Bourbaki

Exercise 19, § 11

Algebra, V, p. 165

It is set in small type as supplementary.

a) Assume the prime number theorem; if π(x)\pi(x) is the number of primes p≤xp \leq x, then π(x)=x/log⁡x\pi(x) = x / \log x as xx tends to +∞+\infty. Deduce that for every odd integer mm, there exists an increasing sequence of prime numbers p1<p2<…<pmp_1 < p_2 < \ldots < p_m such that pm<p1+p2p_m < p_1 + p_2 (consider the prime numbers between xx and 3x/23x/2). b) With the finite sequence of prime numbers pjp_j (1≤j≤m1 \leq j \leq m) determined as in a), put n=p1p2…pmn = p_1 p_2 \ldots p_m. Show that in the cyclotomic polynomial Φn(X)\Phi_n(X) the terms of degree <pm+1< p_m + 1 are the same as those of the polynomial

(1−X)−1∏j=1m(1−Xpj) (1 - X)^{-1} \prod_{j=1}^m (1 - X^{p_j})

Deduce that the coefficient of XpmX^{p_m} in Φn(X)\Phi_n(X) is 1−m1 - m (« I. Schur's theorem »).

Solution

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Transcribed from the printed volume. Not translated.