Bourbaki

Proposition 7

Tag 04QJ, permanent. Cite it as [Bourbaki, Tag 04QJ].

Commutative Algebra, chapter V, § 2. The lift of prime ideals

*Let AA be an integrally closed domain, KK its field of fractions, K′K' a Galois extension of KK, G\mathscr{G} its Galois group, A′A' the integral closure of AA in K′K', p′p' a prime ideal of A′A' and p=A∩p′p = A \cap p'. Finally, let LL be a subfield of K′K' containing KK and let p(L)=p′∩Lp(L) = p' \cap L.

(i) The decomposition field (resp. inertia field) *of p′p' with respect to LL is L(KZ)L(K^{\mathbf{Z}}) (resp. L(KT)L(K^T)); *if further LL is a Galois extension of KK, the decomposition field of p(L)p(L) with respect to KK is L∩KZL \cap K^{\mathbf{Z}}.

(ii) *If LL is contained in KZK^{\mathbf{Z}}, A/pA/p and (A′∩L)/p(L)(A' \cap L)/p(L) have the same field of fractions and in the local ring A′∩LA' \cap L corresponding to the prime ideal p(L)p(L), the maximal ideal is generated by pp. Conversely, if these two conditions hold and further ⋂n>0pnAp′=0\bigcap_{n > 0} p^n A'_p = 0, LL is contained in KZK^{\mathbf{Z}}.

(i) If H\mathscr{H} is the subgroup of G\mathscr{G} leaving invariant the elements of LL, clearly the decomposition group (resp. inertia group) of p′p' with respect to LL is GZ∩HG^{\mathbf{Z}} \cap \mathscr{H} (resp. GT∩HG^T \cap \mathscr{H}) and the first assertion follows from Galois theory if K′K' is a finite Galois extension of KK (Algebra, Chapter V, § 10, no. 6, Corollary 1 to Theorem 3); in the general case it follows from the fact that AZA^{\mathbf{Z}} (resp. ATA^T) is the union of the AZ(p′∩Kα)A^{\mathbf{Z}}(p' \cap K_\alpha) (resp. AT(p′∩Kβ)A^T(p' \cap K_\beta)) in the notation of Remark 2: every element x∈K′x \in K' belongs to some KK, and if it is invariant under GZ(p′)∩HG^{\mathbf{Z}}(p') \cap \mathscr{H} (resp. GT(p′)∩HG^T(p') \cap \mathscr{H}) it is also invariant under GZ(p′∩Kβ′)∩HG^{\mathbf{Z}}(p' \cap K_{\beta'}) \cap \mathscr{H} (resp. GT(p′∩Kβ′)∩HG^T(p' \cap K_{\beta'}) \cap \mathscr{H}) for some suitable β′\beta'; hence it belongs by the beginning of the argument to

L(KZ(p′∩Kα))⊂L(KZ)(resp. L(KT(p′∩Kβ))⊂L(KT)). L(K^{\mathbf{Z}}(p' \cap K_{\alpha})) \subset L(K^{\mathbf{Z}}) \quad (\text{resp. } L(K^T(p' \cap K_{\beta})) \subset L(K^T)).

Suppose now that LL is a Galois extension of KK; the restriction to LL of every σ∈GZ\sigma \in G^{\mathbf{Z}} then leaves invariant p(L)=p′∩Lp(L) = p' \cap L and hence belongs to the decomposition group of p(L)p(L) with respect to KK. Conversely, let τ\tau be an automorphism of LL belonging to this group and let σ\sigma be an extension of τ\tau to a KK-automorphism of K′K'; we write q′=σ.p′q' = \sigma.p'. As p′p' and q′q' both lie above p(L)p(L), there exists an automorphism ρ∈H\rho \in \mathscr{H} such that q′=ρ.p′q' = \rho.p', whence ρ−1σ∈GZ\rho^{-1}\sigma \in G^{\mathbf{Z}} and τ\tau is the restriction of ρ−1σ\rho^{-1}\sigma to LL; in other words, the decomposition group of p(L)p(L) with respect to KK is identical with the group of restrictions to LL of the automorphisms σ∈GZ\sigma \in G^{\mathbf{Z}}, which proves the second assertion.

(ii) To say that L⊂KZL \subset K^{\mathbf{Z}} means that H⊃GZ\mathscr{H} \supset G^{\mathbf{Z}} and the assertions of (ii) are therefore special cases of no. 2, Proposition 4 (ii) and (iii) when [K′:K][K':K] is finite. In the general case the argument is as in the proof of Proposition 6.

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

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Cites

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