a) The $A$-modules $I(\mathfrak{p})$ are indecomposable.

b) Let $I$ be an indecomposable injective $A$-module; the set $\mathrm{Ass}(I)$ consists of a single element.

c) The mapping $\mathfrak{p} \mapsto \mathrm{cl}(I(\mathfrak{p}))$ is a bijection of $\mathrm{Spec}(A)$ onto $\mathscr{J}(A)$. The inverse bijection associates to an element $I$ of $\mathscr{J}(A)$ the unique element of $\mathrm{Ass}(I)$.

Let $\mathfrak{p} \in \mathrm{Spec}(A)$; let us prove that the module $I(\mathfrak{p})$ is indecomposable. By A, X, p. 21, Corollary 2, it is enough to prove that if $a$ and $b$ are ideals of $A$ containing $\mathfrak{p}$ and distinct from $\mathfrak{p}$, the ideal $a \cap b$ is distinct from $\mathfrak{p}$; now if $a$ is an element of $a - \mathfrak{p}$ and $b$ an element of $b - \mathfrak{p}$, the product $ab$ belongs to $(a \cap b) - \mathfrak{p}$.

Let $I$ be an indecomposable injective $A$-module, and let $\mathfrak{p}, \mathfrak{q}$ be elements of $\mathrm{Ass}(I)$. Then $I$ contains a submodule $M$ isomorphic to $A/\mathfrak{p}$ and a submodule $N$ isomorphic to $A/\mathfrak{q}$. One has $M \cap N \neq 0$ (A, X, p. 21, Proposition 14); for every nonzero element $x$ of $M \cap N$, one has $\mathfrak{p} = \mathrm{Ann} x = \mathfrak{q}$. Since $\mathrm{Ass}(I)$ is not empty (IV, § 1, No. 1, Corollary 1 of Proposition 2), it consists of a single element $\mathfrak{p}(I)$.

We have thus defined two mappings $\mathfrak{p} \mapsto \mathrm{cl}(I(\mathfrak{p}))$ from $\mathrm{Spec}(A)$ into $\mathscr{J}(A)$ and $I \mapsto \mathfrak{p}(I)$ from $\mathscr{J}(A)$ into $\mathrm{Spec}(A)$; let us prove that these two mappings are reciprocal bijections. Let $\mathfrak{p} \in \mathrm{Spec}(A)$; then $\mathfrak{p}$ belongs to $\mathrm{Ass}(I(\mathfrak{p}))$, and is therefore the unique element of $\mathrm{Ass}(I(\mathfrak{p}))$. Let $I$ be an indecomposable injective $A$-module, and $\mathfrak{p}$ the unique element of $\mathrm{Ass}(I)$; then $I$ is an injective envelope of $A/\mathfrak{p}$ (A, X, p. 21, Prop. 14). This completes the proof of the proposition.

a) Suppose that the $A$-module $I$ is injective. Then the $A/a^n$-module $I_n$ is injective for every $n \geqslant 0$.

b) Suppose that the ring $A$ is noetherian, and that for every $n \geqslant 0$ the $A/a^n$-module $I_n$ is injective. Then the union of the $I_n$ is an injective $A$-module.

a) The $A/a^n$-module $I_n$ is isomorphic to $\operatorname{Hom}_A(A/a^n, I)$, which is injective (A, X, p. 18, Prop. 11).

b) Let $J$ be the union of the $I_n$, $b$ an ideal of $A$ and $f : b \to J$ an $A$-homomorphism. It is a matter (A, X, p. 16, Prop. 10) of proving that there exists an element $x$ of $J$ such that one has $f(b) = bx$ for every $b \in b$. Since $b$ is of finite type, there exists an integer $n$ such that one has $f(b) \subset I_n$, that is, $f(a^n b) = 0$. By Cor. 2 of Prop. 1 of III, § 3, No. 1, there exists an integer $m \geqslant n$ such that $a^m \cap b \subset a^n b$, hence $f(a^m \cap b) = 0$. Then $f$ induces an $A/a^m$-linear mapping of $b/(a^m \cap b)$ into $I_m$; since the $A/a^m$-module $I_m$ is injective, there exists an element $x$ of $I_m$ such that one has $f(b) = bx$ for every $b \in b$, whence b).

Let $\alpha$ be an ideal of $A$; in what follows we agree to put $\alpha^n = A$ for every integer $n \leq 0$. Let $E$ be an $A$-module. For every $n \in \mathbf{Z}$, let $E_n$ denote the submodule of $E$ consisting of the elements annihilated by $\alpha^n$; let $\mathrm{gr}^\alpha(E)$ be the graded $A$-module of type $\mathbf{Z}$ such that $\mathrm{gr}^\alpha(E)_m = E_{-m+1}/E_{-m}$ for every integer $m$. The module $\mathrm{gr}^\alpha(E)_m$ is zero for $m \geq 1$, and $\mathrm{gr}^\alpha(E)_0$ is identified with $E_1$. Let $\mathrm{gr}(A)$ denote the graded ring associated with $A$ for the $\alpha$-adic filtration: one has $\mathrm{gr}(A)_n = \alpha^n/\alpha^{n+1}$ for every $n \in \mathbf{Z}$. Let $n$ and $m$ be integers. Passing to quotients from the bilinear mapping $(a, x) \mapsto ax$ of $\alpha^n \times E_{-m+1}$ into $E_{-m-n+1}$, one deduces an $A/\alpha$-bilinear mapping

which defines on $\mathrm{gr}^\alpha(E)$ a structure of graded $\mathrm{gr}(A)$-module. For every $n \in \mathbf{Z}$, from the $A/\alpha$-bilinear mapping $\alpha_{n,-n} : \mathrm{gr}(A)_n \times \mathrm{gr}^\alpha(E)_{-n} \longrightarrow E_1$ one deduces an $A/\alpha$-linear mapping $\beta_{E,n} : \mathrm{gr}^\alpha(E)_{-n} \longrightarrow \mathrm{Hom}_{A/\alpha}(\mathrm{gr}(A)_n, E_1)$; the mappings $\beta_{E,n}$ are the components of a homomorphism of graded $A/\alpha$-modules, called canonical,

For $a \in \mathrm{gr}(A)$, $x \in \mathrm{gr}^\alpha(E)$, $\beta_E(x)(a)$ is by definition the component in $\mathrm{gr}^\alpha(E)_0 = E_1$ of the element $ax$ of $\mathrm{gr}^\alpha(E)$. It follows that $\beta_E$ is $\mathrm{gr}(A)$-linear when one endows $\mathrm{Hom}_{\mathrm{gr}_{A/\alpha}}(\mathrm{gr}(A), E_1)$ with the structure of $\mathrm{gr}(A)$-module defined by the formula $(bf)(a) = f(ab)$ for $a, b$ in $\mathrm{gr}(A)$ and $f$ in $\mathrm{Hom}_{\mathrm{gr}_{A/\alpha}}(\mathrm{gr}(A), E_1)$.

Suppose condition (i) is satisfied. The $A/\alpha$-module $E_1$ is injective (Lemma 1, a)), and contains $M$; since every sub-$A/\alpha$-module of $E_1$ is a sub-$A$-module of $E$, $E_1$ is an injective envelope of the $A/\alpha$-module $M$. By Lemma 1, the union of the $E_n$ is an injective sub-$A$-module of $E$ containing $M$, hence equal to $E$. Since $E$ is injective, for every $n \geq 0$ one has an exact sequence

since $\mathrm{Hom}_A(A/\alpha^m, E)$ is identified with $E_m$ for every $m$ and the canonical injection of $\mathrm{Hom}_A(\alpha^n/\alpha^{n+1}, E_1)$ into $\mathrm{Hom}_A(\alpha^n/\alpha^{n+1}, E)$ is bijective, one deduces that the canonical homomorphism $\beta_E$ is bijective, whence (ii).

Suppose (ii) is satisfied. Let $e : M \to I$ be an injective envelope of $M$. Since $I$ is injective, there exists an $A$-linear mapping $\varphi : E \to I$ extending $e$. But $\varphi$ maps $E_n$ into $I_n$ for every $n$, hence induces homomorphisms gr^{a}(\varphi) : \operatorname{gr}^{a}(E) \to \operatorname{gr}^{a}(I) \text{ and } \varphi_1 : E_1 \to I_1 \text{ rendering the diagram commutative}

Since $E_1$ and $I_1$ are injective envelopes of the $A/\alpha$-module $M$, the homomorphism $\varphi_1$ is bijective; since $\beta_E$ and $\beta_I$ are bijective, it follows that $\operatorname{gr}^{a}(\varphi)$ is bijective. This implies, by induction on $n$, that $\varphi$ induces a bijection of $E_n$ onto $I_n$ for every $n \geq 1$; hence $\varphi$ is bijective, which entails (i).

a) There exists on $M$ a unique structure of $\widehat{A}$-module extending the given structure of $A$-module.

b) *The sub-$\widehat{A}$-modules of $M$ are its sub-$A$-modules, and one has $\operatorname{Hom}_A(M, P) = \operatorname{Hom}_{\widehat{A}}(M, P)$ for every $\widehat{A}$-module $P$.

a) Let us identify $\widehat{A}$ with the projective limit of the rings $A/\alpha^n$, and endow $M$ with the discrete topology. Let $a = (a_n)$ be an element of $\widehat{A}$, and $x$ an element of $M$. Since $x$ is annihilated by a power of $\alpha$, the sequence $(a_n x)$ is stationary; let $ax$ denote its limit. The mapping $(a, x) \mapsto ax$ defines on $M$ a structure of $\widehat{A}$-module extending the given structure of $A$-module.

Conversely, suppose given such a structure on $M$; let $a = (a_n)$ be an element of $\widehat{A}$, $x$ an element of $M$ and $m$ an integer such that $\alpha^m x = 0$. For every integer $n$, $a - a_n$ belongs to $\alpha^n$, which is equal to $\alpha^n \widehat{A}$ (III, § 2, n° 12, cor. 2 of prop. 16); hence one has $ax = a_n x$ for $n \geq m$, whence the assertion of uniqueness.

b) It follows from the preceding that one has $Ax = \widehat{A}x$ for every $x \in M$; the $\widehat{A}$-submodules of $M$ are therefore its $A$-submodules. Finally, let $u$ be an $A$-linear homomorphism of $M$ into a $\widehat{A}$-module $P$. Let $a = (a_n)$ be an element of $\widehat{A}$, $x$ an element of $M$, and $m$ an integer such that $\alpha^m x = 0$; one has $\alpha^m u(x) = 0$. Since $a - a_m$ belongs to $\alpha^m \widehat{A}$, one has

so that $u$ is $\widehat{A}$-linear.

a) The $A$-module $I$ is the union of the $I_n$. The injection $A / \mathfrak{p} \to I_1$ extends to an isomorphism of $\kappa(\mathfrak{p})$ onto $I_1$; let us identify $\kappa(\mathfrak{p})$ with $I_1$ by means of this isomorphism. For each integer $n \geqslant 0$, the structure of $A / \mathfrak{p}$-module of $I_{n+1}/I_n$ arises by restriction of scalars from a unique structure of $\kappa(\mathfrak{p})$-vector space; the canonical homomorphism $\beta_{I,-n}: I_{n+1}/I_n \longrightarrow \mathrm{Hom}_{A/\mathfrak{p}}(\mathfrak{p}^n/\mathfrak{p}^{n+1}, \kappa(\mathfrak{p}))$ is an isomorphism of finite-dimensional $\kappa(\mathfrak{p})$-vector spaces.

b) There exists a unique structure of $\widehat{A_p}$-module on $I$ inducing its structure of $A$-module. The canonical homomorphism $\widehat{A_p} \longrightarrow \mathrm{End}_A(I)$ is bijective.

According to A, X, p. 20, Example 1, the $A/\mathfrak{p}$-module $\kappa(\mathfrak{p})$ is an injective envelope of $A/\mathfrak{p}$. It therefore follows from Prop. 2 that $I_1$ identifies with $\kappa(\mathfrak{p})$, that $I$ is the union of the $I_m$, and that for each integer $n \geqslant 0$, $\beta_{I,-n}$ is an isomorphism of $A/\mathfrak{p}$-modules. For every nonzero element $a$ of $A/\mathfrak{p}$, the homothety of ratio $a$ is invertible in $\mathrm{Hom}_{A/\mathfrak{p}}(\mathfrak{p}^n/\mathfrak{p}^{n+1}, \kappa(\mathfrak{p}))$, hence also in $I_{n+1}/I_n$, which completes the proving of a).

Let $s \in A - \mathfrak{p}$. Since the homothety $s_{A/\mathfrak{p}}$ is injective, the trace of $\mathrm{Ker}\, s_I$ on $A/\mathfrak{p}$ is zero, whence it follows that the homothety $s_I$ is injective. Then $s_I$ is a direct factor submodule of $I$ (A, X, p. 19, Cor. 4), hence equal to $I$ since $I$ is indecomposable (No. 1, Prop. 1), so that the homothety $s_I$ is bijective. There therefore exists a unique $A_p$-module structure on $I$ inducing its $A$-module structure; it extends uniquely to a $\widehat{A_p}$-module structure (Lemma 2).

For each integer $n$, one deduces from the canonical ring homomorphism $A_p \longrightarrow \mathrm{End}_A(I)$ an $A$-linear mapping $\alpha_n : A_p/\mathfrak{p}^n A_p \longrightarrow \mathrm{Hom}_A(I_n, I)$. Consider the commutative diagram with exact rows

where $\alpha'_{n+1}$ is the homomorphism induced by $\alpha_{n+1}$. Consider the canonical $\kappa(\mathfrak{p})$-bilinear mapping

(formula (1)). The linear mapping $I_{n+1}/I_n \longrightarrow \mathrm{Hom}_{\kappa(\mathfrak{p})}(\mathfrak{p}^n A_p/\mathfrak{p}^{n+1} A_p, I_1)$ associated with it on the left identifies with $\beta_{I,-n}$, and that associated with it on the right is $\alpha'_{n+1}$. Since $\beta_{I,-n}$ is bijective by a), the same is true of $\alpha'_{n+1}$; one then deduces from the above diagram, by induction on $n$, that $\alpha_n$ is an isomorphism for every $n$. Since $I$ is the union of the $I_n$, the canonical mapping $\mathrm{End}_A(I) \to \varprojlim \mathrm{Hom}_A(I_n, I)$ is bijective; the ring homomorphism $\widehat{A_p} \to \mathrm{End}_A(I)$, which identifies with the projective limit of the mappings $\alpha_n$, is therefore bijective.

a) The canonical mapping J → A_p ⊗_A J is bijective.

b) Let E denote the A/p-module Hom_A(A/p, J). There exists on E a unique structure of κ(p)-vector space extending its structure of A/p-module; the A-module J is isomorphic to I([E:κ(p)]).

Indeed, J is isomorphic to an A-module I^{(c)}, where c is a suitable cardinal (No. 1, th. 1). The corollary follows from the proposition when J = I, and the general case is deduced from it at once.

a) I is a Matlis $\widehat{A}$-module; b) the A-module I is artinian and a cogenerator (A, X, p. 18, def. 3).

Since the A-module I is injective, the A/m_A^n-module I_n is injective for each n (No. 2, lemma 1, a)). As I_n is the set of elements of I annihilated by m_A^n, the $\widehat{A}$-module I is injective (lemma 1, b)). It is indecomposable over $\widehat{A}$ since it is so over A; as it contains the sub-$\widehat{A}$-module I_1 isomorphic to $\kappa_A$, we have $m_{\widehat{A}} \in \operatorname{Ass}_{\widehat{A}}(I)$, hence $\operatorname{Ass}_{\widehat{A}}(I) = \{ m_{\widehat{A}} \}$ (prop. 1), whence a).

Let us prove now that I is artinian. To every sub-A-module M of I, let us associate the graded ideal $a_M$ of gr(A) defined as follows: an element of gr(A)_n belongs to $(\mathfrak{a}_M)_n$ if it is annihilated by all the linear forms $\beta(x)$, where $x$ runs through $((M \cap I_{n+1}) + I_n)/I_n$. Let $M$ and $N$ be submodules of $I$ such that $N \subset M$; then $\mathfrak{a}_M \subset \mathfrak{a}_N$. Suppose $\mathfrak{a}_M = \mathfrak{a}_N$; we have $(M \cap I_{n+1}) + I_n = (N \cap I_{n+1}) + I_n$ for every $n$ since $\beta$ is an isomorphism. By induction on $n$ we deduce from this that $M \cap I_{n+1} = N \cap I_{n+1}$ for every $n$, whence finally $M = N$.

This being so, let $M_0 \supset M_1 \supset \ldots \supset M_n \supset \ldots$ be a decreasing sequence of sub-A-modules of $I$; the increasing sequence $\mathfrak{a}_{M_0} \subset \mathfrak{a}_{M_1} \subset \ldots$ is stationary, since $\mathrm{gr}(A)$ is a finitely generated $\kappa_A$-algebra. The sequence $(M_i)_{i \geq 0}$ is therefore stationary, which implies that the A-module $I$ is artinian. Finally, the A-module $I$ is a cogenerator by virtue of A, X, p. 18, prop. 12.

Let $M$ be an A-module. Recall (A, VIII, § 4, No. 6) that the socle of $M$ is the sum of the simple submodules of $M$, that is to say the set of elements of $M$ annihilated by $m_A$; it is a $\kappa_A$-vector space, canonically isomorphic to $\mathrm{Hom}_A(\kappa_A, M)$.

(i) $M$ is artinian; (ii) every element of $M$ is annihilated by a power of $m_A$, and the socle of $M$ is of finite dimension over $\kappa_A$; (iii) there exist an integer $n \geq 0$ and an injective a-linear mapping of $M$ into $I^n$.

When these conditions are satisfied, every injective envelope of $M$ is isomorphic to $I^s$, where $s$ is the dimension over $\kappa_A$ of the socle of $M$.

(iii) $\Rightarrow$ (i): this is clear since the a-module $I$ is artinian (Lemma 3).

(i) $\Rightarrow$ (ii): suppose that $M$ is artinian. Let $x \in M$; the decreasing sequence of submodules $m_A^n x$ of $M$ is stationary. Let $n$ be an integer such that $m_A^{n+1} x = m_A^n x$; Nakayama's lemma implies that $m_A^n x = 0$. On the other hand, the socle of $M$ is artinian as an a-module, hence also as a $\kappa_A$-vector space, which means that it is of finite dimension.

(ii) $\Rightarrow$ (iii): suppose condition (ii) satisfied; let $e : M \to J$ be an injective envelope of $M$. We have $\mathrm{Ass}(M) \subset \{m_A\}$, hence $\mathrm{Ass}(J) \subset \{m_A\}$ (No. 1, Remark 2), and $J$ is isomorphic to $I^{(c)}$ for a cardinal $c$ (No. 1, Theorem 1). Let $x$ be a nonzero element of $J$ annihilated by $m_A$; since the a-module $Ax$ is simple and its intersection with $e(M)$ is not reduced to 0, $x$ belongs to $e(M)$. Thus $e$ induces an isomorphism of the socle of $M$ onto that of $J$; consequently the socle of $M$ is of dimension $c$, which proves (iii) as well as the last assertion.

Let $M$ be an artinian $\widehat{A}$-module; every element of $M$ is annihilated by a power of $m_{\widehat{A}}$, hence by a power of $m_A$. By Lemma 2 of No. 2, the sub-A-modules of $M$ are its sub-$\widehat{A}$-modules, hence $M$ is artinian as an A-module.

Let us now fix a Matlis $A$-module I. For every $A$-module $M$, let us denote by $D_A(M)$ the $\widehat{A}$-module

The $\widehat{A}$-module $D_A(A)$ is canonically identified with $I$, the $\widehat{A}$-module $D_A(I)$ with $\widehat{A}$ ($n^\circ 2$, prop. 3), and the $\widehat{A}$-module $D_A(\kappa_A)$ with $I_1$ (loc. cit.).

For every $A$-linear mapping $f : M \to N$, we shall denote by

the $\widehat{A}$-linear mapping $\operatorname{Hom}_A(f, 1_I)$. Since the $A$-module $I$ is injective, the sequence $(D_A(g), D_A(f))$ is exact for every exact sequence $(f, g)$ of $A$-linear mappings.

We shall apply these definitions to the ring $\widehat{A}$ endowed with the Matlis module I (lemma 3, a)) ; for every $\widehat{A}$-module $P$, $D_{\widehat{A}}(P)$ is therefore the $\widehat{A}$-submodule $\operatorname{Hom}_{\widehat{A}}(P, I)$ of $D_A(P)$. It amounts to the same thing to say that $P$ is artinian as an $A$-module or as a $\widehat{A}$-module (lemma 5) ; if this is the case, one has $D_{\widehat{A}}(P) = D_A(P)$ (loc. cit.).

Let $M$ be an $A$-module. For $m \in M$, the mapping $f \mapsto f(m)$ of $D_A(M)$ into $I$ is $\widehat{A}$-linear ; let us denote it by $\alpha_M(m)$. Thus one defines an $A$-linear homomorphism

We denote by $\widehat{\alpha}_M : \widehat{A} \otimes_A M \longrightarrow D_{\widehat{A}}(D_A(M))$ the $\widehat{A}$-linear mapping deduced from $\alpha_M$.

a) In order that $M$ be artinian, it is necessary and sufficient that the $\widehat{A}$-module $D_A(M)$ be of finite type. When this is so, the homomorphism $\alpha_M$ is bijective.

b) For $M$ to be of finite type, it is necessary and sufficient that $D_A(M)$ be artinian (as an $A$-module or as a $\widehat{A}$-module). In this case the homomorphism $\widehat{\alpha}_M$ is an isomorphism.

c) For $M$ to be of finite length, it is necessary and sufficient that $D_A(M)$ be of finite length (as an $A$-module or as a $\widehat{A}$-module). In this case $\alpha_M$ is an isomorphism of $M$ onto $D_A(D_A(M))$, and one has $\operatorname{long}_A(D_A(M)) = \operatorname{long}_A(M)$.

Let us first prove that the homomorphism $\alpha_M$ is injective for every $A$-module $M$. Let $m$ be a nonzero element of $M$; its annihilator is contained in $m_A$. There therefore exists a surjective $A$-homomorphism of $Am$ onto $\kappa_A$, and consequently a nonzero homomorphism of $Am$ into $I$. Since $I$ is injective, this extends to a homomorphism $f : M \to I$ such that $f(m) \neq 0$. This proves the injectivity of $\alpha_M$.

Suppose that the $A$-module $M$ is artinian. By lemma 4, there exist an integer $r$ and an injective $A$-linear mapping $f : M \to I^r$. The homomorphism $D_A(f) : D_A(I^r) \to D_A(M)$ is then surjective; since $D_A(I^r)$ is identified with $\widehat{A}^r$, this proves that the $\widehat{A}$-module $D_A(M)$ is of finite type. Similarly, if $M$ is of finite type, there exist an integer $n$ and a surjective homomorphism $u : A^n \to M$; the homomorphism $D_A(u) : D_A(M) \to I^n$ is injective, so that $D_A(M)$ is artinian (as an $A$-module or as a $\widehat{A}$-module).

Suppose that $M$ is artinian. There exist an integer $r$ and an injective $A$-linear mapping $f : M \to I^r$; since $I$ is artinian (lemma 3), the $A$-module $\operatorname{Coker}(f)$ is also so, and one can find an integer $s$ and an exact sequence of $A$-modules

Hence one deduces a commutative diagram with exact rows

The $\widehat{A}$-module $D_{\widehat{A}}(D_A(I))$ is identified with $I$ and $\alpha_I$ with the identity mapping; hence $\alpha_{I^r}$ and $\alpha_{I^s}$ are bijective, and analogously so is $\alpha_M$ (A, X, p. 7, Corollary 3).

If the $A$-module $M$ is finitely generated, there exist integers $m$ and $n$ and an exact sequence of $A$-modules

from which one deduces a commutative diagram with exact rows

Since $\widehat{\alpha}_A$ is equal to $1_{\widehat{A}}$, it follows that $\widehat{\alpha}_M$ is an isomorphism.

It remains to prove the equality $\operatorname{long}_A(M) = \operatorname{long}_A(D_A(M))$ when $M$ is of finite length. We may suppose $M \neq 0$; there then exists an exact sequence

whence one deduces an exact sequence

One has $\operatorname{long}_A(M) = \operatorname{long}_A(N) + 1$ and

one concludes by induction on the integer $\operatorname{long}_A(M)$.