Bourbaki

Exercise 17, § 2

Commutative Algebra, IX, p. 74

Let A be a complete discrete valuation ring, whose residue field kk is perfect of characteristic pp and whose field of fractions is of characteristic 0. Let ff be the unique homomorphism from W(k)W(k) into A that induces the identity on the residue field kk. Let finally π\pi be a uniformizer of A.

a) There exists an Eisenstein polynomial P with coefficients in f(W(k))f(W(k)) such that P(π)=0P(\pi) = 0 (use VIII, § 5, No. 4). b) Let e be the valuation of p. For every integer n≥0n \geq 0, denote by UnU_n the multiplicative group 1+mA1 + m_A, and put λ(n)=inf⁡(np,n+e)\lambda(n) = \inf(np, n + e), e1=e/(p−1)e_1 = e/(p - 1). Let u be the class in k∗k^* of pπ−ep\pi^{-e}. Let φ\varphi be the endomorphism of U1U_1 which associates xpx^p to x. We then have, for every integer n≥1n \geq 1, φ(Un)⊂Uλ(n)\varphi(U_n) \subset U_{\lambda(n)}, φ(Un+1)⊂Uλ(n)+1\varphi(U_{n+1}) \subset U_{\lambda(n)+1} and the induced homomorphism φn:Un/Un+1→Uλ(n)/Uλ(n)+1\varphi_n : U_n/U_{n+1} \to U_{\lambda(n)}/U_{\lambda(n)+1} is an isomorphism if n≠e1n \neq e_1. If n=e1n = e_1, φn\varphi_n is injective if and only if the equation xp−1+u=0x^{p-1} + u = 0 has no solution in kk, and surjective if and only if the equation xp+ux=yx^p + ux = y has a solution in kk for every y∈ky \in k. c) The following two conditions are equivalent: (i) A contains p p-th roots of unity; (ii) e1e_1 is an integer and the equation xp−1+u=0x^{p-1} + u = 0 has a solution in kk. d) The topology induced on U1U_1 by that of A coincides with the p-adic topology of U1U_1. e) Let S be the set of integers n≥1n \geq 1 such that A contains pnp^n pnp^n-th roots of unity. Then s=Card(S)s = \mathrm{Card}(S) is finite. f) Let I be the set of integers i prime to p and satisfying 1≤i<e+e11 \leq i < e + e_1. Then we have Card(I)=e\mathrm{Card}(I) = e and the image of N−{0}\mathbf{N} - \{0\} by λ\lambda is N−(I∪{0})\mathbf{N} - (I \cup \{0\}). g) If e1e_1 is an integer and φe1\varphi_{e_1} is not surjective, put I′=I∪{e+e1}I' = I \cup \{e + e_1\}; otherwise, put I′=II' = I. For each integer i∈I′i \in I', let πi\pi_i be an element of A of valuation i. Then the mapping from W(k)I′W(k)^{I'} into U1U_1 which associates ∏i∈I′(1+πi)ai\prod_{i \in I'} (1 + \pi_i)^{a_i} to (ai)i∈I′(a_i)_{i \in I'} is surjective; it is a homomorphism of ZpZ_p-modules. h) Let nn be an integer >e1> e_1 and let I(n)I(n) be the interval [n,λ(n)−1][n, \lambda(n) - 1] of N\mathbf{N}. For every integer i∈I(n)i \in I(n), let πi\pi_i be an element of A of valuation i. Then the mapping from W(k)I(n)W(k)^{I(n)} into UnU_n which associates ∏i∈I(n)(1+πi)ai\prod_{i \in I(n)} (1 + \pi_i)^{a_i} to (ai)i∈I(n)(a_i)_{i \in I(n)} is an isomorphism of ZpZ_p-modules. i) Suppose that k is finite and denote by d the degree over Qp\mathbf{Q}_p of the field of fractions of A. Then the ZpZ_p-module 1+mA1 + m_A is isomorphic to (Z/psZ)×Zpd(\mathbf{Z}/p^s\mathbf{Z}) \times \mathbf{Z}_p^d. (One may use the preceding questions, or else use exerc. 13, p. 73.)

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.