§ 2. Primary decomposition · Commutative Algebra

2. Primary decomposition

1. Primary s bmodules

Proposition 1

Let $A$ be a Noetherian ring and $M$ an $A$ -module. The following conditions are equivalent: (a) $\operatorname{Ass}(M)$ is reduced to a single element.

(b) $M \neq 0$ and every homothety of $M$ is either injective or almost nilpotent ( $§ 1$ , no. 4). If these conditions are fulfilled and $p$ is the set of $a \in A$ such that the homothety $a_M$ is almost nilpotent, then $\operatorname{Ass}(M) = \{p\}$ .

This follows immediately from $§ 1$ , no. 4, Corollary to Proposition 9 and no. 1, Corollary 2 to Proposition 2.

Definition 1

Let $A$ be a Noetherian ring, $N$ an $A$ -module and $Q$ a submodule of $N$ . If the module $M = N/Q$ satisfies the conditions of Proposition 1, $Q$ is called p-primary with respect to $N$ (or in $N$ ).

When there is no ambiguity, we shall simply say that $Q$ is "p-primary" or "primary"; clearly for every submodule $N' \neq Q$ of $N$ containing $Q$ , $Q$ is p-primary in $N'$ .

Definition 1 applies in particular to the case $N = A$ ; the submodules of $N$ are then the ideals of $A$ and hence an ideal $q$ of $A$ is called primary if $\operatorname{Ass}(A/q)$ has a single element or, what amounts to the same, if $A \neq q$ and every divisor of zero in the ring $A/q$ is nilpotent. If $q$ is p-primary, it follows from Definition 1 that $p$ is the radical (Chapter II, $§ 2$ , no. 6) of the ideal $q$ .

Remark

Let $q$ be a p-primary submodule of an $A$ -module $N$ . If $N/Q$ is finitely generated, there exists an integer $k > 0$ such that $p^k N \subset Q$ by $§ 1$ , no. 4, Proposition 9.

Examples

Example 1

If $p$ is a prime ideal of $A$ , $p$ is p-primary ( $§ 1$ , no. 1, Proposition 1).

Example 2

Let $q$ be an ideal of $A$ such that there exists a single prime ideal $m$ (necessarily maximal) containing $q$ ; then, if $M$ is an $A$ -module such that $qM \neq M$ , $qM$ is m-primary with respect to $M$ . For every element of $\operatorname{Ass}(M/qM)$ contains $q$ , hence is equal to $m$ and $\operatorname{Ass}(M/qM) \neq 0$ ( $§ 1$ , no. 1, Corollary 1 to Proposition 2). In particular $q$ is an m-primary ideal in $A$ .

Example 3

Let $m$ be a maximal ideal of $A$ ; the m-primary ideals are then the ideals $q$ of $A$ for which there exists an integer $n \geq 1$ such that $m^n \subset q \subset m$ . For if $m^n \subset q \subset m$ , $m$ is the only prime ideal containing $q$ (Chapter 11, $§ 1$ , no. 1, Corollary to Proposition 1) and the conclusion follows from Example 2; conversely, if $q$ is m-primary, $m$ is the radical of $q$ and there therefore exists $n \geq 1$ such that $m^n \subset q$ (Chapter II, $§ 2$ , no. 6, Proposition 15).

Example 4

In a principal ideal domain $A$ , the primary ideals are $(0)$ and the ideals of the form $Ap^n$ , where $p$ is an extremal element and $n \geq 1$ ; this follows immediately from Example 3.

Example 5

The powers of any prime ideal are not necessarily primary ideals (Exercise 1). On the other hand, there exist primary ideals which are not powers of prime ideals (Exercise 1).

Proposition 2

Let M be a module over a Noetherian ring A, p a prime ideal of A and $(\mathscr{Q}_i)_{i \in I}$ a non-empty finite family of submodules of M which are p-primary with respect to M. Then $\bigcap_{i \in I} \mathscr{Q}_i$ is p-primary with respect to M.

$M / (\bigcap_{i \in I} \mathscr{Q}_i)$ is isomorphic to a submodule $\neq 0$ of the direct sum $\bigoplus_{i \in I} (M / \mathscr{Q}_i)$ . Now

\operatorname{Ass}\left( \bigoplus_{i \in I} (M / \mathscr{Q}_i) \right) = \bigcup_{i \in I} \operatorname{Ass}(M / \mathscr{Q}_i) = \{p\}

(§ 1, no. 1, Corollary 1 to Proposition 3). Hence $\operatorname{Ass}\left(M / (\bigcap_{i \in I} \mathscr{Q}_i)\right) = \{p\}$ (§ 1, no. 1, Proposition 3 and Corollary 1 to Proposition 2).

Proposition 3

Let A be a Noetherian ring, S a multiplicative subset of A, p a prime ideal of A, M an A-module, N a submodule of M and $i = i_A^S$ the canonical mapping of M to $S^{-1}M$ .

(i) Suppose that $p \cap S \neq \varnothing$ . If N is p-primary with respect to M, then $S^{-1}N = S^{-1}M$ .

(ii) Suppose that $p \cap S = \varnothing$ . For N to be p-primary with respect to M, it is necessary and sufficient that N be of the form $i_1(N')$ , where $N'$ is a sub- $S^{-1}A$ -module of $S^{-1}M$ which is $(S^{-1}p)$ -primary with respect to $S^{-1}M$ ; then $N' = S^{-1}N$ .

(i) If $p \cap S \neq \varnothing$ and N is p-primary with respect to M, then

\operatorname{Ass}_{S^{-1}A}(S^{-1}(M/N)) = \varnothing

(§ 1, no. 2, Corollary to Proposition 5) and hence $S^{-1}(M/N) = 0$ (§ 1, no. 1, Corollary 1 to Proposition 2), whence $S^{-1}M/S^{-1}N = 0$ .

(ii) Suppose that $p \cap S = \varnothing$ . If N is p-primary with respect to M, then $\operatorname{Ass}_{S^{-1}A}(S^{-1}(M/N)) = \{S^{-1}p\}$ (§ 1, no. 2, Corollary to Proposition 5) and hence the submodule $N' = S^{-1}N$ of $S^{-1}M$ is $(S^{-1}p)$ -primary; moreover, if $s \in S$ and $m \in M$ are such that $sm \in N$ , then $m \in N$ , for the homothety with ratio s in M/N is injective, whence $N = i_1(N')$ (Chapter II, § 2, no. 4, Proposition 10). Conversely, let $N'$ be a submodule of $S^{-1}M$ which is $(S^{-1}p)$ -primary with respect to $S^{-1}M$ ; let us write $N = i_1(N')$ ; then $N' = S^{-1}N$ (Chapter II, § 2, no. 4, Proposition 10) and $\operatorname{Ass}_{S^{-1}A}(S^{-1}(M/N)) = \operatorname{Ass}_{S^{-1}A}((S^{-1}M)/N') = \{S^{-1}p\}$ . As the canonical mapping $M/N \to S(M/N)$ is injective, no prime ideal of A associated with M/N meets S (§ 1, no. 2, Proposition 6); it follows that $\operatorname{Ass}(M/N) = \{p\}$ (§ 1, no. 2, Corollary to Proposition 5), so that N is p-primary with respect to M.

2. The existence of a primary decomposition

Definition 2

Let $A$ be a Noetherian ring, $M$ an $A$ -module and $N$ a submodule of $M$ . A finite family $(\mathscr{Q}_i)_{i \in I}$ of submodules of $M$ which are primary with respect to $M$ and such that $N = \bigcap_{i \in I} \mathscr{Q}_i$ is called a primary decomposition of $N$ in $M$ .

Example

Let us take $A = \mathbf{Z}$ , $M = \mathbf{Z}$ , $N = n\mathbf{Z}$ for some integer $n > 0$ . If $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ is the decomposition of $n$ into prime factors,

n\mathbf{Z} = (p_1^{\alpha_1}\mathbf{Z}) \cap \ldots \cap (p_k^{\alpha_k}\mathbf{Z})

is a primary decomposition of

n\mathbf{Z}

\mathbf{Z}

by Example 4 of no. 1.

By an abuse of language, the relation $N = \bigcap_{i \in I} \mathscr{Q}_i$ is called a primary decomposition of $N$ in $M$ . It amounts to the same to say that $\{0\} = \bigcap_{i \in I} (\mathscr{Q}_i/N)$ is a primary decomposition of $\{0\}$ in $M/N$ . If $(\mathscr{Q}_i)_{i \in I}$ is a primary decomposition of $N$ in $M$ , the canonical mapping from $M/N$ to $\bigoplus_{i \in I} (M/\mathscr{Q}_i)$ is injective. Conversely let $N$ be a submodule of $M$ and $f$ an injective homomorphism from $M/N$ to a finite direct sum $P = \bigoplus_{i \in I} P_i$ , where each set $\operatorname{Ass}(P_i)$ is reduced to a single element $p_i$ ; let $f_i$ be the homomorphism $M/N \to P_i$ obtained by taking the composition off with the projection $P \to P_i$ , and let $Q_i/N$ be the kernel off.; then the $Q_i$ distinct from $M$ are primary with respect to $M$ (no. 1, Definition 1) and $N = \bigcap_{i \in I} Q_i$ . Moreover, $\operatorname{Ass}(M/N) \subset \bigcup_{i \in I} \{p_i\}$ by virtue of § 1, no. 1, Proposition 3.

Theorem 1

Let $M$ be a finitely generated module over a Noetherian ring and let $N$ be a submodule of $M$ . There exists a primary decomposition of $N$ of the form

N = \bigcap_{p \in \operatorname{Ass}(M/N)} Q(p)

where, for all

p \in \operatorname{Ass}(M/N)

Q(p)

p

-primary with respect to

M

We may replace $M$ by $M/N$ and therefore suppose that $N = 0$ . By § 1, no. 4, Corollary to Theorem 2, $\operatorname{Ass}(M)$ is finite; by § 1, no. 1, Proposition 4, there exists, for each $p \in \operatorname{Ass}(M)$ , a submodule $Q(p)$ of $M$ such that $\operatorname{Ass}(M/Q(p)) = \{p\}$ and $\operatorname{Ass}(Q(p)) = \operatorname{Ass}(M) - \{p\}$ . Let us write $P = \bigcap_{p \in \operatorname{Ass}(M)} Q(p)$ ; for all $p \in \operatorname{Ass}(M)$ , $\operatorname{Ass}(P) \subset \operatorname{Ass}(Q(p))$ and hence $\operatorname{Ass}(P) = \varnothing$ , which implies $P = 0$ (§ 1, no. 1, Corollary 1 to Proposition 2) and therefore proves the theorem.

3. Uniqueness properties in the primary decomposition

Definition 3

Let $M$ be a module over a Noetherian ring and $N$ a submodule of $M$ . A primary decomposition $N = \bigcap_{i \in I} Q_i$ of $N$ in $M$ is called reduced if the following conditions are fulfilled:

(a) there exists no index $i \in I$ such that $\bigcap_{j \neq i} Q_j \subset Q_i;$

(b) if $\operatorname{Ass}(M/Q_i) = \{p_i\}$ , the $p_i$ (i $\in I$ ) are distinct.

From every primary decomposition $N = \bigcap_{i \in I} Q_i$ of $N$ in $M$ a reduced primary decomposition of $M$ in $N$ can be deduced as follows: let $J$ be a minimal element of the set of subsets $I'$ of $I$ such that $N = \bigcap_{i \in I'} Q_i$ . Clearly $(Q_i)_{i \in J}$ satisfies condition (a). Then let $\Phi$ be the set of $p_i$ for $i \in J$ ; for all $p \in \Phi$ , let $H(p)$ be the set of $i \in J$ such that $p_i = p$ and let $Q(p) = \bigcap_{i \in H(p)} Q_i$ ; it follows from Proposition 2 of no. 1 that $Q(p)$ is $p$ -primary with respect to $M$ ; further $N = \bigcap_{p \in \Phi} Q(p)$ and the family $Q((p))_{p \in \Phi}$ is therefore a reduced primary decomposition of $N$ in $M$ .

We shall see that the primary decomposition defined in the proof of Theorem 1 of no. 2 is reduced; this follows from the following proposition:

Proposition 4

Let* $M$ *be a module over a Noetherian ring, N a submodule of M, $N = \bigcap_{i \in I} Q_i$ a primary decomposition of N in M and, for all $i \in I$ , let $\{p_i\} = \operatorname{Ass}(M/Q_i)$ . For this decomposition to be reduced, it is necessary and sufficient that the $p_i$ be distinct and belong to $\operatorname{Ass}(M/N)$ ; *then

(2)

\operatorname{Ass}(M/N) = \bigcup_{i \in I} \{p_i\}

(3)

\operatorname{Ass}(Q_i/N) = \bigcup_{j \neq i} \{p_j\} \quad \text{for all } i \in I.

If the condition of the statement is fulfilled, $N = \bigcap_{j \neq i} Q_j$ cannot hold, for we would deduce that $\operatorname{Ass}(M/N) \subset \bigcup_{j \neq i} \{p_j\}$ (§ 1, no. 1, Corollary 2 to Proposition 3) contrary to the hypothesis; the primary decomposition $(Q_i)_{i \in I}$ of $N$ is then certainly reduced. Conversely, $\operatorname{Ass}(M/N) \subset \bigcup_{i \in I} \{p_i\}$ always holds (§ 1, no. 1, Corollary 2 to Proposition 3); on the other hand, for all $i \in I$ , let us write $P_i = \bigcap_j Q_j$ ; then $P_i \cap Q_i = N$ and $P_i \neq N$ if $(Q_i)_{i \in I}$ is reduced, hence $P_i/N$ is non-zero and is isomorphic to the submodule $(P_i + Q_i)/Q_i$ of $M/Q_i$ , whence $\{p_i\} = \operatorname{Ass}(P_i/N)$ (§ 1, no. 1, Proposition 3 and Corollary 1 to Proposition 2); as $P_i/N \subset M/N, p_i \in \operatorname{Ass}(M/N)$ , which completes the proof of the necessity of the condition in the statement and formula (2). Finally, as $N = \bigcap_{j \neq i} (Q_j \cap Q_i), \operatorname{Ass}(Q_i/N) \subset \bigcup_{j \neq i} \operatorname{Ass}(Q_i/(Q_j \cap Q_i))$ (§ 1, no. 1, Corollary 2 to Proposition 3); but $Q_i/(Q_j \cap Q_i)$ is isomorphic to the submodule $(Q_i + Q_j)/Q_j$ of $M/Q_j$ , hence $\operatorname{Ass}(Q_i/(Q_j \cap Q_i)) \subset \{p_j\}$ and

\operatorname{Ass}(Q_i/N) \subset \bigcup_{j \neq i} \{p_j\};

Corollary

Let $A$ be a Noetherian ring, $M$ an $A$ -module, $N$ a submodule of $M$ and $(Q_i)_{i \in I}$ a primary decomposition of $N$ in $M$ . Then $\mathrm{Card}(I) \geq \mathrm{Card}(\mathrm{Ass}(M/N))$ ; for $(Q_i)_{i \in I}$ to be a reduced primary decomposition, it is necessary and sufficient that $\mathrm{Card}(I) = \mathrm{Card}(\mathrm{Ass}(M/N))$ .

It follows from the remarks preceding Proposition 4 that there exists a reduced primary decomposition $(R_j)_{j \in J}$ of $N$ in $M$ such that $\mathrm{Card}(J) \leq \mathrm{Card}(I)$ ; the first assertion then follows from the second and the latter is a consequence of Proposition 4.

Proposition 5

Let $A$ be a Noetherian ring, $M$ an $A$ -module, $N$ a submodule of $M$ , $N = \bigcap_{i \in I} Q_i$ a reduced primary decomposition of $N$ in $M$ and, for all $i \in I$ , let $\{p_i\} = \mathrm{Ass}(M/Q_i)$ . If $p_i$ is a minimal element of $\mathrm{Ass}(M/N)$ , $Q_i$ is equal to the saturation on $N$ with respect to $p_i$ (Chapter 11, § 2, no. 4) (cf. Exercise 2).

We can obviously restrict our attention to the case where $N = 0$ , replacing if need be $M$ by $M/N$ . If $p_i$ is minimal in $\mathrm{Ass}(M)$ , the set of elements of $\mathrm{Ass}(M)$ which do not meet $A - p_i$ reduces to $p_i$ ; the proposition then follows from formula (3) above and § 1, no. 2, Proposition 6, the kernel of the canonical mapping $M \to M_0$ , being equal to the saturation of 0 with respect to $p_i$ (Chapter 11, § 2, no. 4).

Remark

The prime ideals $p_i \in \mathrm{Ass}(M/N)$ which are not minimal elements of this set are sometimes called the immersed prime ideals associated with $M/N$ ; if $M/N$ is finitely generated, for $p_0 \in \mathrm{Ass}(M/N)$ to be immersed, it is necessary and sufficient that $V(p_0)$ be not an irreducible component of $\mathrm{Supp}(M/N)$ (Chapter II, § 4, no. 3, Corollary 2 to Proposition 14); if $(Q(p))_{p \in \mathrm{Ass}(M/N)}$ and $(Q'(p))_{p \in \mathrm{Ass}(M/N)}$ are two reduced primary decompositions of $N$ in $M$ , it may be that $Q'(p_0) \neq Q(p_0)$ (Exercise 24 (c)); a canonical reduced primary decomposition of $N$ in $M$ may always be defined by imposing supplementary condition on the primary submodules which appear in it (Exercise 4).

4. The localization of a primary decomposition

Given a submodule $N$ of a module $M$ over a Noetherian ring $A$ , to simplify we shall denote by $D_I(M/N)$ , in this no., the set of reduced primary decompositions of $N$ in $M$ whose indexing set is $I$ (equipotent to $\mathrm{Ass}(M/N)$ ).

Proposition 6

Let $A$ be a Noetherian ring, $M$ an $A$ -module, $N$ a submodule of $M$ and $I = \mathrm{Ass}(M/N)$ . Let $S$ be a multiplicative subset of $A$ and $J$ the subset of $I$ consisting of the indices i such that $S \cap p_i = \varnothing$ . Let $N'$ be the saturation of $N$ with respect to $S$ in $M$ . Then:

(i) If $(Q_i)_{i \in I}$ is an element of $D_I(M/N)$ , the family $(Q_i)_{i \in J}$ is an element of $D_J(M/N')$ and the family $(S^{-1}Q_i)_{i \in J}$ is an element of $D_J(S^{-1}M/S^{-1}N)$ .

(ii) The mapping $(Q_i)_{i \in J} \to (S^{-1}Q_i)_{i \in J}$ is a bijection of $D_J(M/N')$ onto $D_J(S^{-1}M/S^{-1}N)$ .

(iii) If $(Q_i)_{i \in J}$ is an element of $D_J(M/N')$ and $(R_i)_{i \in I}$ an element of $D_I(M/N)$ , the family $(T_i)_{i \in I}$ such that $T_i = Q_i$ for $i \in J$ and $T_i = R_i$ for $i \in I - J$ is an element of $D_I(M/N)$ .

(i) We know (no. 1, Proposition 3) that for $i \in J, S^{-1}Q_i$ is primary for $S^{-1}p_i$ and that for $i \in I - J, S^{-1}Q_i = S^{-1}M$ ; as $S^{-1}N = \bigcap S^{-1}Q_i$ (Chapter II, § 2, no. 4), then also $S^{-1}N = \bigcap_{i \in J} S^{-1}Q_i$ . The $S^{-1}p_i$ for $i \in J$ are distinct and their set is $\operatorname{Ass}(S^{-1}M/S^{-1}N)$ ( $§ 1$ , no. 2, Corollary to Proposition 5); then (Proposition 4) $(S^{-1}Q_i)_{i \in J}$ is a reduced primary decomposition of $S^{-1}N$ . Moreover, $Q_i = (i_M^S)^{-1}(S^{-1}Q_i)$ (no. 1 Proposition 3), hence $N' = (i_M^S)^{-1}(S^{-1}N) = \bigcap_{i \in J} Q_i$ and $(Q_i)_{i \in J}$ is obviously a reduced primary decomposition of $N'$ in $M$ .

(ii) As $S^{-1}N' = S^{-1}N$ , we may replace $N$ by $N'$ , that is suppose that $J = I$ . Let $(P_i)_{i \in I}$ be a reduced primary decomposition of $S^{-1}N$ in $S^{-1}M$ and let us write $Q_i = (i_M^S)^{-1}(P_i)$ ; it follows from no. 1, Proposition 3 that $Q_i$ is primary for $p_i$ ( $i \in I$ ) and $(Q_i)_{i \in I}$ is then a reduced primary decomposition of $N$ in $M$ by virtue of no. 3, Corollary to Proposition 4. Finally, as, for all $i \in I$ and every submodule $Q'_i$ of which $M$ is $p_i$ -primary with respect to $M, Q'_i = (i_M^S)^{-1}(S^{-1}Q'_i)$ by virtue of no. 1, Proposition 3 and the hypothesis $J = I$ , we see that two mappings $D_I(M/N) \to D_I(S^{-1}M/S^{-1}N)$ and $D_I(S^{-1}M/S^{-1}N) \to D_I(M/N)$ have been defined whose compositions are the identities on $D_I(M/N)$ and $D_I(S^{-1}M/S^{-1}N)$ , which proves (ii).

(iii) By virtue of (i), $N' = \bigcap_{i \in J} R_i$ , whence

N = N' \cap \bigcap_{i \in I - J} R_i = \left( \bigcap_{i \in J} Q_i \right) \cap \left( \bigcap_{i \in I - J} R_i \right)

and it follows immediately from no. 3, Corollary to Proposition 4 that this primary decomposition is reduced.

Corollary

The mappings

D_I(M/N) \to D_J(M/N') \quad \text{and} \quad D_I(M/N) \to D_J(S^{-1}M/S^{-1}N)

defined in Proposition 6 (i) are surjective.

Proposition 6 (iii) shows that the mapping $D_I(M/N) \to D_J(M/N')$ is surjective and Proposition 6 (ii) then shows that the mapping

D_I(M/N) \to D_J(S^{-1}M/S^{-1}N)

is surjective.

5. Rings and modules of finite length

If an $A$ -module $M$ is of finite length, we shall denote this length by $\operatorname{long}_A(M)$ or $\operatorname{long}(M)$ . Recall that every Artinian ring is Noetherian (Algebra, Chapter VIII, § 6, no. 5, Corollary 3 to Proposition 12) and that every finitely generated module over an Artinian ring is of finite length (loc. cit., Corollary 1 to Proposition 12). Moreover, every Artinian integral domain is a field (Algebra, Chapter VIII, § 6, no. 4, Proposition 9).

Proposition 7

Let $M$ be a finitely generated module over a Noetherian ring $A$ . The following properties are equivalent:* (a) $M$ is of finite length. (b) Every ideal $p \in \operatorname{Ass}(M)$ is a maximal ideal of $A$ . (c) *Every ideal $p \in \operatorname{Supp}(M)$ is a maximal ideal of $A$ .

Let $(M_i)_{0 \leq i \leq n}$ be a composition series of $M$ such that, for $0 \leq i \leq n - 1$ , $M_i/M_{i+1}$ is isomorphic to $A/p_i$ , where $p_i$ is prime ( $§ 1$ , no. 4, Theorem 1). If $M$ is of finite length, so is each of the $A$ -modules $A/p_i$ , which implies that each of the rings $A/p_i$ is Artinian; but as $A/p_i$ is an integral domain, it is therefore a field, in other words $p_i$ is maximal; we conclude that (a) implies (b) ( $§ 1$ , no. 4, Theorem 2). Condition (b) implies (c) by $§ 1$ , no. 3, Proposition 7. Finally, if all the ideals of $\operatorname{Supp}(M)$ are maximal, so are the $p_i$ ( $§ 1$ , no. 4, Theorem 2), hence the $A/p_i$ are simple $A$ -modules and $M$ is of finite length, which completes the proof.

Corollary 1

For every module of finite length $M$ over a Noetherian Zing $A$ , $\operatorname{Ass}(M) = \operatorname{Supp}(M)$ .

Every element of $\operatorname{Supp}(M)$ is then minimal in $\operatorname{Supp}(M)$ and the conclusion follows from $§ 1$ , no. 3, Corollary 1 to Proposition 7.

Corollary 2

Let $M$ be a finitely generated module over a Noetherian ring $A$ and $p$ a prime ideal of $A$ . For $M_p$ to be a non-zero $A_p$ -module of finite length, it is necessary and sufficient that $p$ be a minimal element of $\operatorname{Ass}(M)$ .

By $§ 1$ , no. 2, Corollary to Proposition 5, $\operatorname{Ass}_{A_p}(M_p)$ is the set of ideals $q_p$ , where $q$ runs through the set of ideals of $\operatorname{Ass}(M)$ which are contained in $p$ . On the other hand, $p_p$ is the unique maximal ideal of $A_p$ ; by Proposition 7, for $M_p$ to be an $A_p$ -module of finite length, it is necessary and sufficient that no element of $\operatorname{Ass}(M)$ be strictly contained in $p$ . On the other hand, for $M_p \neq 0$ , it is necessary and sufficient by definition that $p \in \operatorname{Supp}(M)$ (Chapter 11, § 4, no. 4), that is that $p$ contain an element of $\operatorname{Ass}(M)$ ( $§ 1$ , no. 3, Proposition 7). This proves the corollary.

Remark (1). Let $M$ be a finitely generated module over a Noetherian ring $A$ ; let $(M_1), \ldots$ be a composition series of $M$ such that, for $0 \leq i \leq n - 1$ , $M_i / M_{i+1}$ is isomorphic to $A/p_i$ , where $p_i$ is a prime ideal of $A$ ( $§ 1$ , no. 4, Theorem 1). If $p$ is a minimal element of $\operatorname{Ass}(M)$ , the length $\operatorname{long}_{A_p}(M_p)$ is equal to the number of indices $i$ such that $p_i = p$ . For the $(M_i)_p$ form a composition series of $M$ , and $(M_i)_p / (M_{i+1})_p$ is isomorphic to $(A/p_i)_p$ and hence to $\{0\}$ if $p_i \neq p$ (since $p$ is minimal in the set of $p_i$ by $§ 1$ , no. 4, Theorem 2) and to $(A/p)_p$ which is a field, if $p_i = p$ .

Proposition 8

Let $M$ be a module of finite length over a Noetherian ring $A$ .

(i) There only exists a single primary decomposition of $\{0\}$ with respect to $M$ indexed by $\operatorname{Ass}(M)$ (necessarily reduced) — let $\{0\} = \bigcap_{p \in \operatorname{Ass}(M)} Q(p)$ be this decomposition, where $Q(p)$ is $p$ -primary with respect to $M$ .

(ii) There exists an integer $n_0$ such that, for all $n \geq n_0$ and all $p \in \operatorname{Ass}(M)$ , $Q(p) = p^n M$ .

(iii) For all $p \in \operatorname{Ass}(M)$ , the canonical mapping of $M$ to $M_p$ is surjective and its kernel is $Q(p)$ .

(iv) The canonical injection of $M$ into $\bigoplus_{p \in \operatorname{Ass}(M)} (M/Q(p))$ is bijective.

As every element $p \in \operatorname{Ass}(M)$ is minimal in $\operatorname{Ass}(M)$ (Proposition 7), assertion (i) follows from no. 3, Proposition 5. As $M$ is finitely generated, there exists $n_0$ such that $p^n M \subset Q(p)$ for all $p \in \operatorname{Ass}(M)$ and all $n \geq n_0$ (no. 1, Remark); but as $p$ is a maximal ideal, $p^n M$ is $p$ -primary with respect to $M$ (no. 1, Examples 2 and 3) and, as $\bigcap_{p \in \operatorname{Ass}(M)} p^n M = \{0\}$ , it follows from (i) that necessarily $p^n M = Q(p)$ for all $p \in \operatorname{Ass}(M)$ ; whence (ii). As the $p^n$ , for $p \in \operatorname{Ass}(M)$ , are relatively prime in pairs (Chapter 11, § 1, no. 2, Proposition 3), the canonical mapping $M \to \bigoplus_{p \in \operatorname{Ass}(M)} (M/p^n M)$ is surjective (Chapter 11, § 1, no. 2, Proposition 6), whence (iv). Then $\operatorname{Ass}(Q(p)) = \operatorname{Ass}(M) - \{p\}$ and $\operatorname{Ass}(M/Q(p)) = \{p\}$ (no. 3, Proposition 4); as the elements of $\operatorname{Ass}(M)$ are maximal ideals, $p$ is the only element of $\operatorname{Ass}(M)$ which does not meet $A - p$ ; $Q(p)$ is therefore the kernel of the canonical mapping $j : M \to M_p$ ( $§ 1$ , no. 2, Proposition 6). If $s \in A - p$ , the homothety of $M/Q(p)$ with ratio $s$ is injective by virtue of the relation $\operatorname{Ass}(M/Q(p)) = \{p\}$ (no. 1, Proposition 1); since $M/Q(p)$ is Artinian, this homothety is bijective (Algebra, Chapter VIII, § 1, no. 2, Lemma 3). The canonical mapping $M \to M/Q(p)$ is then written $f \circ j$ , where $f : M_p \to M/Q(p)$ is an $A$ -homomorphism (Chapter II, § 2, no. 2, Proposition 3); as $\operatorname{Ker}(j) = \operatorname{Ker}(f \circ j) = Q(p)$ , $f$ is injective; we conclude that $j$ is surjective and $f$ bijective.

Corollary

If M is a module of finite length over a Noetherian ring A, then

\operatorname{long}_A(M) = \sum_{p \in \operatorname{Ass}(M)} \operatorname{long}_{A_p}(M_p).

This will follow from Proposition 8 (iv) if we prove that

\operatorname{long}_A(M/\mathscr{Q}(p)) = \operatorname{long}_{A_p}(M_p).

Now, it follows from Proposition 1 of no. 1 that for all $s \in A - p$ the homothety with ratio s on $M/\mathscr{Q}(p)$ is injective; the homothety with ratio s on every submodule R of $M/\mathscr{Q}(p)$ is therefore injective and, as R is Artinian, it is bijective (Algebra, Chapter VIII, § 2, no. 2, Lemma 3); we conclude that the sub-A-modules of $M/\mathscr{Q}(p)$ are the images under the bijection $f : M, \to M/\mathscr{Q}(p)$ of the sub-A,-modules of M, (Chapter 11, § 2, no. 3), whence our assertion.

Proposition 9

Let A be a Noetherian ring. The following conditions are equivalent :* (a) A is Artinian. (b) All the prime ideals $\mathfrak{a}$ in A are maximal ideals. (c) *All the elements of Ass(A) are maximal ideals.

If these conditions are fulfilled, A has only a finite number of prime ideals, which are all maximal and associated with the A-module A; further, A is a semi-local ring and its Jacobson radical is nilpotent.

To say that A is Artinian is equivalent to saying that A is an A-module of finite length; hence (a) and (c) are equivalent by Proposition 7. Clearly (b) implies (c). Finally, (a) implies (b) since every Artinian integral domain is a field. The properties (a), (b) and (c) are therefore equivalent.

Suppose they hold. As every prime ideal of A belongs to Supp(A) and every element of Supp(A) contains an element of Ass(A) ( $\S$ 1, no. 3, Proposition 7), it follows from (c) that Ass(A) is the set of all prime ideals of A; then A has only a finite number of prime ideals, all of them maximal and associated with the A-module A. This obviously implies that A is semi-local; finally, we know that the Jacobson radical of an Artinian ring is nilpotent (Algebra, Chapter VIII, § 6, no. 4, Theorem 3).

Remark (2) The conditions of Proposition 9 for a Noetherian ring A imply that the spectrum of A is finite and discrete, every point of Spec(A) being therefore closed (Chapter 11, § 4, no. 3, Corollary 6 to Proposition 11). Conversely, for a Noetherian ring A, to say that every point $\mathfrak{a}$ of Spec(A) is closed means that every prime ideal of A is maximal (loc. cit..) and hence this condition is equivalent to that of Proposition 9.

Corollary 1

Every Artinian ring A is isomorphic to the direct composition of a finite family of Artinian local rings.

It follows from Proposition 9 and Proposition 8 (iii) and (iv) that, if $(\mathfrak{m}_i)_{1 \leq i \leq n}$ is the family of maximal ideals of $A$ , the canonical mapping $A \to \prod_i A_{\mathfrak{m}_i}$ is bijective.

Remark (3). This corollary can also be deduced from the fact that $\operatorname{Spec}(A)$ is finite and discrete and Chapter 11, § 4, no. 3, Proposition 15.

Corollary 2

Let $A$ be a Noetherian ring and $m$ an ideal of $A$ . The following conditions are equivalent:

(a) $A$ is a semi-local ring and $m$ is a defining ideal of $A$ . (b) $A$ is a Zariski ring with the $m$ -adic topology and $A/m$ is Artinian.

If (a) holds, $A$ is a Zariski ring with the $m$ -adic topology (Chapter III, § 3, no. 3, Example 3); further, as by hypothesis $m$ contains a power of the Jacobson radical $r$ of $A$ , every prime ideal of $A$ which contains $m$ also contains $r$ (Chapter II, § 1, no. 1, Proposition 1); it is therefore maximal, since $r$ is a finite intersection of maximal ideals (\emph{loc. cit.}); Proposition 9 then shows that $A/m$ is Artinian. Conversely, if (b) holds, every maximal ideal $p$ of $A$ contains the Jacobson radical of $A$ and hence contains $m$ (Chapter III, § 3, no. 3, Proposition 6); as $A/m$ is Artinian, the ideals $p/m$ are finite in number (Proposition 9) and hence $A$ has only a finite number of maximal ideals, which implies that it is semi-local.

Corollary 3

Let $A, A'$ be two rings and $h$ a homomorphism from $A$ to $A'$ . Suppose that $A$ is semi-local and Noetherian and that $A'$ is a finitely generated $A$ -module. Then the ring $A'$ is semi-local and Noetherian; if $m$ is a defining ideal of $A$ , $mA'$ is a defining ideal of $A'$ .

We know that $A'$ is a Zariski ring with the $mA'$ -adic topology (Chapter III, § 3, no. 3, Proposition 7). As $A/m$ is Artinian (Corollary 2) and $A'/mA'$ is a finitely generated $(A/m)$ -module, $A'/mA'$ is an Artinian ring, hence $A'$ is semi-local and $mA'$ is a defining ideal of $A'$ (Corollary 2).

Corollary 4

Let $A$ be a complete semi-local Noetherian ring, $m$ a defining ideal of $A$ , $E$ a finitely generated $A$ -module and $(F_n)$ a decreasing sequence of submodules of $E$ such that $\bigcap_n F_n = 0$ . Then, for all $p > 0$ , there exists $n > 0$ such that $F_n \subset m^p E$ .

As $A$ is a Zariski ring, $E$ is Hausdorff and the $F_n$ are closed under the $m$ -adic topology. On the other hand, $E$ is complete (Chapter III, § 2, no. 12, Corollary 1 to Proposition 16). Finally, $E/m^p E$ is a finitely generated module over the ring $A/m^p$ , which is Artinian (Corollary 2); then $E/m^p E$ is an Artinian $(A/m^p)$ -module and hence an Artinian $A$ -module. The corollary then follows from Chapter III, § 2, no. 7, Proposition 8.

Corollary 5

In a complete semi-local Noetherian ring every decreasing sequence $\phi$ ideals whose intersection is 0 is a filter base which converges to 0.

It is sufficient to apply Corollary 4 to the A-module $\mathbf{A}$ .

Proposition 10

Let $A$ be a Noetherian ring and $p_1, \ldots, p_n$ the prime ideals associated with the A-module $A$ , where $p_i \neq p_j$ for $i \neq j$ .

(i) The set $S = \bigcap_{i=1}^n (A - p_i)$ is the set of elements which are not divisors of 0 in $A$ .

(ii) If all the $p_i$ are minimal elements of $\operatorname{Ass}(A)$ , the total ring of fractions $S^{-1}A$ of $A$ is Artinian.

(iii) If the ring $A$ is reduced, all the $p_i$ are minimal elements of $\operatorname{Ass}(A)$ (and therefore are the minimal elements of $\operatorname{Spec}(A)$ ) and each of the $A_{p_i}$ is a field; for each index $i$ , the canonical homomorphism $S^{-1}A \to A_{p_i}$ (Chapter II, § 2, no. 1, Corollary 1 to Proposition 2) is surjective and its kernel is $S^{-1}p_i$ ; finally the canonical homomorphism from $S^{-1}A$ to $\prod_{i=1}^n (S^{-1}A/S^{-1}p_i)$ is bijective.

The fact that $S$ is the set of elements which are not divisors of 0 in $\mathbf{A}$ has already been seen (§ 1, no. 1, Corollary 3 to Proposition 2). The prime ideals of $S^{-1}A$ are of the form $S^{-1}p$ , where $p$ is a prime ideal of $A$ contained in $\bigcup_{i=1}^n p_i$ (Chapter 11, § 2, no. 5, Proposition 10) that is contained in one of the $p_i$ (Chapter II, § 1, no. 1, Proposition 2). If $p_i$ is a minimal element of $\operatorname{Ass}(A)$ , it is a minimal element of $\operatorname{Spec}(A)$ (§ 1, no. 3, Corollary to Proposition 7); if each of the $p_i$ is a minimal element of $\operatorname{Ass}(A)$ , we then see that the prime ideals of $S^{-1}A$ are the $S^{-1}p_i$ and they are therefore all maximal, which proves that $S^{-1}A$ is Artinian (Proposition 9).

Suppose finally that the ring $A$ is reduced. Then $\bigcap_{i=1}^n p_i = \{0\}$ (§ 1, no. 3, Corollary 2 to Proposition 7). We deduce that $\{0\} = \bigcap_{i=1}^n p_i$ is a reduced primary decomposition of the ideal $\{0\}$ (no. 3, Corollary to Proposition 4); in particular, none of the $p_i$ can contain a $p_j$ of index $j \neq i$ and therefore the $p_i$ are all minimal elements of $\operatorname{Ass}(A)$ . The ring $S^{-1}A$ is then Artinian by (ii). The $S^{-1}p_i$ are prime ideals associated with the $S^{-1}A$ -module $S^{-1}A$ (§ 1, no. 2, Corollary to Proposition 5) and $\{0\} = S^{-1} \left( \bigcap_{i=1}^n p_i \right) = \bigcap_{i=1}^n S^{-1}p_i$ (Chapter II, § 2, no. 4); as the $S^{-1}p_i$ are distinct, $(S^{-1}p_i)_{1 \leq i \leq n}$ is a reduced primary decomposition of $\{0\}$ in $S^{-1}A$ (no. 3, Corollary to Proposition 4). Proposition 8 then shows that the canonical homomorphismg,: $S^{-1}A \to (S^{-1}A)_{p_i}$ is surjective and has kernel $S^{-1}p_i$ and the canonical homomorphism $S^{-1}A \to \prod_{i=1}^n (S^{-1}A/S^{-1}p_i)$ is bijective. We know moreover that the canonical homomorphism $S^{-1}A \to A$ , is composed of g, and an isomorphism $(S^{-1}A)_{S^{-1}p_i} \to A$ , (Chapter II, § 2, no. 3, Proposition 7). Finally, it follows from Proposition 8 that $(S^{-1}A)_{S^{-1}p_i}$ is isomorphic to $S^{-1}A/S^{-1}p_i$ and hence is a field since $S^{-1}p_i$ is a maximal ideal.

6. Primary decomposition and extension of scalars

In this no., A and B will denote two rings and we shall consider a ring homomorphism $\rho : A \to B$ which makes $B$ into an A-algebra; recall that, for every B-module F, $\rho_*(F)$ is the commutative group F with the A-module structure defined by $a.y = \rho(a)y$ for all $a \in A, y \in F$ .

Lemma 1

Let A be a Noetherian ring, $p$ a prime ideal of A, E an A-module whose annihilator contains a power of $p$ and such that $\operatorname{Ass}(E) = \{p\}$ and F a B-module such that $\rho_*(F)$ is a flat A-module. The condition $\mathfrak{P} \in \operatorname{Ass}_B(E \otimes_A F)$ then implies $\rho^{-1}(\mathfrak{P}) = p$ .

If n is such that $p^nE = 0$ , then $p^nB \subset \operatorname{Ann}(E \otimes_A F)$ , whence $p^nB \subset \mathfrak{P}$ , which implies $p^n \subset \rho_1(\mathfrak{P})$ and therefore $p \subset \rho_1(\mathfrak{P})$ since $\rho_1(\mathfrak{P})$ is prime. Moreover, if $a \in A - p$ , the homothety h with ratio a on E is injective ( $§ 1$ , no. 1, Corollary 2 to Proposition 2); as $h \otimes 1_F$ is the homothety $h'$ with ratio $\rho(a)$ on $E \otimes_A F$ and $\rho_*(F)$ is flat, $h'$ is injective (Chapter I, § 2, no. 2, Definition 1); this proves that $\rho(a) \notin \mathfrak{P}$ , whence $\rho^{-1}(\mathfrak{P}) = p$ .

Theorem 2

Let $\rho : A \to B$ be a ring homomorphism, E an A-module and F a B-module such that $\rho_*(F)$ is a flat A-module. Then

\operatorname{Ass}_B(E \otimes_A F) \supset \bigcup_{p \in \operatorname{Ass}_A(E)} \operatorname{Ass}_B(F/pF).

When A is Noetherian, the two sides of (5) are equal.

Let $p \in \operatorname{Ass}_A(E)$ ; by definition there exists an exact sequence

0 \to A/p \to E.

Since F is a flat A-module we derive an exact sequence

0 \to F/pF \to E \otimes_A F

whence $\operatorname{Ass}_B(F/pF) \subset \operatorname{Ass}_B(E \otimes_A F)$ , which proves the inclusion (5).

Suppose now that A is Noetherian and let us prove the opposite inclusion.

We proceed in stages:

(i) Suppose first that E is a finitely generated A-module and that Ass_A(E) is reduced to a single element $p$ . By § 1, no. 4, Theorem 1 there exists a composition series $(E_i)_{0 \leq i \leq n}$ of E such that $E_i/E_{i+1}$ is isomorphic to $A/p_i$ , where $p_i$ is a prime ideal of A; moreover ( $§ 1$ , no. 4, Theorem 2 and no. 3, Proposition 7) all the $p_i$ contain $p$ . As F is a flat A-module, the $E_i \otimes_A F$ form a composition series of $E \otimes_A F$ and $(E_i \otimes_A F)/(E_{i+1} \otimes_A F)$ is identified with

(A/p_i) \otimes_A F = F/p_i F.

Then by virtue of

§ 1

, no. 1, Proposition 3

\operatorname{Ass}_B(E \otimes_A F) \subset \bigcup_{i=0}^{n-1} \operatorname{Ass}_B(F/p_i F).

We know that E is annihilated by a power of

p

(no. 1, Remark); Lemma 1 then shows that, for all

\mathfrak{P} \in \operatorname{Ass}_B(E \otimes_A F)

\rho^{-1}(\mathfrak{P}) = p

. As

F/p_i F

is isomorphic to

(A/p_i) \otimes_A F

\rho^{-1}(\mathfrak{P}') = p_i

for all

\mathfrak{P}' \in \operatorname{Ass}_B(F/p_i F)

by Lemma 1, whence

\operatorname{Ass}_B(E \otimes_A F) \cap \operatorname{Ass}(F/p_i F) = \varnothing

p_i \neq p

, which proves the theorem in the case considered.

(ii) Suppose only that E is a finitely generated A-module. Let $p_i$ ( $1 \leq i \leq n$ ) be the elements of $\operatorname{Ass}_A(E)$ and let $\{0\} = \bigcap^n Q_i$ be a corresponding reduced primary decomposition (no. 3); E is then isomorphic to a submodule of the direct sum of the $E_i = E/Q_i$ and, as F is a flat A-module, $E \otimes_A F$ is isomorphic to a submodule of the direct sum of B-modules $E_i \otimes_A F$ . We deduce ( $§ 1$ , no. 1, Proposition 3 and Corollary 1 to Proposition 3) that

\operatorname{Ass}_B(E \otimes_A F) \subset \bigcup_{i=1}^n \operatorname{Ass}_B(E_i \otimes_A F).

But

E_i

is a finitely generated A-module such that

\operatorname{Ass}_A(E_i)

is reduced to a single element

p_i

(no. 1, Definition 1). By (i),

\operatorname{Ass}_B(E_i \otimes_A F) = \operatorname{Ass}_B(F/p_i F)

, whence the theorem in this case.

(iii) General case. The B-module $E \otimes_A F$ is the union of the submodules $E' \otimes_A F$ , where $E'$ runs through the set of finitely generated submodules of the A-module E. If $\mathfrak{P}$ belongs to $\operatorname{Ass}_B(E \otimes_A F)$ , then there exists a finitely generated submodule $E'$ of E such that $\mathfrak{P} \in \operatorname{Ass}_B(E' \otimes_A F)$ . By (ii), there exists $p \in \operatorname{Ass}_A(E')$ such that $\mathfrak{P} \in \operatorname{Ass}_B(F/pF)$ ; as $\operatorname{Ass}_A(E') \subset \operatorname{Ass}_A(E)$ , this completes the proof of Theorem 2.

Corollary 1

If A is Noetherian and $\mathfrak{P} \in \operatorname{Ass}_B(E \otimes_A F)$ , then $\rho^{-1}(\mathfrak{P}) \in \operatorname{Ass}_A(E)$ and $\rho^{-1}(\mathfrak{P})$ is the only prime ideal $p$ of A such that $\mathfrak{P} \in \operatorname{Ass}_B(F/pF)$ .

Corollary 2

Suppose that $A$ and $B$ are Noetherian and that $B$ is a flat $A$ -module. Let $p$ be a prime ideal of $A$ , $Q \subset E$ a $p$ -primary submodule and $\mathfrak{P}$ a prime ideal of $B$ . For $Q \otimes_A B$ to be a $q$ -primary submodule of $E \otimes_A B$ , it is necessary and sufficient that $pB$ be a 'p-primary ideal' of $B$ .

Let us apply Theorem 2 to the $A$ -module $E/Q$ and the $B$ -module $B$ ; then $\operatorname{Ass}_A(E/Q) = \{p\}$ and $(E/Q) \otimes_A B$ is isomorphic to $(E \otimes_A B)/(Q \otimes_A B)$ and hence $\operatorname{Ass}_B((E \otimes_A B)/(Q \otimes_A B)) = \operatorname{Ass}_B(B/pB)$ . To say that $Q \otimes_A B$ is $\mathfrak{P}$ -primary in $E \otimes_A B$ therefore means that $\operatorname{Ass}_B(B/pB)$ is reduced to $\mathfrak{P}$ , whence the corollary.

Remark

Suppose that $A$ and $B$ are Noetherian. Let $\mathfrak{P}$ be a prime ideal of $B$ and let $p = \overline{\rho}^{-1}(\mathfrak{P})$ ; let us write $S = A - p$ and let $k(p) = S^{-1}(A/p)$ be the field of fractions of $A/p$ . Since $\mathfrak{P}$ contains $pB$ , $\mathfrak{P}/pB$ is a prime ideal of $B/pB$ . If $\rho'$ is the composite homomorphism $A \xrightarrow{\rho} B \to B/pB$ , we know that $S^{-1}(B/pB)$ is identified with the ring $(\rho'(S))^{-1}(B/pB)$ and $\mathfrak{P}' = S^{-1}(\mathfrak{P}/pB)$ with an ideal of this ring (Chapter II, § 2, no. 2, Proposition 6); as $\mathfrak{P}/pB$ does not meet $\rho'(S)$ , $\mathfrak{P}'$ is a prime ideal of $S^{-1}(B/pB)$ (Chapter 11, § 2, no. 5, Proposition 11); moreover there are canonical isomorphisms between $S^{-1}(B/pB)$ , $S^{-1}((A/p) \otimes_A B)$ and $(S^{-1}(A/p)) \otimes_A B = k(p) \otimes_A B$ ; similarly $S^{-1}(F/pF)$ is canonically identified with $k(p) \otimes_A F$ . This being so, under the hypotheses of Theorem 2, in order that $\mathfrak{P} \in \operatorname{Ass}_B(E \otimes_A F)$ , it is necessary and sufficient that $p \in \operatorname{Ass}_A(E)$ and

\mathfrak{P}' \in \operatorname{Ass}_{k(p) \otimes_A B}(k(p) \otimes_A F).

For by Theorem 2 and its Corollary 1, it amounts to verifying that the conditions

\text{``}\mathfrak{P} \in \operatorname{Ass}_B(F/pF)\text{''} \quad \text{and} \quad \text{``}\mathfrak{P}' \in \operatorname{Ass}_{k(p) \otimes_A B}(k(p) \otimes_A F)\text{''}

are equivalent; but, as

B

is Noetherian, this follows from § 1, no. 2, Corollary to Proposition 5 and the above identifications.

Proposition 11

Suppose that $A$ and $B$ are Noetherian and that $B$ is a flat $A$ -module. Let $E$ be an $A$ -module and $E'$ a submodule of $E$ such that, for every ideal $p \in \operatorname{Ass}_A(E/E')$ , $pB$ is a prime ideal of $B$ or equal to $B$ . Let $E' = \bigcap_{p \in \operatorname{Ass}(E/E')} Q(p)$ be a reduced primary decomposition of $E'$ in $E$ , $Q(p)$ being $p$ -primary for all $p \in \operatorname{Ass}(E/E')$ . (i) If $p \in \operatorname{Ass}(E/E')$ and $pB = B$ , then $Q(p) \otimes_A B = E \otimes_A B$ . (ii) If $p \in \operatorname{Ass}(E/E')$ and $pB$ is prime, $Q(p)$ is $pB$ *-primary* in $E \otimes_A B$ . (iii) If $\Phi$ is the set $\mathfrak{p} \in \operatorname{Ass}(E/E')$ such that $pB$ is prime, then

E' \otimes_A B = \bigcap_{p \in \Phi} (Q(p) \otimes_A B)

and this relation is a reduced primary decomposition of

E' \otimes_A B

E \otimes_A B

If $pB = B$ , Theorem 2 applied to $E/\mathbf{Q}(p)$ and $B$ shows that

\operatorname{Ass}_B((E/\mathbf{Q}(p)) \otimes_A B) = \varnothing

and, as

B

is Noetherian and is a flat

A

-module, we conclude (§ 1, no. 1, Corollary 1 to Proposition 2) that

\mathbf{Q}(p) \otimes_A B = E \otimes_A B

. Assertion (ii) follows from Corollary 2 to Theorem 2, taking

\mathfrak{P} = pB

. Finally the relation

E' \otimes_A B = \bigcap_{p \in \Phi} (\mathbf{Q}(p) \otimes_A B)

follows from the fact that

B

is a flat

A

-module (Chapter I, § 2, no. 6, Proposition 6); as

p = \overline{\rho}^{-1}(pB)

for

p \in \Phi

(Lemma 1),

pB \neq p'B

for two distinct ideals

p, p'

of the set

\Phi

; on the other hand,

\operatorname{Ass}((E \otimes_A B)/(E' \otimes_A B)) = \Phi

by Theorem 2; we conclude from no. 3, Proposition 4 that

E' \otimes_A B = \bigcap_{p \in \Phi} (\mathbf{Q}(p) \otimes_A B)

is a reduced primary decomposition.

Corollary

Suppose that* $pB$ is prime for all $p \in \operatorname{Ass}_A(E/E')$ . Then, if $p_1, \ldots, p_n$ are the minimal elements of $\operatorname{Ass}_A(E/E')$ , the $p_iB$ are minimal elements of

\operatorname{Ass}_A((E \otimes_A B)/(E' \otimes_A B)).

It follows from Proposition 11 that in this case

p_iB \neq p_jB

for

i \# j

Examples (1) Let us take $B = S^{-1}A$ , where $S$ is a multiplicative subset of $A$ ; if $A$ is Noetherian, the hypotheses of Proposition 11 are satisfied and we recover a part of Proposition 6 of no. 4. (2) Let $A$ be a Noetherian ring, $m$ an ideal of $A$ and $B$ the Hausdorff completion of $A$ with respect to the $m$ -adic topology; then $B$ is a flat $A$ -module and Theorem 2 may be applied with $F = B$ ; but in general the hypotheses of Proposition 11 are not satisfied for the prime ideals of $A$ (Chapter III, § 2, Exercise 15 (b)). (3) Let $A$ be a Noetherian ring and $B$ the polynomial algebra $A[X_1, \ldots, X_n]$ ; $B$ is Noetherian and is a free $A$ -module and therefore flat. Also, if $p$ is a prime ideal of $A$ , $B/pB$ is isomorphic to $(A/p)[X_1, \ldots, X_n]$ , which is an integral domain, and hence $pB$ is prime; the hypotheses of Proposition 11 are therefore satisfied for every $A$ -module $E$ and every submodule $E'$ of $E$ . (4) Let $A$ be a finitely generated algebra over a field $k$ , $K$ an extension of $k$ and $B = A \otimes_k K$ the algebra over $K$ obtained by extension of scalars; $A$ and $B$ are Noetherian and $B$ is a free $A$ -module and hence Theorem 2 may be applied to $F = B$ . In certain cases (for example if $k$ is algebraically closed) it can be shown that for every prime ideal $p$ of $A$ , $pB$ is prime or equal to $B$ ; we shall return later to this example.

Exercises

See the exercises for § 2.