Bourbaki

Exercise 23, § 11

Algebra, V, p. 166

Let nn and pp be odd and distinct prime numbers. a) Let Q∈Fp[X]Q \in \mathbf{F}_p[X] be the image of the cyclotomic polynomial Φn\Phi_n. Show that the number ii of irreducible factors of QQ is even if and only if pp is a square mod nn. (Let KK be the splitting field of QQ. Show that Gal(K/Fp)\mathrm{Gal}(K/\mathbf{F}_p) may be identified with the subgroup of (Fn)∗(\mathbf{F}_n)^* generated by the image of pp. Study the parity of the index ((Fn)∗:Gal(K/Fp))((\mathbf{F}_n)^*: \mathrm{Gal}(K/\mathbf{F}_p)).) b) Show that the discriminant of Xn−1∈Fp[X]X^n - 1 \in \mathbf{F}_p[X] is

(−1)n(n−1)2+n−1nn. (-1)^{\frac{n(n-1)}{2} + n-1} n^n.
c) Show that ii is even if and only if (−1)n(n−1)2n(-1)^{\frac{n(n-1)}{2}} n is a square modp (apply En. 22 to the polynomial Xn−1X^n - 1). d) Put (np)=1\left( \frac{n}{p} \right) = 1 if nn is a square modp and (np)=−1\left( \phantom{\frac{n}{p}} \right) = -1 otherwise. Deduce from c) that
(np)(pn)=(−1)(n−1)(p−1)4(a Law of quadratic reciprocity »). \left( \frac{n}{p} \right) \left( \frac{p}{n} \right) = (-1)^{\frac{(n-1)(p-1)}{4}} \quad \text{(a Law of quadratic reciprocity »)}.

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Transcribed from the printed volume. Not translated.