Theorem 3

a) The A-module I is a Matlis module. For every integer $m \geqslant 0$, $I_m$ is the a-submodule of I consisting of the elements annihilated by $m_A^m$.

b) For every A-module M of finite length, the a-linear mapping $\theta_M : T(M) \to \operatorname{Hom}_A(M, I)$ is bijective.

c) For every a-linear mapping $f : M \to N$ between A-modules of finite length, one has $\theta_M \circ T(f) = \operatorname{Hom}_A(f, 1_I) \circ \theta_N$.

Let us prove c). Let n be an integer and M, N A-modules of finite length annihilated by $m_A^n$. Let $f : M \to N$ be an a-linear mapping. For $u$ in $T(N)$ and x in M, one has by FD 3)

This proves c).

Let us prove b). First consider the particular case $M = A/m_A^n$. Then $T(M)$ is equal by definition to $I_n$. If $a$ is an element of A, of class $\bar{a}$ in M, one has $\varphi_{M,\bar{a}}^n = a 1_M$, whence $T(\varphi_{M,\bar{a}}^n) = a 1_{I_n}$; thus $\theta_M : I_n \to \operatorname{Hom}_A(A/m_A^n, I)$ is the canonical isomorphism which sends an element $x$ of $I_n$ to the mapping $\bar{a} \mapsto a x$. This proves b) in this case.

Now suppose given an exact sequence

of A-modules of finite length annihilated by $m_A^n$. Consider the diagram

it is commutative by the beginning of the proof, and its rows are exact by FD 4). We deduce that $\theta_M$ is bijective if $\theta_P$ and $\theta_N$ are so. Applying this to a presentation

of the $A/\mathfrak{m}_A^n$-module $M$, we deduce that $\theta_M$ is bijective for every $A$-module of finite length annihilated by $\mathfrak{m}_A^n$, whence b).

Let us prove a). It follows from the foregoing applied to the $A$-module $A/\mathfrak{m}_A^n$ that $I_n$ is the set of elements of $I$ which are annihilated by $\mathfrak{m}_A^n$. By FD 5) the $A$-module $I_1 = T(\kappa_A)$ is isomorphic to $\kappa_A$; by Prop. 2 of No. 2, it is enough for us to prove that, for every integer $n \geqslant 0$, the canonical $A$-linear mapping

is bijective. Now from the exact sequence one deduces an exact sequence By composing $T(u)$ with $\theta_{\mathfrak{m}_A^n/\mathfrak{m}_A^{n+1}}$, one thus obtains a surjective homomorphism with kernel $I_n$. By c), $\gamma$ is the composite of the arrows $\theta_{I_{n+1}} : I_{n+1} \to \mathrm{Hom}_A(A/\mathfrak{m}_A^{n+1}, I)$ and $\mathrm{Hom}(u, 1) : \mathrm{Hom}_A(A/\mathfrak{m}_A^{n+1}, I) \to \mathrm{Hom}_A(\mathfrak{m}_A^n/\mathfrak{m}_A^{n+1}, I)$; since $\theta_{I_{n+1}}$ is the linear mapping associated with the multiplication $A/\mathfrak{m}_A^{n+1} \times I_{n+1} \to I$, the isomorphism $I_{n+1}/I_n \to \mathrm{Hom}_A(\mathfrak{m}_A^n/\mathfrak{m}_A^{n+1}, I)$ deduced from $\gamma$ coincides with $\beta$, which completes the proof.