counting multiples of $a$ in intervals? Is that really very similar? – Olivier Apr 4 '19 at 8:31
• Salut Olivier, I checked in the Disquisitiones and Gauss' proof uses induction on the primes $p,q$. Interestingly in n°139, Gauss introduces a symbol $[x,y]$ which is almost the Hilbert symbol! (more precisely $[x,y]$ is the product of the $(x,y)_p$ where $p$ runs through the prime factors of $y$). But I don't know whether this is Gauss' first proof. – François Brunault Apr 4 '19 at 8:57
• The excerpt from Milnor's book is available at people.reed.edu/~ormsbyk/kgroup/resources/GaussQuadratic.pdf. It has Tate's calculation of $K_2(\mathbf Q)$ on p. 101 and says on p. 102 "Tate remarks that his proof of this theorem is lifted directly from the argument which was used by Gauss in his first proof of the quadratic reciprocity law." – KConrad Apr 4 '19 at 9:13
• The proof using Gauss's lemma is known as Gauss's third proof. – Victor Protsak Apr 4 '19 at 12:45

There are multiple proofs of quadratic reciprocity via finite fields, including Gauss's third proof. One of my favorites is Zolotareff's proof based on his interpretation of Legendre symbol as the sign of a permutation ("Zolotarev's lemma"). This approach seems to fit your criteria, because $$\Bbb{F}_p\simeq\Bbb{Z}_p/p\Bbb{Z}_p$$ as rings ("we understand unit group $$\Bbb{Z}^{\times}_p$$ well") and one may inflate functions from $$\Bbb{F}_p$$ to $$\Bbb{Z}_p$$.

• I like the proof, but I think this proof is not the kind of proof I want (I know that my question is too ambiguous, so it is hard to explain what I want in my mind). But anyway, thank you for introducing this proof! I saw this lemma as an exercise in Neukirch. – Seewoo Lee Apr 7 '19 at 7:49