Quadratic Reciprocity and the Theta Function

Argument plot of the classical Jacobi theta function.

The law of quadratic reciprocity is a beautiful result in elementary number theory which relates the Legendre symbols (\frac{p}{q}) and (\frac{q}{p}). For p and q odd primes, it states that

\displaystyle \Big(\frac{p}{q}\Big)\Big(\frac{q}{p}\Big) = \begin{cases}1, \quad & p \text{ or } q \equiv 1 \bmod 4, \\ -1, \quad & p \text{ and } q \equiv 3 \bmod 4. \end{cases}

One of the better-known proofs of quadratic reciprocity involves the quadratic Gauss sum, an exponential sum defined by

\displaystyle g_p := \sum_{a=1}^p \Big(\frac{a}{p}\Big) e\Big(\frac{a}{p}\Big),

in which e(x) := e^{2\pi i x}. In essence, this proof works by computing g_p^q in two ways:

  1. Euler’s criterion: g_p^q \equiv g_p \cdot (g_p^2)^{\frac{q-1}{2}} \equiv g_p \cdot (\frac{g_p^2}{q}) \bmod q. Note that this is well-defined because p^* := g_p^2 = (\frac{-1}{p}) p is an integer.
  2. Fermat’s Little Theorem (for algebraic integers): Working in the cyclotomic field \mathbb{Q}(\zeta_p), we find that g_p^q \equiv \sum_{a(p)} (\frac{a}{p}) e(\frac{aq}{p}) \bmod q. Changing variables a \mapsto q^{-1} a \bmod p gives g_p^q \equiv (\frac{q}{p}) g_p \bmod q.

It follows that (\frac{p^*}{q}) \equiv (\frac{q}{p}) \bmod q, which implies quadratic reciprocity.

In this note, we give a second proof of quadratic reciprocity using the analytic properties of the Jacobi theta function \theta(z), which is a modular form of weight \frac{1}{2}. In particular, we produce two formulas for the behavior of \theta(z) as z \to p/q, and derive quadratic reciprocity by comparing them. As in the previous proof, Gauss sums play a prominent role. Otherwise, this proof is distinct; in particular, it avoids Euler’s criterion and any semblance of algebraic number theory.


The Jacobi theta function \theta(z) is defined on the upper half-plane by the series \theta(z) = \sum_{n \in \mathbb{Z}} e(n^2 z). It is a modular form of weight \frac{1}{2} on the congruence subgroup \Gamma_0(4), transforming via

\displaystyle \theta(\gamma z) = j_\theta(\gamma, z) \theta(z) = \epsilon_d^{-1} \Big(\frac{c}{d}\Big) (cz+d)^{\frac{1}{2}} \theta(z), \quad \gamma = \Big( \begin{matrix} a & b \\ c & d \end {matrix} \Big) \in \Gamma_0(4),

in which \epsilon_d = 1 for d \equiv 1 \bmod 4, \epsilon_d =i for d \equiv 3 \bmod 4, and (\frac{c}{d}) denotes Kronecker’s extension of the Legendre symbol. In addition, \theta(z) satisfies the functional equation \theta(-1/4z) = (-2iz)^{1/2} \theta(z), which follows from Poisson summation and analytic continuation. (In both functional equations, we use the principal square root on \mathbb{C}.)

We only need the second functional equation in our proof, but we will use it several times. For starters, it implies that \theta(iy) \sim (2y)^{-1/2} as y \to 0 from above. And more generally, this asymptotic implies that

\displaystyle \sum_{n \equiv a (p)} e^{-2\pi n^2 y} \sim \frac{1}{p} \cdot (2y)^{-\frac{1}{2}}

as y \to 0, for any choice of a \bmod p. (Hint: Use the squeeze theorem to compare the asymptotics with different choices of a.)


We now begin our proof in earnest. Let p and q be odd primes, and consider the limit of \theta(\frac{q}{p} + iy) as y \to 0. We have

\displaystyle \theta(\tfrac{q}{p}+iy) = \sum_{n \in \mathbb{Z}} e\Big(n^2\Big(\frac{q}{p}+iy\Big)\Big) = \sum_{a (p)} e\Big(\frac{a^2 q}{p} \Big) \sum_{n \equiv a \bmod p} e^{-2\pi n^2 y}.

In the limit as y \to 0, each inner sum grows as p^{-1} (2 y)^{-1/2}. For the outer sum over a \bmod p, we let \overline{q} denote q^{-1} \bmod p and observe that

\displaystyle \sum_{a(p)} e\Big(\frac{a^2 q}{p} \Big) = \sum_{b(p)} \Big(1+ \Big(\frac{b}{p}\Big)\Big) e\Big(\frac{bq}{p}\Big) = \sum_{b(p)} \Big(\frac{b}{p}\Big) e\Big(\frac{bq}{p}\Big)
\displaystyle = \sum_{b(p)} \Big(\frac{b\overline{q}}{p}\Big) e\Big(\frac{b}{p}\Big) = \Big(\frac{q}{p}\Big) \sum_{b(p)} \Big(\frac{b}{p}\Big) e\Big(\frac{b}{p}\Big) = \Big(\frac{q}{p}\Big) g_p.

It follows that \theta(\tfrac{q}{p} + iy) \sim p^{-1} (2 y)^{-\frac{1}{2}} (\frac{q}{p}) g_p as y \to 0. We will use this asymptotic a few times in what follows.

From the functional equation \theta(-1/4z) = (-2iz)^{1/2} \theta(z), we see that

\displaystyle \theta\Big(\frac{-1}{4(\frac{p}{q}+iy)}\Big) = (-2i (\tfrac{p}{q}+iy))^{\frac{1}{2}} \theta(\tfrac{p}{q}+iy).

As y \to 0, the right-hand side tends to (-2ip/q)^{1/2} q^{-1} (2y)^{-\frac{1}{2}} (\frac{p}{q}) g_q. Meanwhile, the left-hand side equals \theta(\frac{-q}{4p} + i \frac{q^2}{4p^2} y + O(y^2)). Ignoring the non-dominant O(y^2) term, this tends to

\displaystyle \sum_{a(4p)} e\Big(\frac{-a^2 q}{4p}\Big) \cdot \frac{1}{4p} \Big(\frac{2q^2 y}{4p^2}\Big)^{-\frac{1}{2}} = (2 y)^{-\frac{1}{2}} \frac{1}{2q} \sum_{a(4p)} e\Big(\frac{-a^2 q}{4p}\Big).

The sum over a \bmod 4p is a generalized Gauss sum. By the Chinese remainder theorem, we have

\displaystyle \sum_{a(4p)} e\Big(\frac{-a^2 q}{4p}\Big) = \sum_{a(p)} e\Big(\frac{-4a^2 q}{p}\Big) \sum_{a(4)} e\Big(\frac{-a^2 pq}{4}\Big).

The first exponential sum is (\frac{-4q}{p}) g_p = (\frac{-1}{p})(\frac{q}{p}) g_p. The second depends only on pq \bmod 4 and can be computed exactly; we see that it equals 2-2i when pq \equiv 1 \bmod 4 and 2+2i when pq \equiv 3 \bmod 4. It follows that

\displaystyle \theta\Big(\frac{-1}{4(\frac{p}{q}+iy)}\Big) = (2 y)^{-\frac{1}{2}} \frac{1}{2q} \Big(\frac{-1}{p} \Big)\Big(\frac{q}{p}\Big) g_p \cdot \begin{cases} 2-2i, \quad & pq \equiv 1 \bmod 4 \\ 2+2i, \quad & pq \equiv 3 \bmod 4 \end{cases}

in the limit as y \to 0.

Of course, these two expressions for the limiting behavior must agree. After some simplification, we conclude that

\displaystyle \Big(\frac{p}{q}\Big)\Big(\frac{q}{p}\Big) = \Big(\frac{-1}{p}\Big) \frac{g_p}{\sqrt{p}} \cdot \frac{\sqrt{q}}{g_q} \cdot \begin{cases} 1, \quad & pq \equiv 1 \bmod 4 \\ i, \quad & pq \equiv 3 \bmod 4 \end{cases}

Note that the right-hand side has absolute value 1 because \vert g_p \vert = \sqrt{p} as a consequence of g_p^2 = (\frac{-1}{p}) p, and the same for q. (Note that the cases in the previous line could be written \epsilon_{pq}.)


Recall that the Gauss sum g_p is one of the complex square roots of (\frac{-1}{p}) p. At this point, we could prove quadratic reciprocity if we knew which square-root (depending on p) to choose. The correct choice of sign is of course well-known now, but it took Gauss a few years to determine the sign after first introducing the Gauss sums. Fortunately, we find ourselves in a position to determine the sign of g_p immediately.

The key observation is that none of our formulas so far have required p and q to be odd primes. In fact, it is enough for p and q to be odd, coprime, and square-free. Taking p prime and q=1 in the previous offset equation gives

\displaystyle 1 = \Big(\frac{-1}{p}\Big) \frac{g_p}{\sqrt{p}} \epsilon_p \quad \Longrightarrow \quad g_p = \epsilon_p^{-1} \Big(\frac{-1}{p}\Big) \sqrt{p} = \epsilon_p \sqrt{p}.

Of course, g_q = \epsilon_q \sqrt{q} as well. Returning to the same offset equation from before, it follows that

\displaystyle \Big(\frac{p}{q}\Big)\Big(\frac{q}{p}\Big) = \frac{\epsilon_{pq}}{\epsilon_p \epsilon_q}.

The right-hand side only depends on the values of p and q mod 4. And, as expected, it equals 1 unless both p \equiv 3 \bmod 4 and q \equiv 3 \bmod 4, when it equals -1. This completes our proof of quadratic reciprocity.


Exercise: Prove the second supplement to quadratic reciprocity by considering \theta(\frac{2}{p}+iy) as y \to 0.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s