# What is the intuition behind the law of quadratic reciprocity?

Law of quadratic reciprocity states as follows:

Law of quadratic reciprocity — Let $$p$$ and $$q$$ be distinct odd prime numbers, and define the Legendre symbol as:

$$\left(\frac {q}{p}\right)=\left\{\begin{array}{rl} 1 & \text{if } n^2\equiv q \pmod p \text{ for some integer } n, \\ -1 & \text{otherwise.} \end{array} \right.$$

Then:

$$\left({\frac {p}{q}}\right)\left({\frac {q}{p}}\right)=(-1)^{{\frac {p-1}{2}}{\frac {q-1}{2}}}.}$$

I have heard from class that there are hundreds of proof of this theorem. But the proof that I have learned in class is a very elementary one. As we know that many theorems in number theory have some very nice explanations using abstract algebra. Is there a proof of this theorem from the perspective of abstract algebra? And what is the intuition behind it? Thank you!

• I think it's a result too deep for it to be "intuitive." Of course, Gauss's intuition could say something about it but it is not just any intuition but the intuition of a genius. Feb 6 '20 at 1:02
• You may find of interest Wyman's Monthly exposition What is a Reciprocity Law? Feb 6 '20 at 1:32
• Gauss, who discovered it, gave 6 different proofs. For me, it's not intuitive; it's just a fact. Feb 6 '20 at 6:47
• Perhaps of interest is this excellent youtube video containing an elementary proof (based on this blog post) that gives more intuition than any other proof I've seen. Mar 16 '20 at 10:04
• I think all theorems become more intuitive the more proofs of them you see. Apr 1 '20 at 3:12

It would help if you told us what proof or proofs you already know.

From the perspective of algebraic number theory, the quadratic reciprocity law can be described using the cyclotomic field $$\mathbf Q(\zeta_p)$$ and its unique quadratic subfield, which is $$\mathbf Q(\sqrt{p^*})$$ for $$p^* = (-1)^{(p-1)/2}p$$, where $$p$$ is an odd prime.

Write the quadratic reciprocity law (for two different odd primes $$p$$ and $$q$$) as $$\left(\frac{q}{p}\right) = \left(\frac{p^*}{q}\right).$$ The intuition behind this formula is that, using the field extension $$\mathbf Q(\zeta_p) \supset \mathbf Q(\sqrt{p^*})$$ mentioned above, the two sides of this equation describe in different ways the Frobenius element associated to $$q$$ in $${\rm Gal}(\mathbf Q(\sqrt{p^*})/\mathbf Q)$$, which as a quotient $${\rm Gal}(\mathbf Q(\zeta_p)/\mathbf Q) \cong (\mathbf Z/(p))^\times$$ is $$(\mathbf Z/(p))^\times$$ modulo its squares. For further details about this approach you need to learn algebraic number theory; this proof can't be described at an intuitive level. That's part of what makes the quadratic reciprocity law so mysterious.

• It's possible to keep things down to earth by looking only at $p=3,5$, it isn't too hard to guess that it generalizes to every $p$ even if the detailed proof needs good knowledge of number fields Mar 1 '20 at 19:07
• @reuns do you mean run through this proof with $p = 3, 5$ and arbitrary $q$? Because $\mathbf Q(\zeta_3) = \mathbf Q(\sqrt{3^*}) = \mathbf Q(\sqrt{-3})$, I think that example is too simple. There's also a quick elementary way to show $(q|3) = (-3|q)$: $(-3|q) = 1$ for an odd prime $q$ if and only if $\mathbf Z/q\mathbf Z$ contains a cube root of unity (thinking of the formula $(-1+\sqrt{-3})/2$ in $\mathbf C$), and that's equivalent to $3\mid (q-1)$ since $(\mathbf Z/q\mathbf Z)^\times$ is cyclic. So $(3^*|q) = 1$ if and only if $q \equiv 1 \bmod 3$, which is the same as $(q|3) = 1$.
– KCd
Mar 2 '20 at 0:15

The proof that best fits the intuition represented by this formula is Eisenstein's one.

Briefly, Eisenstein uses Gauss' lemma and finds that $$(\frac{p}{q})(\frac{q}{p})$$ equals $$-1$$ or $$+1$$ depending on the parity (odd/even) of the number of lattice points in the square with as boundary lines the $$x$$-axis, the $$y$$-axis, and the lines $$x=\frac{p}{2}$$ and $$y=\frac{q}{2}$$.

And that number is clearly equal to $$\frac{p-1}{2}\cdot\frac{q-1}{2}$$