Bourbaki

Exercise 36, § 1

Commutative Algebra, IX, p. 52

Let AA be a ring. Let n,mn, m be in JJ and let d=pgcd⁡(n,m)d = \operatorname{pgcd}(n, m). a) The mapping a↦Fn(a)=(Fn,r(a))r∈Ja \mapsto F_n(a) = (F_{n,r}(a))_{r \in J} is an endomorphism of the ring U(A), and one has Fn∘Fm=FmnF_n \circ F_m = F_{mn}, Φn∘Fm=Φnm\Phi_n \circ F_m = \Phi_{nm}. b) The mapping a↦Vn(a)=(Vn,r(a))r∈Ja \mapsto V_n(a) = (V_{n,r}(a))_{r \in J} is an endomorphism of the additive group of U(A), and one has Vn∘Vm=VnmV_n \circ V_m = V_{nm}. Moreover, Φn∘Vq\Phi_n \circ V_q is equal to 0 if qq does not divide nn and to qΦn/qq \Phi_{n/q} if qq divides nn. c) One has Fn∘Vm=d×Vm/d∘Fn/dF_n \circ V_m = d \times V_{m/d} \circ F_{n/d}; in other words Fn(Vm(a))F_n(V_m(a)) (for a∈U(A)a \in U(A)) is the sum in U(A) of dd terms equal to Vm/d(Fn/d(a))V_{m/d}(F_{n/d}(a)). In particular, one has Fn(Vn(a))=n×aF_n(V_n(a)) = n \times a and Fn∘Vm=Vm∘FnF_n \circ V_m = V_m \circ F_n if d=1d = 1. d) For any a,ba, b in U(A), one has

Vn(a)×Vm(b)=d×Vnm/d(Fm/d(a)×Fn/d(b)) V_n(a) \times V_m(b) = d \times V_{nm/d}(F_{m/d}(a) \times F_{n/d}(b))
nd×Vn(a×Fm(b))=Vn(a)×Vn/d(Fm/d(b)) \frac{n}{d} \times V_n(a \times F_m(b)) = V_n(a) \times V_{n/d}(F_{m/d}(b))
nd×Vn(Fm(b))=Vn(1)×Vn/d(Fm/d(b)). \frac{n}{d} \times V_n(F_m(b)) = V_n(1) \times V_{n/d}(F_{m/d}(b)).
In particular, one has
Vn(a)×Vn(b)=n×Vn(a×b), V_n(a) \times V_n(b) = n \times V_n(a \times b),
and
Vn(a)×Vm(b)=Vnm(Fm(a)×Fn(b)) V_n(a) \times V_m(b) = V_{nm}(F_m(a) \times F_n(b))
if d=1d = 1. (Reduce to the case where A=Z[X,Y]A = \mathbf{Z}[X, Y], a=Xa = X, b=Yb = Y and use Exercise 33, d), p. 51 and Exercise 28, p. 50.)

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