Proof of Quadratic Reciprocity

We are now ready to prove Theorem 4.1.7 using Gauss sums.

Proof. Let $q$ be an odd prime with $q\neq p$ . Set $p^* = (-1)^{(p-1)/2}p$ and recall that Proposition 4.4.5 asserts that $p^* = g^2$ , where $g = g_1 = \sum_{n=0}^{p-1} \left(\frac{n}{p}\right)\zeta^n$ .

Proposition 4.2.1 implies that

$$(p^*)^{(q-1)/2} \equiv \left(\frac{p^*}{q}\right) \pmod{q}.$$

We have $g^{q-1} = (g^2)^{(q-1)/2} = (p^*)^{(q-1)/2}$ , so multiplying both sides of the displayed equation by $g$ yields a congruence

$$g^q \equiv g \left(\frac{p^*}{q}\right) \pmod{q}.$$ (4.4.3)

But wait, what does this congruence mean, given that $g^q$ is not an integer? It means that the difference $g^q - g \left(\frac{p^*}{q}\right)$ lies in the ideal $(q)$ in the ring $\mathbb{Z}[\zeta]$ of all polynomials in $\zeta$ with coefficients in  $\mathbb {Z}$ .

The ring $\mathbb{Z}[\zeta]/(q)$ has characteristic $q$ , so if  $x, y\in\mathbb{Z}[\zeta]$ , then $(x+y)^q \equiv x^q + y^q \pmod{q}$ . Applying this to (4.4.3), we see that

$$g^q = \left(\sum_{n=0}^{p-1} \left(\frac{n}{p}\right)\zeta^n \rig... ...equiv \sum_{n=0}^{p-1} \left(\frac{n}{p}\right) \zeta^{nq} \equiv g_q\pmod{q}.$$

By Lemma 4.4.10,

$$g^q \equiv g_q \equiv \left(\frac{q}{p}\right) g\pmod{q}.$$

Combining this with (4.4.3) yields

$$\left(\frac{q}{p}\right) g \equiv \left(\frac{p^*}{q}\right)g \pmod{q}.$$

Since $g^2 = p^*$ and $p\neq q$ , we can cancel $g$ from both sides to find that $\left(\frac{q}{p}\right) \equiv \left(\frac{p^*}{q}\right) \pmod{q}$ . Since both residue symbols are $\pm 1$ and $q$ is odd, it follows that $\left(\frac{q}{p}\right) = \left(\frac{p^*}{q}\right)$ . Finally, we note using Proposition 4.2.1 that

$$\left(\frac{p^*}{q}\right) = \left(\frac{(-1)^{(p-1)/2}p}{q}\right) = \left(\frac{-1}{q}\rig... ...ft(-1\right)^{\frac{q-1}{2}\cdot \frac{p-1}{2}} \cdot \left(\frac{p}{q}\right).$$

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