RINGS WHOSE MODULES REQUIRE AN INVARIANT NUMBER OF MINIMAL GENERATORS

William H. Rant

Abstract. We examine rings having the property that, for each minimally generated module, the number of elements in a minimal generator set is invariant.

Introduction. Let R be a ring. A left R–module M is a minimally generated module if M contains a subset S = {m_{i} : i ∈ I} that generates M, and for each i, m_{i} ∉ span{m_{j} : j ≠ i}. The set S is a minimal generator set (abbreviated as “mgs”) of M. Every left module of a left perfect ring has a minimal generator set. A ring R has the left invariant minimality property (IMP) and is call an IMP ring if, for each minimally generated left R-module M, the number of elements in each minimal generator set of M is invariant. The prefix “left” (IMP) is presumed throughout. Unless stated to the contrary, a module is presumed to be a left module. Every module is presumed to be unitary and every ring has an identity. Throughout, “R” denotes a ring and “J” denotes the Jacobson radical of R.

\underline{Theorem 1}. A ring R has IMP if and only if R is a local ring.

\underline{Proof}. Assume R is a local ring with J = Rad(R). Let S = \{m_{i} : i ∈ I\} and T = {t_{k} : k ∈ K} be minimal generator sets of a minimally generated R–module M. In M/JM we let S^{∧} = {m_{i} + JM : i ∈ I} and T^{∧} = {t_{k} + JM : k ∈ K}. We will show that the canonical map p: S → S^{∧}, where p(m_{i}) = m_{i} + JM, for all i ∈ I, is a bijection. Suppose that p(m_{k}) = p(m_{n}) for some k ≠ n, then m_{k}+JM = m_{n}+JM, so m_{k} = m_{n} + ∑ r_{i}m_{i} for some r_{i} ∈ J, thus (1 − r_{k})m_{k} = ∑_{i≠k} r_{i}m_{i} + m_{n}. Since r_{k} ∈ J = Rad(R), r_{k} is a unit. m_{k} = ∑_{i≠k}(1 − r_{k})^{−1}r_{i}m_{i} + (1 − r_{k})^{−1}m_{n}. This is a contradiction since S is a minimal generator set of M. So p is injective (and clearly surjective). So card(S^{∧}) = card(S).

Now we show that S^{∧} is a basis for the vector space M/JM over the division ring R/J. Clearly S^{∧} generates M/JM. Suppose that S^{∧} is not a minimal generator set of M/JM. Then for some k, m_{k} + JM = ∑_{i≠k}(a_{i} + J)(m_{i} + JM), so m_{k} = ∑_{i≠k} a_{i}m_{i} + ∑_{i} r_{i}m_{i} for some r_{i} ∈ J. Hence, m_{k} = ∑_{i≠k}(1 − r_{k})^{−1}(a_{i} + r_{i})m_{i}, a contradiction since S = {m_{i} : i ∈ I} is a minimal generator set of M. We have shown that S^{∧} is a basis of M/JM and that card(S^{∧}) = card(S). In the same manner it can be shown that the mapping t_{j} → t_{j} + RM is a bijection from T to T^{∧} (so card(T^{∧}) = card(T)), and that T^{∧} is a basis of M/JM over R/J.

Since any two bases of a vector space have the same cardinality, we know that card(S^{∧}) = card(T^{∧}). Thus, card(S) = card(T). Hence, the ring R has IMP.

For the converse, assume that R has IMP. To show that R is a local ring we will show that R has only one maximal left ideal. Let M_{1} be a maximal left ideal of R. Suppose x ∈ R−M_{1}. We will show that x is a unit. Since R = M_{1} +Rx, 1 = r +ax for some r ∈ M_{1}, a ∈ R; so {r, x} generates R. Now {1} is a minimal generator set of R with one element, and since R has IMP, every mgs of R must have exactly one element. Thus, {r} or {x} is a mgs of R. Now {r} does not generate R. So {x} is a mgs of R. Hence, 1 = yx for some y ∈ R. Suppose xy ≠ 1. Then 1 − xy = e is a non–trivial idempotent and R = Re ⊕ R(1 − e). Hence, {e, 1 − e} is a minimal generator set of R with two elements, a contradiction. Hence, xy = 1; so x is a unit. Therefore, R has only one maximal left ideal. This proves that R is a local ring [1].

\underline{Definition}. Let C be a class of R–modules. A ring R has the invariant minimality property for modules in C if, for each minimally generated R–module in C, the number of elements in a minimal generator set is invariant.

\underline{Comments}. (A) The proof of Theorem 1 shows that the following are equivalent:

(1) Ring R has IMP;

(2) Ring R has IMP for the class of cyclic R–modules;

(3) Ring R has IMP for the class of free R–modules;

(4) Every minimal generator set of the module _{R}R contains exactly one element.

(B) Ring R has the left IMP if and only if ring R has the right IMP (since both

conditions are equivalent to R being a local ring). (C) The existence of a minimal generator set for each R–module does not imply

that the ring R has IMP. Let R be a left perfect ring that is not a local ring. Every left R–module of a left perfect ring has a mgs [5], but R doesn’t have IMP.

\underline{Corollary 1}. A ring R has IMP if and only if the ring R/J has IMP.

\underline{Proof}. R is a local ring if and only if R/J is a local ring. The statement follows from Theorem 1.

\underline{Corollary 2}. Let R be a ring such that each left R–module has a minimal generator set. Then, R has IMP if and only if R is a left perfect local ring.

\underline{Proof}. Assume R has IMP. Then R is a local ring. Let M be a non–zero R–module. The non–zero, minimally generated module M contains a maximal submodule [5], thus, JM ⊆ Rad(M) ≠ M ⇒ M ≠ JM. Hence, J = Rad(R) is left T –nilpotent [1]. Since ring R/J is semisimple (being a division ring) and J is left T –nilpotent, R is a left perfect ring [1]. Conversely, if we assume that R is a local ring then R has IMP by Theorem 1.

\underline{Corollary 3}. Let R be a semisimple ring. Then R has IMP if and only if R is a division ring.

\underline{Proof}. Assume R is a semisimple ring with IMP. Since R is a semisimple ring, J = Rad(R) = 0. Since R has IMP, R is a local ring; thus, R/J = R/{0} is a division ring. Hence, R is a division ring.

For the converse, if we assume that R is a division ring, then R is a local ring, which implies R has IMP.

In [4] a ring is defined to be an invariant basis number (IBN) ring if for every free module, the number of elements in a basis is invariant. It is proven as a consequence of [4] that a local ring is an invariant basis number ring. This result follows directly from Theorem 1.

\underline{Corollary 4}. If R is a local ring, then R is an invariant basis number ring.

\underline{Proof}. If R is a local ring then R has IMP by Theorem 1. Since a basis for a free left R–module F is also a minimal generator set of F, it follows that any two bases of F must have the same cardinality. So R is an IBN–ring.

\underline{Definition}. Let M be an R–module. M is finitely related if there exist an exact sequence 0 → G → F → M → 0 of finitely generated R–modules F and G, with F free.

\underline{Definition}. Let M be a minimally generated R–module and let S = \{m_{i} : i ∈ I\} be a mgs of M. Let F be a free R–module with basis {f_{i} : i ∈ I}. Consider the epimorphism p: F → M defined by ∑ a_{i}f_{i} → ∑ a_{i}m_{i} and let K = ker(f). If for some mgs S of M, K is a minimally generated R–module, then we define M to be a minimally related A–module.

It is known that every finitely related flat module is projective [4]. Over a local ring this extends to a minimally related flat module.

\underline{Theorem 2}. Every non–zero, flat, minimally related module of a local ring is a free module.

\underline{Proof}. Let M be a non–zero, flat, minimally related R–module of a local ring R. Let S = {m_{i} : i ∈ Q} be a mgs of M and let F be a free module with basis {f_{i} : i ∈ Q} such that the homomorphism f: F → M defined by ∑ a_{i}f_{i} → ∑ a_{i}m_{i} has a minimally generated kernel K. Let x = ∑ a_{i}f_{i} ∈ K. Since S is a mgs of M, each a_{i} is a non–unit, and since R is a local ring, each a_{i} must be in J = Rad(R), so K is contained in JF. Since M is a flat module, K∩JF = JK, so K = JK. Assume that K ≠ 0. Because K is minimally generated, K has a maximal submodule, so K ≠ RadK, and thus, K ≠ JK; a contradiction. Therefore, K = 0, so F ≈ M, and it follows that M is a free module.

\underline{References}

1. F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 1st ed., Springer–Verlag, New York, 1974.

2. H. Bass, “Finitistic Dimension and a Homological Generalization of Semiprimary Rings,” Transactions of the American Mathematical Society, 95 (1960), 466–488.

3. C. Faith, Algebra II: Ring Theory, 1st ed., Springer–Verlag, New York, 1976.

4. C. Faith, Algebra I: Rings, Modules, and Categories, corrected reprint, 1st ed., Springer–Verlag, New York, 1981.

5. W. H. Rant, “Minimally Generated Modules,” Canad. Math. Bull., 23 (1980), 103–105.

6. W. H. Rant, “Left Perfect Rings That are Right Perfect and a Characterization of Steinitz Rings,” Proceedings of the American Mathematical Society, 32 (1972), 81–84.

William H. Rant Department of Natural Sciences and Mathematics Lincoln University Jefferson City, MO 65101 email: rantw@lincolnu.edu