Bourbaki

Proposition 5

Tag 0DB0, permanent. Cite it as [Bourbaki, Tag 0DB0].

Commutative Algebra, chapter X, § 8. Dualité des modules de longueur finie

Let ρ:A→B\rho : A \to B be a local homomorphism of noetherian local rings, such that the residual extension κA→κB\kappa_A \to \kappa_B induced by ρ\rho is of finite degree. Let IAI_A be a Matlis AA-module.

a) Let IBI_B denote the sub-BB-module of HomA(B,IA)\mathrm{Hom}_A(B, I_A) formed by the AA-homomorphisms from BB into IAI_A whose kernel contains a power of mBm_B. Then IBI_B is a Matlis BB-module.

b) Let MM be a BB-module. The canonical mapping

α:HomB(M,HomA(B,IA))⟶HomA(M,IA) \alpha : \mathrm{Hom}_B(M, \mathrm{Hom}_A(B, I_A)) \longrightarrow \mathrm{Hom}_A(M, I_A)

defined by α(u)(m)=u(m)(1)\alpha(u)(m) = u(m)(1) induces a BB-isomorphism of DB(M)=HomB(M,IB)D_B(M) = \mathrm{Hom}_B(M, I_B) onto the sub-BB-module HomAcont(M,IA)\mathrm{Hom}_A^{cont}(M, I_A) of DA(M)=HomA(M,IA)D_A(M) = \mathrm{Hom}_A(M, I_A) consisting of mappings f:M→IAf : M \to I_A such that for every element mm of MM, there exists an integer n⩾0n \geqslant 0 such that f(mBnm)=0f(m^n_B m) = 0.

The above condition on ff means that ff is continuous when IAI_A is given the discrete topology and MM the finest topology inducing on each finitely generated submodule the mBm_B-adic topology, which justifies the notation. Analogously, the condition g∈IBg \in I_B means that gg is continuous when IAI_A is given the discrete topology and BB the mBm_B-adic topology.

Let us prove a). For every BB-module of finite length MM, denote by T(M)T(M) the BB-module HomA(M,IA)\mathrm{Hom}_A(M, I_A); for every BB-linear mapping f:M→Nf : M \to N between BB-modules of finite length, denote by T(f):T(N)→T(M)T(f) : T(N) \to T(M) the BB-linear mapping HomA(f,1IA)\mathrm{Hom}_A(f, 1_{I_A}). The verification of conditions FD 1) to FD 4) of No. 5 is immediate. Moreover, for every BB-module of finite length NN, one has longA(N[A])=longB(N) [κB:κA]\mathrm{long}_A(N_{[A]}) = \mathrm{long}_B(N) \ [\kappa_B : \kappa_A]; since one has longA(T(M))=longA(M)\mathrm{long}_A(T(M)) = \mathrm{long}_A(M) (No. 3, th. 2), one deduces that longB(T(M))=longB(M)\mathrm{long}_B(T(M)) = \mathrm{long}_B(M), which implies FD 5). We may therefore apply theorem 3 of No. 5; one has

T(B/mBn)=HomA(B/mBn,IA), T(B/m_B^n) = \mathrm{Hom}_A(B/m_B^n, I_A),

so that the Matlis BB-module lim←⁡T(B/mBn)\varprojlim T(B/m_B^n) is identified with the sub-BB-module IBI_B of HomA(B,IA)\mathrm{Hom}_A(B, I_A), which proves a).

Let us prove b). The mapping α\alpha is the inverse of the canonical isomorphism

β:HomA(M,1A)⟶HomB(M,HomA(B,IA)) \beta : \mathrm{Hom}_A(M, 1_A) \longrightarrow \mathrm{Hom}_B(M, \mathrm{Hom}_A(B, I_A))

which associates to v∈HomA(M,IA)v \in \mathrm{Hom}_A(M, I_A) the map v′v' of MM into HomA(B,IA)\mathrm{Hom}_A(B, I_A) such that v′(m)(b)=v(bm)v'(m)(b) = v(bm) (A, II, p. 74, prop. 1). In order that v′v' take its values in IBI_B, it is necessary and sufficient that vv belong to HomAcont(M,IA)\mathrm{Hom}_A^{cont}(M, I_A), whence b).

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

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Translated with machine help, not yet reviewed by a person.