Bourbaki

Exercise 2, Appendix 0

Lie Groups and Lie Algebras, II, 207

Let D be the total algebra of the monoid N∗\mathbf{N}^* (Algebra, Chapter III, § 2, no. 10). If n∈N∗n \in \mathbf{N}^*, let nωn^\omega denote its image in D, so that 1ω=11^\omega = 1 and (nm)ω=nωmω(nm)^\omega = n^\omega m^\omega for n,mn, m in N∗\mathbf{N}^*.† Every element ff of D can be written uniquely as a series ∑n=1∞annω\sum_{n=1}^\infty a_n n^\omega, where an∈Ka_n \in K.

(a) Let f=∑n=1∞annωf = \sum_{n=1}^\infty a_n n^\omega be an element of D. Show that ff is invertible in D if and only if a1a_1 is invertible in K. In particular, if K is a local ring, so is D.

(b) We write ζ=∑n=1∞nω\zeta = \sum_{n=1}^\infty n^\omega and μ=∑n=1∞μ(n)nω\mu = \sum_{n=1}^\infty \mu(n) n^\omega. Show that ζ\zeta and μ\mu are inverses of one another. Deduce that, if s=∑ns(n)nωs = \sum_n s(n) n^\omega and t=∑nt(n)nωt = \sum_n t(n) n^\omega are two elements of D, the relations s=ζ⋅ts = \zeta \cdot t and t=μ⋅st = \mu \cdot s are equivalent (variant of the Möbius inversion formula).

(c) Let P be the set of prime numbers. Show that the family of (1−pω)(1 - p^\omega) for p∈Pp \in P is multipliable in D and that

μ=∑p∈P(1−pω)andζ=∏p∈P11−pω. \mu = \sum_{p \in P} (1 - p^\omega) \quad \text{and} \quad \zeta = \prod_{p \in P} \frac{1}{1 - p^\omega}.

† We often write −s-s instead of ω\omega; we then say that the elements of D are the Dirichlet formal power series with coefficients in K.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.