Commutative Algebra, IX, p. 54
Let A be a ring. Let J=P×Q be a decomposition of the monoid J into a product of submonoids P and Q (which are therefore generated by the prime numbers they contain). Suppose that every q∈Q is invertible in A, hence also in U(A) (p. 52, Exercise 35, e)). a) For every a∈U(A), the series
εQ(a)=q∈Q∑qμ(q)VqFq(a), where
μ denotes the Möbius function (Lie, II, p. 71), is convergent for the product topology of
U(A)=AJ. Thus one defines the additive endomorphism
εQ=∑q∈Qqμ(q)VqFq of
U(A). For every
n∈J, the mapping
Φn∘εQ of
U(A) into
A is equal to 0 if
n∈/P and to
Φn if
n∈P. b) For every
q∈Q,q=1, one has
εQVq=0=FqεQ. (Use Exercise 36, c), p. 52.) c) Now prove that
εQ is an idempotent whose image is the intersection of the kernels of the
Fq (
q∈Q,q=1). For
m∈J and
a∈A, calculate
εQVmτ(a). Deduce that the kernel of
εQ is the set of elements of the form
∑n∈QVnτ(an), with
an∈A. The subset
εQU(A) of
U(A) is stable under addition and multiplication. Endowed with these two operations, it is a commutative ring, with unit
εQ(1U(A)), and the mapping
εQ:U(A)→εQU(A) is a ring homomorphism. d) For every
q∈Q, let us put
eq=q1VqεQFq. Now prove that one has
eq2=eq and
eqeq′=0 if
q=q′, and that, for every
a∈U(A) one has
(∗)a=q∈Q∑eq(a), where the sum is convergent for the product topology of
U(A). (For (
) reduce to the case where a=X∈U(R[X]), where R is the subring of Q consisting of rational numbers with denominator in Q. It is then enough to apply Φ and to verify the analogue of () in
R[X]J, which results from the formula
Φpq∘eq′=Φpqδqq′ if
p∈P,q∈Q,q′∈Q. The subset
eqU(A) of
U(A) is stable under addition and multiplication. Endowed with these two operations, it is a commutative ring, with unit
eq(1U(A)) and the mapping
eq:U(A)→eqU(A) is a ring homomorphism. e) For every
q∈Q, one has
Fqeq=εQFq and
(q1Vq)εQ=eq(q1Vq). Conclude that
a↦q1Vq(a) is an isomorphism of rings
eqU(A)→εQU(A), with inverse
a↦q1Vq(a). f) Now prove that the canonical projection
πP:U(A)→UP(A) (Exercise 39) defines an isomorphism of rings
εQU(A)→UP(A). Deduce an isomorphism of rings of
U(A) onto
UP(A)Q transforming
a into
(πPεQFq(a))q∈Q for every
a∈U(A). g) If
A is a
Q-algebra, one can take
P={1} and
Q=J. One deduces an isomorphism of rings
U(A)→AJ, which is none other than
Φ.
h) Suppose that P={pn∣n∈N}, where p is a prime number. Then every prime number q=p is invertible in A. One obtains an isomorphism of rings U(A)→W(A)Q (cf. exerc. 39, d)). We shall denote by ωA the mapping of W(A) into U(A) which, by the isomorphism of U(A) and W(A)Q, corresponds to the inclusion of W(A) in W(A)Q according to the first component.