Bourbaki

Example 4

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

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

If AA is a Gorenstein ring, for example a regular ring, one can apply Example 3 by taking Ω=A\Omega = A and d=dim⁡(A)d = \dim(A) (§ 3, No. 7, Prop. 11).

For every integer n⩾0n \geqslant 0, set In=T(A/mAn)I_n = T(A/\mathfrak{m}_A^n). For m⩾nm \geqslant n, denote by pmn:A/mAm⟶A/mAnp_{mn} : A/\mathfrak{m}_A^m \longrightarrow A/\mathfrak{m}_A^n the canonical surjection and by imn:T(A/mAn)⟶T(A/mAm)i_{mn} : T(A/\mathfrak{m}_A^n) \longrightarrow T(A/\mathfrak{m}_A^m) the AA-linear mapping T(pmn)T(p_{mn}). It is injective by FD 4), and one has imn∘inp=impi_{mn} \circ i_{np} = i_{mp} for m⩾n⩾pm \geqslant n \geqslant p by FD 3). Let I=lim←⁡T(A/mAn)I = \varprojlim T(A/\mathfrak{m}_A^n) be the inductive limit AA-module of the system ((In),(imn))((I_n), (i_{mn})). For n⩾0n \geqslant 0, the canonical mapping In→II_n \to I is injective; we identify InI_n with its image in II, so that II is the increasing union of the InI_n.

Let M be an A-module of finite length, and n an integer ⩾0\geqslant 0 such that mAnM=0m_A^n M = 0. For x∈Mx \in M, let us denote by φM,xn\varphi_{M,x}^n the a-linear mapping of A/mAnA/m_A^n into M which maps the class of 1 to x. The mapping T(φM,xn):T(M)→InT(\varphi_{M,x}^n) : T(M) \to I_n is a-linear, and one has T(φM,axn)=aT(φM,xn)T(\varphi_{M,a x}^n) = a T(\varphi_{M,x}^n) for a∈Aa \in A by FD 1). Hence the mapping (x,u)↦T(φM,xn)(u)(x, u) \mapsto T(\varphi_{M,x}^n)(u) of M×T(M)M \times T(M) into I is A-bilinear. It does not depend on the choice of the integer n: in fact, for every integer q⩾nq \geqslant n and every element x of M, one has φM,xq=φM,xn∘pqn\varphi_{M,x}^q = \varphi_{M,x}^n \circ p_{q n}, whence by FD 3) T(φM,xq)=iqn∘T(φM,xn)T(\varphi_{M,x}^q) = i_{q n} \circ T(\varphi_{M,x}^n). We deduce an a-linear mapping

θM:T(M)⟶Hom⁡A(M,I) \theta_M : T(M) \longrightarrow \operatorname{Hom}_A(M, I)
satisfying θM(u)(x)=T(φM,x)(u)\theta_M(u)(x) = T(\varphi_{M,x})(u) for u∈T(M), x∈Mu \in T(M),\ x \in M.

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

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