Bourbaki

Example 2

Tag 0CQ1, permanent. Cite it as [Bourbaki, Tag 0CQ1].

Commutative Algebra, chapter VIII, § 5. Anneaux locaux réguliers

Put A = \mathbf{Z} and L = R_{pf}(\mathbf{Q}) where p is a prime number and f an integer \geqslant 1 (cf. A, V, p. 78). Thus L = \mathbf{Q}(\zeta) with \zeta = \exp(2\pi i/p^f). Let B denote the integral closure of \mathbf{Z} in L and P the polynomial of \mathbf{Z}[T] such that

P(T−1)=(Tpf−1)/(Tpf−1−1). P(T - 1) = (T^{p^f} - 1)/(T^{p^{f-1}} - 1) .

Put d = p^f - p^{f-1}. One has P(\zeta - 1) = 0, P(0) = p and

P(T−1)≡(T−1)dmod  .pZ[T], P(T - 1) \equiv (T - 1)^d \mod. p\mathbf{Z}[T] ,

whence P(T) \equiv T^d. It follows that P is an Eisenstein polynomial for p\mathbf{Z}_{(p)}. There is therefore a unique prime ideal q of B above p\mathbf{Z}, and one has B_q = \mathbf{Z}_{(p)}[\zeta] ; moreover, \zeta - 1 is a uniformizer of B_q and one has

[L:K]=d=pf−pf−1. [L : K] = d = p^f - p^{f-1} .

If v is the unique normalised valuation of \mathbf{Q}(\zeta) such that v(p) > 0, one has v(p) = d. Moreover, the field B/q has p elements. One can prove (cf. p. 96, exerc. 13) that B is equal to \mathbf{Z}[\zeta].

\* 5. Structure of complete regular noetherian local rings

In this number, we use definitions and results from chapter IX.

Let A be a complete regular noetherian local ring; let pp denote the characteristic of its residue field κA\kappa_A, and distinguish two cases.

A) p=0p = 0. Then (IX, § 3, no 3, th. 1), there exists a subfield K of A such that the canonical projection of A onto κA\kappa_A induces an isomorphism of K onto κA\kappa_A. Applying then cor. 3 of th. 1 of no 2 to the K-algebra A, one deduces:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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