Bourbaki

Exercise 11, § 1

Commutative Algebra, IX, p. 43

Let AA be a ring of characteristic pp. a) Let n∈Nn \in \mathbf{N}; prove that pn−1/n!∈Z(p)p^{n-1}/n! \in \mathbf{Z}_{(p)}. Prove that (nm)!/n!(m!)n(nm)!/n!(m!)^n is an integer for n,mn, m in N\mathbf{N}. b) For n∈Nn \in \mathbf{N}, let γn:V1(A)→W(A)\gamma_n : V_1(A) \to W(A) be the mapping defined by

γn(Vx)=1if n=0,γn(Vx)=(pn−1/n!)V(xn)if n⩾1. \begin{align*} \gamma_n(Vx) &= 1 & \text{if } n = 0, \\ \gamma_n(Vx) &= (p^{n-1}/n!) V(x^n) & \text{if } n \geqslant 1. \end{align*}
(i) Prove that one has γn(x)∈V1(A)\gamma_n(x) \in V_1(A) if n⩾1n \geqslant 1 and x∈V1(A)x \in V_1(A). (ii) For x,y∈V1(A)x, y \in V_1(A) and n∈Nn \in \mathbf{N}, one has
γn(x+y)=∑i=0nγi(x)γn−i(y). \gamma_n(x + y) = \sum_{i=0}^n \gamma_i(x) \gamma_{n-i}(y).
(iii) For λ∈W(A), x∈V1(A), n∈N\lambda \in W(A), \ x \in V_1(A), \ n \in \mathbf{N}, one has
γn(λx)=λnγn(x). \gamma_n(\lambda x) = \lambda^n \gamma_n(x).
(iv) For x∈V1(A)x \in V_1(A) and n,mn, m in N\mathbf{N}, one has
γn(x)γm(x)=(m+nn)γn+m(x) \gamma_n(x) \gamma_m(x) = \binom{m+n}{n} \gamma_{n+m}(x)
and
γn(γm(x))=(nm)!n!(m!)nγnm(x). \gamma_n(\gamma_m(x)) = \frac{(nm)!}{n!(m!)^n} \gamma_{nm}(x).
(v) One has Fγn(Vx)=(pn/n!)xnF \gamma_n(Vx) = (p^n/n!) x^n and γn(px)=(pn/n!)xn\gamma_n(px) = (p^n/n!) x^n, for x∈W(A)x \in W(A).

Solution

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