Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 36 (1995), No. 1, 9-15.

A Note on Unions of Ideals and Cosets of Ideals

Michael Zandt

Fakultät für Mathematik und Informatik, Universität Leipzig Augustusplatz, 04109 Leipzig, Germany E-mail: zandt@aix520.informatik.uni-leipzig.de

Abstract. The paper gives a refinement to a well known and easy-to-prove result of basic ideal theory: If an ideal I is contained in the union of a given finite collection of ideals, with at most two of them not prime, I must be contained in one of them. The proposition, stated in this way, leaves out the special structure of the ring under consideration. We show that the lower bound implicit in the proposition can be replaced by the minimum over cardinalities of residue fields of the given ring. MSC 1991: 13A15

Closely related to the fact that an ideal can not be contained in the union of two arbitrary and a finite number of prime ideals unless it is contained in one of them (cf. [1], (1.B)), a vector space over a field k can not be covered by a finite collection of proper subspaces unless their number exceeds the cardinality of k (cf.[2], p.23. A similar proposition holds for residue classes of subspaces.). Since a nontrivial field contains at least two elements, the aforesaid might suggest that the field case is inherent in the general case treated by the above quoted proposition. And indeed, the basic results of this paper are

Proposition A: Let I, I_{1}, ..., I_{r} be ideals, I_{r+1}, ..., I_{r+s} prime ideals of A, where r, s ∈ N,

r + s ≥ 1. If r ≤ min{card (^{A}/_{m}) | m ∈ m-spec (A)}, then I ⊂ ⋃_{i=1}^{r+s} I_{i} implies I ⊂ I_{i} for

some i ∈ {1, ..., r + s}.

Proposition B: Let I, I_{1}, ..., I_{t} be ideals of A, t ∈ N, a_{ij} ∈ A, j = 1, ..., n_{i}, n_{i} ∈ N for

i = 1, ..., t. Suppose I ⊄ I_{i} for all i and ∑_{i=1}^{t} n_{i} < inf {card (^{A}/_{m}) | m ∈ m-spec (A)}. Then

I ⊄ ⋃_{i=1}^{t} ⋃_{j=1}^{n_{i}} (a_{ij} + I_{i}).

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1995 Heldermann Verlag, Berlin

In the proofs, the above quoted basic lemma will, without special reference, frequently be part of the argument. A will always denote a commutative ring with 1, M will be an arbitrary A-module, m-spec (A) the set of maximal ideals of A, m-supp (M) :=

m-spec (A) ∩ supp (M). An equality M = ⋃_{i=1}^{r} M_{i} or an inclusion I ⊂ ⋃_{i=1}^{r} I_{i} with submod-

ules M_{i} or ideals I_{i}, I, respectively,will be called irredundant if none of the M_{i} or I_{i} can be omitted without the relation under consideration turning false.

I. A preparatory lemma and its consequences

If P is a subset of spec (A), \overline{P} will denote the closure of P in the Zariski topology. If L is an arbitrary set and {M_{α}}_{α∈L} a system of submodules of an A-module M, any mapping s : L → ⨆_{α∈L} supp (^{M}/_{M_{α}}) with s(α) ∈ supp (^{M}/_{M_{α}}) for all α ∈ L will be called a section for {M_{α}}_{α∈L}.

The proof of the following lemma is clear.

Lemma 1: A subset P ⊂ spec (A) the number of minimal elements of which is finite is disconnected in the Zariski topology if and onlyif there exists a partition P = P_{1} ∪ P_{2} such that, for any p ∈ P_{1}, q ∈ P_{2}, p ⊄ q and q ⊄ p.

The following lemma and its corollary are crucial to all subsequent proving.

Lemma 2: Let r > 1 be an integer, M_{1}, ..., M_{r} submodules of M, X = {1, ..., r}. If either of the following two conditions is satisfied:

(i) There exists a section s for {M_{i}}_{i∈X} such that s(X) is disconnected and

⋃_{i=1}^{r}((M_{i} + s(i)M)_{s(i)} ∩ M) is irredundant;

(ii) there exists a section s for \{M_{i}\}_{i∈X} such that \overline{s(X)} is disconnected and

⋃_{i=1}^{r}(M_{i} + s(i)M) is irredundant,

then we have M ≠ ⋃_{i=1}^{r} M_{i}.

Proof: (i) Having chosen a partition X = X_{1} ∪ X_{2} according to Lemma 1, we can find an element a_{i} ∈ s(i)\ ⋃_{j∈X_{2}} s(j), for each i ∈ X_{1}. Setting a = ∏_{i∈X_{1}} a_{i}, we obtain a ∈

( ⋂_{i∈X_{1}} s(i))\( ⋃_{j∈X_{2}} s(j)), and likewise, we find an element b ∈ ( ⋂_{j∈X_{2}} s(j))\( ⋃_{i∈X_{1}} s(i)). Now

let N_{i} = (M_{i} + s(i)M)_{s(i)} ∩ M, for i ∈ X, and let x_{1} ∈ M\ ⋃_{i∈X_{1}} N_{i}, x_{2} ∈ M\ ⋃_{i∈X_{2}} N_{i}.

Obviously, ax_{2} ∈ ⋂_{i∈X_{1}} N_{i}, while ax_{2} ∉ N_{j}, for all j ∈ X_{2}. The latter becomes clear by

localization at s(j) and by the fact that \frac{a}{1} is a unit in A_{s(j)} and (N_{j})_{s(j)} = (M_{j}+s(j)M)_{s(j)}.

Thus, we get ax_{2} ∈ ( ⋂_{i∈X_{1}} N_{i})\( ⋃_{j∈X_{2}} N_{j}) and, by analogy, bx_{1} ∈ ( ⋂_{j∈X_{2}} N_{j})\( ⋃_{i∈X_{1}} N_{i}) ,

whereby we have proved ax_{2} + bx_{1} ∈ M\ ⋃_{i∈X} N_{i} ⊂ M\ ⋃_{i∈X} M_{i}.

(ii) Let \overline{s(X)} = P_{1} ∪ P_{2} be a partition according to Lemma 1. For each i ∈ X, we have either V (s(i)) ⊂ P_{1} or V (s(i)) ⊂ P_{2}, thus obtaining a partition X = X_{1} ∪ X_{2} with V (s(i)) ⊂ P_{1}, for all i ∈ X_{1},and V (s(i)) ⊂ P_{2}, for all i ∈ X_{2}. With N_{i} = M_{i} + s(i)M, we can find elements x_{1} ∈ M\ ⋃_{i∈X_{1}} N_{i}, x_{2} ∈ M\ ⋃_{i∈X_{2}} N_{i}. Quite generally, we will denote the

residue class of an element x by \overline{x}, the actual meaning of this depending on the factorization under consideration. There are ideals m_{i} ∈ m-supp (^{M}/_{N_{i}}) and m_{j} ∈ m-supp (^{M}/_{N_{j}}) such that \overline{x}_{1} does not vanish in (^{M}/_{N_{i}})_{m_{i}}, nor does \overline{x}_{2} in (^{M}/_{N_{j}})_{m_{j}}, for each i ∈ X_{1} and j ∈ X_{2}, respectively. As, by Lemma 1, no element of s(X_{1}) is contained in any element of s(X_{2}) and vice versa, we can again find elements a ∈ ( ⋂_{i∈X_{1}} s(i))\( ⋃_{j∈X_{2}} m_{j}) and

b ∈ ( ⋂_{j∈X_{2}} s(j))\( ⋃_{i∈X_{1}} m_{i}). An argument similar to the proof of (i) leads us to ax_{2} + bx_{1} ∈

M\ ⋃_{i=1}^{r} N_{i} ⊂ M\ ⋃_{i=1}^{r} M_{i}.

Corollary: Let M_{ij} be maximal submodules of M, i = 1, ..., t, t ∈ N, t > 1, j = 1, ..., n_{i}, n_{i} ∈ N, n_{i} > 0, with ideals m_{i} ∈ m-spec (A) such that m_{i} ≠ m_{j}, for i ≠ j, and such that

there exists an A-isomorphism ^{M}/_{M_{ij}} ≃ ^{A}/_{m_{i}}, for all i and j. Let M′ = ⋃_{i=1}^{t} ⋃_{j=1}^{n_{i}} M_{ij} be

irredundant. Then M ≠ M′.

Proof: Since supp (^{M}/_{M_{ij}}) = \{m_{i}\}, for all i and j, the only section s for \{M_{ij}\}{t \atop i=1} {n_{i} \atop j=1} is defined by s((i, j)) = m_{i}. s{(i, j)|i = 1, ..., t; j = 1, ..., n_{i}} being apparently disconnected, M_{ij} + m_{i}M = M_{ij} renders the assertion a simple application of Lemma 1 (ii).

If M = ⋃_{i=1}^{r} M_{i} is irredundant with proper submodules M_{i}, we can easily find a finitely

generated submodule N of M (namely N = ∑_{i=1}^{r} Ax_{i}, where x_{i} ∈ M_{i}\ ⋃_{j≠i} M_{j} chosen by

virtue of the irredundancy assumption) such that N = ⋃_{i=1}^{r}(M_{i} ∩ N) is irredundant, too.

Note that a finitely generated A-module always contains maximal (proper) submodules containing a given submodule. To prove our main proposition, we need the first part of the following lemma the general boundary of which can be improved upon in case certain submodules of M are sure to occur in a given covering of M (parts (ii) and (iii)).

Lemma 3: Let M = ⋃_{i=1}^{r} M_{i} be irredundant with submodules M_{1}, ..., M_{r} of M, r ∈ N,

r > 1, and let X = {1, ..., r}. Then

(i) there exists an ideal m ∈ m-supp (M) with card (^{A}/_{m}) < r; (ii) sup {card (^{A}/_{m})|m ∈ m-supp (^{M}/_{M_{i}}) for some i ∈ X} < r. (iii) If m ∈ m-supp (^{M}/_{M_{i}}) for some i ∈ X, then at least s = card (^{A}/_{m}) + 1 among the

submodules M_{j} satisfy m ∈ m-supp (^{M}/_{M_{j}}).

Proof: (i) We choose a finitely generated submodule N of M such that N = ⋃_{i=1}^{r}(M_{i} ∩ N) is irredundant, and a maximal (proper) submodule N_{i} ⊂ N satisfying N ∩ M_{i} ⊂ N_{i}, for each i ∈ {1, ..., r}. Let m_{1}, ..., m_{r} be the corresponding maximal ideals of A, i.e.

N/_{N_{i}} ≃ ^{A}/_{m_{i}}, i = 1, ..., r. Without loss of generality, N = ⋃_{i=1}^{r′} N_{i} is irredundant (r′ ≤ r).

As an immediate consequence of the Corollary, we obtain m_{1} = ... = m_{r′}, and we will

denote this ideal by m. Thus, ^{N}/_{N_{i}} ≃ ^{A}/_{m}, i = 1, .., r′, implying mN ⊂ ⋂_{i=1}^{r′} N_{i}. Applying

the initially mentioned result concerning vector spaces to ^{A}/_{m} and \overline{N} = ⋃_{i=1}^{r′} \overline{N}_{i}, where

\overline{N} = ^{N}/_{⋂_{i=1}^{r′} N_{i}}, \overline{N}_{i} = ^{N_{i}}/_{⋂_{i=1}^{r′} N_{i}}, (\overline{N}/_{\overline{N}_{i}} ≃ ^{N}/_{N_{i}} ≠ 0),

we get r ≥ r′ > card (^{A}/_{m}), and m ∈ m-supp (N) ⊂ m-supp (M), as asserted. (ii) W.l.o.g., we consider m ∈ m-supp (^{M}/_{M_{1}}). Our assumptions ensure the existence

of elements x ∈ M\ ⋃_{i=2}^{r} M_{i} ⊂ M_{1} and x′ with ax′ ∉ M_{1} + mx′, for all a ∈ A\m. Let us

now examine the finite submodule N = Ax + Ax′ more closely.

We can assume that, for each i ∈ X, either x′ ∈ M_{i}, or \overline{x}′ is not in the kernel of the canonical A-homomorphism ^{M}/_{M_{i}} → (^{M}/_{M_{i}})_{m}. For otherwise we could replace x′ by ax′, with a ∈ A\m and ax′ ∈ M_{i}, whereby whatever part of the above alternative remains true or untrue, respectively, for any M_{j} with j ≠ i. Thus, a partition X = X_{1} ∪ X_{2} ensues, where x′ ∈ M_{i}, for i ∈ X_{1}, and x′ ∉ M_{i} + mx′, for i ∈ X_{2}. The choice of x′ implies X_{1} ≠ ∅ and X_{2} ≠ ∅. For i ∈ X_{1}, we have M_{i} ∩ N = (M_{i} ∩ Ax) + Ax′, and there exists a maximal (proper) submodule N_{i} ⊂ N with M_{i} ∩ N ⊂ N_{i}, as well as an ideal m_{i} ∈ m-supp (^{Ax}/_{M_{i}∩Ax}) such that ^{N}/_{N_{i}} ≃ ^{A}/_{m_{i}}. Obviously, x ∉ N_{i}, i ∈ X_{1}. For i ∈ X_{2}, m ∈ m-supp (^{N}/_{M_{i}∩N}) since x′ ∉ (M_{i} ∩ N) + mx′; hence there exists a

maximal submodule N_{i} with M_{i} ∩ N ⊂ N_{i} ⊂ N and ^{N}/_{N_{i}} ≃ ^{A}/_{m}. W.l.o.g., N = ⋃_{i=1}^{r′} N_{1}

is irredundant (r′ ≤ r). There must be an index i ∈ {1, ..., r′} such that ^{N}/_{N_{i}} ≃ ^{A}/_{m} as x ∉ N_{i} for any i ∈ X_{1}. The Corollary in the first section shows that ^{N}/_{N_{i}} ≃ ^{A}/_{m} for i = 1, ..., r′. Thus, we can consider ^{N}/_{⋂_{i=1}^{r′} N_{i}} as a vector space over ^{A}/_{m}, and, as above,

we obtain r ≥ r′ > card (^{A}/_{m}). This proves (ii).

Since the above construction can be carried out for any maximal ideal m satisfying our condition, (iii) is settled by a reminder of m-supp (^{N}/_{N_{i}}) ⊂ m-supp (^{N}/_{N∩M_{i}}), ^{N}/_{N∩M_{i}} ≃ M_{i}+N/_{M_{i}} and m-supp (^{M_{i}+N}/_{M_{i}}) ⊂ m-supp (^{M}/_{M_{i}}).

Proof of Proposition A: The case of ^{A}/_{m} being infinite for all m ∈ m-spec (A) is now easily

settled by noting that the irredundancy of I ⊂ ⋃_{i=1}^{r+s} I_{i} is inherited by I = ⋃_{i=1}^{r+s}(I_{i} ∩ I), and

that I ⊄ I_{i}, i = 1, ...r + s, leads to a contradiction to Lemma 3(i).

For s = 0, the finite-cardinality case is an equally simple consequence of Lemma 3(i). The induction for s ≥ 1 is carried out in exactly the same way as is usually employed in the case of r ≤ 2.

We collect some immediate consequences of Lemma 3 in the following

Corollary: For r ∈ N, r > 1, let I, I_{1}, ..., I_{r} be ideals of A such that I ⊂ ⋃_{i=1}^{r} I_{i} is

irredundant. Then

(i) r > sup{card (^{A}/_{m})|m ∈ m-supp (^{I}/_{I∩I_{i}}) , i = 1, ..., r}; (ii) there exist ideals m_{i} ∈ V (I_{i}) ∩ m-spec (A) such that r > max_{i=1}^{r}{card (^{A}/_{m_{i}})}; in par-

ticular, r > max_{i=1}^{r} min_{V_{i}}{card (^{A}/_{m_{i}})}, where V_{i} = V (I_{i}) ∩ m-spec (A);

(iii) if m ∈ m-supp (^{I}/_{I∩I_{i}}) for some i ∈ {1, ..., r}, there are at least s := card (^{A}/_{m}) + 1 ideals among I_{1}, ..., I_{r}, say I_{i_{1}}, ..., I_{i_{s}} , such that m ∈ V (I_{i_{ν}}), ν = 1, ..., s.

Proof: (i) is an immediate consequence of Lemma 3(ii). Since, by assumption, ^{I}/_{I∩I_{i}} ≠ 0, there exist ideals m_{i} ∈ m-supp (^{I}/_{I∩I_{i}}) ⊂ V (I_{i}). Thus, (ii) follows from (i). Lemma 3(iii) and m-supp (^{I}/_{I∩I_{iν}}) ⊂ V (I_{i_{ν}}), for all ν, ensure the existence of ideals I_{i_{1}}, ..., I_{i_{s}} in (iii).

II. Residue classes and proof of Proposition B

To get some results concerning residue classes, we will now make use of what we know about submodules (ideals).

Lemma 4: For t ∈ N, t > 1, let M_{1}, ..., M_{t} ⊂ M be proper submodules, x_{ij} ∈ M, j = 1, ..., n_{i}, n_{i} ∈ N, i = 1, ..., t, such that {x_{i1}, ..., x_{in_{i}}} does not contain a full residue system mod M_{i} for any i ∈ {1, ..., t}.

(i) If ∑_{i=1}^{t} n_{i} < min{card (^{A}/_{m})|m ∈ m-supp (M)}, then M ≠ ⋃_{i=1}^{t} ⋃_{j=1}^{n_{i}}(x_{ij} + M_{i}).

If, moreover, M = ⋃_{i=1}^{t} ⋃_{j=1}^{n_{i}}(x_{ij} + M_{i}) is irredundant, then

(ii) ∑_{k≠i} n_{k} ≥ inf{card (^{A}/_{m})|m ∈ m-supp (M_{i})}, for all i ∈ {1, ..., t};

(iii) ∑_{k≠i} n_{k} ≥ inf{card (^{A}/_{m})|m ∈ m-supp (^{M_{i}}/_{M_{i}∩M_{k}}), k ≠ i}, for all i ∈ {1, ..., t}.

Proof: (i) W.l.o.g., we assume the union of residue classes to be irredundant. Lemma 3(i)

furnishes an element x ∈ M\ ⋃_{i=1}^{t} M_{i}. Thus, there exist ideals m_{i} ∈ m-supp (^{M}/_{M_{i}}) ⊂

m-supp (M) such that \frac{\overline{x}}{1} ≠ 0 in (^{M}/_{M_{i}})_{m_{i}}, i = 1, ..., t, which, in particular, means that a ≢ a′ mod m_{i}, a, a′ ∈ A, implies ax ≢ a′x mod M_{i}. Hence, for each x_{ij}, there is at most one residue class mod m_{i} such that ax + M_{i} = x_{ij} + M_{i} for some a in this class. Let m_{1} = ... = m_{i_{1}}, m_{i_{1}+1} = ... = m_{i_{2}}, ..., m_{i_{t′−1}+1} = ... = m_{i_{t′}}, 0 < i_{1} < i_{2} < ... < i_{t′} = t,

t′ ≤ t. Since ∑_{i=i_{ν}+1}^{i_{ν+1}} n_{i} < card (^{A}/_{m_{i_{ν+1}}} ), ν = 0, ..., t′−1 (i_{0} = 0), we can find residue classes

a_{i_{ν}} mod m_{i_{ν}}, ν = 1, ..., t′, such that none of the residue classes a_{i_{ν}} x+M_{i}, i = i_{ν−1}+1, ..., i_{ν}, occurs among the x_{ij} + M_{i}. According to the Chinese Remainder Theorem, there is an element a ∈ A satisfying a ≡ a_{i_{ν}} mod m_{i_{ν}}, ν = 1, ..., t′. But then, none of the x_{ij} + M_{i} contains ax. (ii) Let us say i = 1. Suppose x_{1} ∈ M such that x_{1} + M_{1} does not occur in

{x_{11} +M_{1}, ..., x_{1n_{1}} +M_{1}}. Hence x_{1} +M_{1} ⊂ ⋃_{i=2}^{t} ⋃_{j=1}^{n_{i}}(x_{ij} +M_{i}), i.e. M_{1} ⊂ ⋃_{i=2}^{t} ⋃_{j=1}^{n_{i}}(x_{ij} −x_{1} +

M_{i}). If (x_{ij} −x_{1} +M_{i})∩M_{1} ≠ ∅, then (x_{ij} −x_{1} +M_{i})∩M_{1} = x′_{ij} +(M_{i} ∩M_{1}), with suitable elements x′_{ij} ∈ M_{1}. Thus, reordering our set of submodules as well as the sets of residue

classes in case of necessity, we get M_{1} = ⋃_{i=2}^{t′} ⋃_{j=1}^{n_{i}′}(x′_{ij} +M_{i} ∩M_{1}), where t′ ≤ t, n_{i}′ ≤ n_{i}, and,

by the irredundancy assumption, M_{1} ≠ x′_{ij} + M_{i} ∩ M_{1}, for i = 2, ..., t′. (ii) now follows from (i). The proof of (i) shows that the general lower bound is determined by those ideals of m-supp (M) contained in m-supp (^{M}/_{M_{i}}) for some i ∈ {1, ..., t}. Tracking the proof of (ii), we can apply this to M_{1} and ^{M_{1}}/_{M_{1}∩M_{i}}, i ≠ 1, to prove (iii).

Proof of Proposition B: W.l.o.g., (a_{ij} + I_{i}) ∩ I ≠ ∅, for all i and j. We have (a_{ij} + I_{i}) ∩ I = a′_{ij} + (I_{i} ∩ I) with suitable elements a′_{ij} ∈ I, as is easily checked. Since m-supp (I) ⊂ m-spec (A), the assertion is an immediate consequence of Lemma 4(i).

Remark: (i) I ⊄ I_{i} in Proposition B is not actually a restriction as I ⊂ I_{i} implies (a + I_{i}) ∩ I = ∅ for a ∉ I_{i}, thus rendering I_{i} completely irrelevant. (ii) Generally, the lower bound of Lemma 4(i) can not be improved upon since for any m ∈ m-supp (M), M finite, there exists a maximal submodule N ⊂ M with ^{M}/_{N} ≃ ^{A}/_{m}, and so M is covered by the s = card (^{A}/_{m}) residue classes of N. (iii) The application of Lemma 4(ii) and (iii) to the ideal case is straightforward and requires just a fairly clear reformulation.

III. Examples. Exactness of Lower Bounds

To show that the lower bound in Proposition A (Lemma 3) can not be improved upon, we give

Example 1: Let M be a finite A-module, m ∈ m-supp (M). If ^{M}/_{mM} is non-cyclic, there exist submodules M_{0}, M_{α} ⊂ M, α ∈ ^{A}/_{m}, such that M = ( ⋃_{α∈^{A}/_{m}} M_{α})∪M_{0} is irredundant.

Proof: Assume that ^{M}/_{mM} has an ^{A}/_{m}-basis {x_{1}, ..., x_{n}}, n ≥ 2. If n > 2, we consider the A/_{m}-space \overline{M} = ^{M}/_{mM+∑_{i=3}^{n} Ax_{i}}, otherwise, we set \overline{M} = ^{M}/_{mM}. With x_{1} and x_{2} now

being considered as elements of \overline{M}, we have \overline{M} = (^{A}/_{m})x_{1} ⊕ (^{A}/_{m})x_{2}. If \overline{M}_{0} = (^{A}/_{m})x_{2} and \overline{M}_{α} = (^{A}/_{m}) (x_{1} + αx_{2}), for each α ∈ ^{A}/_{m}, then \overline{M} = ( ⋃_{α∈^{A}/_{m}} \overline{M}_{α}) ∪ \overline{M}_{0}. Indeed,

β ≠ 0 implies βx_{1} + γx_{2} = β(x_{1} + β^{−1}γx_{2}) ∈ \overline{M}_{β^{−1}γ}, whereas γx_{2} ∈ \overline{M}_{0} for β = 0, hence the above equality holds. \overline{M}_{α} ∩ \overline{M}_{0} = 0, for arbitrary α ∈ ^{A}/_{m}, being obvious, let x ∈ \overline{M}_{α} ∩ \overline{M}_{β}, α ≠ β, so that x = γ(x_{1} + αx_{2}) = γ(x_{1} + βx_{2}) implying γ = 0 and x = 0. Thus, the above equality is irredundant. M_{0} and M_{α}, α ∈ ^{A}/_{m}, of our initial assertion are obtained by taking inverse images in M of \overline{M}_{0}, \overline{M}_{α}, α ∈ ^{A}/_{m}.

To get an idea how this works, one may consider Z-modules of the form M = ^{Z}/_{nZ} ⊕ Z/_{nZ}, with n ∈ N, n > 1. We have m-supp (M) = {n_{1}Z, ..., n_{s}Z} if n_{1}, ..., n_{s} are the prime divisors of n. Setting N = (n_{i}Z)M, i ∈ {1, ..., s}, we get M_{0} = N + Z(0, 1), M_{1} = N + Z(1, 1), ..., M_{n_{i}} = N + Z(1, n_{i}) = N + Z(1, 0).

For the ideal case, consider A = k[x, y] with a field k and indeterminates x, y, I = (x, y). We have (x, y) ∈ supp (I), and ^{I}/_{I^{2}} is non-cyclic.

Finally, an equally simple example proves the lower bounds of Lemma 4 to be exact.

Example 2: Let A = Z, M = ^{Z}/_{6Z} ⊕^{Z}/_{6Z}, M_{1} = (2Z)M, M_{2} = {(0, a)|a ∈ ^{Z}/_{6Z}}. No inclusion relation holds between M_{1} and M_{2}, and we have M_{1} = {(0, 0), (0, 2), (0, 4), (2, 0), ..., (4, 4)}; M_{1} + (0, 1) = {(0, 1), (0, 3), ..., (4, 5)}; M_{2} + (2, 0) = {(2, a)|a ∈ ^{Z}/_{6Z}}; M_{2} + (4, 0) = {(4, a)|a ∈ ^{Z}/_{6Z}}. M_{1} ∪(M_{1} +(0, 1)) = M_{2} ∪(M_{2} +(2, 0))∪(M_{2} +(4, 0)) is easily checked. Hence M = (M_{1} + (1, 0))∪(M_{1} +(1, 1))∪M_{2} ∪(M_{2} +(2, 0))∪(M_{2} +(4, 0)) is irredundant. supp (M_{1}) = {3Z} and supp (M_{2}) = {2Z, 3Z} show that the present example hits the infimum of Lemma 4(ii).

References

[1] Matsumura, H.: Commutative Algebra. 2nd Ed. London: Benjamin–Cummings Publ. Comp. 1980.

[2] Storch, U., Wiebe, H.: Lehrbuch der Mathematik. Vol. II: Lineare Algebra. Mannheim: BI 1990.

Received January 15, 1995.