Bourbaki

Exercise 1, § 12

Algebra, V, p. 166

a) Let qq be a power of a prime number. Show that for every integer n>1n > 1, if hih_i (1≤i≤r1 \leq i \leq r) are the distinct prime divisors of the integer nn, then the number of elements ζ∈Fqn\zeta \in \mathbf{F}_{q^n} such that Fqn=Fq(ζ)\mathbf{F}_{q^n} = \mathbf{F}_q(\zeta) is equal to

v=qn−∑iqn/hi+∑i<jqn/hihj−∑i<j<kqn/hihjhk+⋯+(−1)rqn/h1h2…hr v = q^n - \sum_i q^{n/h_i} + \sum_{i < j} q^{n/h_i h_j} - \sum_{i < j < k} q^{n/h_i h_j h_k} + \cdots + (-1)^r q^{n/h_1 h_2 \ldots h_r}
(observe that such an element is characterized by the property of not belonging to any of the fields Fqn/hi\mathbf{F}_{q^{n/h_i}}). If h1≤…≤hrh_1 \leq \ldots \leq h_r, we have
qn−∑iqn/hi≤v≤qn−qn/h1. q^n - \sum_i q^{n/h_i} \leq v \leq q^n - q^{n/h_1}.
Examine the case where nn is a prime power. b) Let bn(q)b_n(q) be the number of monic irreducible polynomials of degree nn over Fq\mathbf{F}_q. Deduce from a) an expression for bn(q)b_n(q). In particular, if q≥3q \geq 3, show that the number of polynomials (not necessarily monic) of degree nn in Fq[X]\mathbf{F}_q[X] which are irreducible is at least equal to qn(q−2)/n≥qn+1/3nq^n(q-2)/n \geq q^{n+1}/3n. c) Prove that ∑d∣ndbd(q)=qn\sum_{d|n} d b_d(q) = q^n (establish the identity 1/(1−qT)=∏(1−Tn)−bn(q)1/(1-qT) = \prod (1-T^n)^{-b_n(q)}). \* Using the Mobius inversion formula (Lie, II, p. 176, Appendix), show that
nbn(q)=∑d∣nμ(n/d)qd. nb_n(q) = \sum_{d|n} \mu(n/d) q^d.
Compare this result with that obtained in b). d) Let an(q)a_n(q) be the number of monic polynomials without multiple factors in Fq[X]\mathbf{F}_q[X]. Show that a0(q)=1, a1(q)=q, an(q)=qn−qn−1a_0(q) = 1,\ a_1(q) = q,\ a_n(q) = q^n - q^{n-1} for n>1n > 1 (establish the identity qn=∑an−2i(q) qiq^n = \sum a_{n-2i}(q)\ q^i).

Solution

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Transcribed from the printed volume. Not translated.