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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:

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

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!

Martin Sleziak
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Bach
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3 Answers3

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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 of ${\rm Gal}(\mathbf Q(\zeta_p)/\mathbf Q) \cong (\mathbf Z/(p))^\times$ 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.

KCd
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  • 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 – reuns Mar 01 '20 at 19:07
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    @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 02 '20 at 00:15
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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 $\frac{p-1}{2}\cdot\frac{q-1}{2}$

Maestro13
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I think the most natural way to understand how quadratic reciprocity works and why it is true in terms of prime splitting. It follows from a constructive proof of a special case of the Kronecker-Weber theorem.

Here is how prime splitting works with the Kronecker-Weber theorem. If $K$ is an abelian Galois extension of $\mathbb Q$ of degree $d$, the Kronecker-Weber theorem says that there is an integer $m$ such that $K$ is contained in the cyclotomic extension $L = \mathbb Q(e^{2 \pi i /m})$. If $\Gamma$ denotes the Galois group of $L/K$, then $\Gamma$ is a subgroup of $\operatorname{Gal}(L/\mathbb Q)$, which itself can be identified with the group of units modulo $m$.

Granting all this (which is granting A LOT), the rest is very basic algebraic number theory: suppose $p$ is a rational prime number which does not divide $m$, and $f(p)$ is the order of the image of $p$ in the quotient group $(\mathbb Z/m\mathbb Z)^{\ast}/\Gamma$, then $p$ splits in the ring $\mathcal O_K$ into a product of $\frac{n}{f(p)}$ distinct prime ideals.

This relates to quadratic reciprocity in the following way: if $q$ is any odd prime number, and $q^{\ast} = q(-1)^{\frac{q-1}{2}}$, it is a result of Gauss that the field $K = \mathbb Q(\sqrt{q^{\ast}})$ is contained in the cyclotomic field $L = \mathbb Q(e^{2\pi i /q})$. The Galois group of $L/K$ is the subgroup of $(\mathbb Z/q\mathbb Z)^{\ast}$ consisting of the quadratic residues modulo $q$. Now if $p \neq q$ is any odd prime number, then we see that (just looking at the field $K$) $q^{\ast}$ is a quadratic residue modulo $p$ if and only if $p$ splits in $\mathcal O_K$, if and only if (now applying the inclusion $K \subset L$) $p$ is a square modulo $q$.

D_S
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