Open Problems
Pere Ara
Problem. Determine the kernel of the natural map GL1(R) --t K1(R) when R is a separative exchange ring. This has been solved for unit-regular rings and regular right self-injective rings by Menal and Moncasi [Po Menal and J. Moncasi, Kl of von Neumann regular rings, J. Pure and Applied Algebra 33 (1984), 295-312], and for exchange rings with primitive factor rings Artinian by Chen and Li [H. Chen and F.-U. Li, Whitehead groups of exchange rings with primitive factor rings artinian, Preprintj. In all cases, one gets K1(R) = GL1(R)ab provided that 1/2 E R.
Yoshitomo Baba Problem. For given modules M and N, we say that Mis N-injective if there exists an extension homomorphism f' from N to M for each sub• module N' of N and each homomorphism f from N' to M. Also we say M is simple-N-injective if there exists an extension homomorphism f' from N to M for each submodule N' of N and each homomorphism f from N' to M with the image of f simple. Let R be a perfect ring and M, N given R-modules with the sode of M simple. If M is simple-N-injective, then is ManN-injective module? If R is a semiprimary ring, then we know that any simple-N-injective module is N-injective for given R-modules M and N with the so de of M simple by Proposition 2 in [Y. Baba and K. Oshiro, On a theorem of Fuller, J. Algebra 154 (1993), 86-94j. This open problem was first given by M. Hoshino of the University of Tsukuba to Y. Baba.
Gary F. Birkenmeier, Jae Keol Park and Jin Yong Kim Problem. In [G. F. Birkenmeier, Idempotents and completely semiprime ideals, Comm. Algebra 11 (1983),567-580], an idempotent e E R is called left (right) semicentral if Re = eRe (eR = eRe). We use S£(R) and Sr(R) for the sets of all left and right semicentral idempotents, respectively. For an idempotent e E R, observe that S£(eRe) = {O, e} if and only if Sr (eRe) = {O, e}; when this occurs e is said to be semicentral reduced. If 1, the unity of R, is semicentral reduced, then R is said to be semicentral reduced [G. F. Birkenmeier, H. E. Heatherly, J. Y. Kim and J. K. Park, Open Problems 442
Triangular matrix representations, To appear in J. Algebra]. In the paper "Semicentral reduced algebras" by G. F. Birkenmeier, J. Y. Kim and J. K. Park in this Proceedings, it is shown that for any positive integer n, R is semicentral reduced if and only if the n-by-n full matrix ring Mn{R) is semicentral reduced. Now we ask: Is the semicentral reduced property a Morita invariant property?
Yasuyuki Hirano Problem. A ring is called a quasi-Baer ring, if the left annihilator of every left ideal is generated by an idempotent. Let R be a quasi-Baer ring, G a finite group and suppose that the order of G is invertible in R. Then is the group ring R[GJ quasi-Baer?
Hidetoshi Marubayashi Problem 1. Find necessary and sufficient conditions for the generalized crossed product algebras R * G to be semi-hereditary and Priifer in terms of G and R. Problem 2. Let P be a prime ideal of a fully bounded Priifer order in a simple Artinian ring. Is R/P a Goldie ring? (If R is a Dubrovin valuation ring, then the answer is affirmative.) Problem 3. Does the approximation theorem hold in the case that the order is a semi-local Bezout order without the assumption that the order is a PI-ring? We may assume that R;, are all fully bounded Dubrovin valuation rings. For definitions and relevant references, see the paper "Non• Commutative valuation rings and their global theories" by H. Marubayashi in this Proceedings.
Kaoru Motose Problem. Let Recently it has been shown that w. K. Nicholson and M. F. Yousif Problem. Following Harada, an associative ring R is called right simple• injective if, for each right ideal T of R, every R-linear map from T into R with simple-image extends to R, i.e., the map is given by left multiplication by an element c of R. In [W. K. Nicholson and M. F. Yousif, On perfect simple-injective rings, Proc. Amer. Math. Soc. 125 (1997), 979-985]' it was shown that every semiprimary right simple-injective ring R is right self-injective, every left perfect left and right simple-injective ring R is a quasi-Frobenius ring, and an example of a left perfect left simple-injective ring was provided. It is an open question whether every left perfect right simple-injective ring is right self-injective. K. M. Rangaswamy Problem. Is a Butler module over a valuation domain completely decom• posable? (For the definition and relevant references, see the paper "On torsion-free modules over valuation domains" by K. M. Rangaswamy in this Proceedings.) s. Tariq Rizvi If x, yare elements in a modular lattice L with x ::; y, then y/x will denote the interval [x, y], i.e., y/x = {a ELI x::; a::; y}. The Krull dimension of a partially ordered set (P, ::;), denoted by "'( P), is an ordinal number defined recursively as follows: ",(P) = -1 if and only if P is a trivial poset, where -1 is assumed to be the predecessor of zero. Let a 2 0 be an ordinal number, and assume that we have already defined which posets have Krull dimension (3 for any ordinal (3 < a. Then we define: ",(P) = a if and only if ",(P) ~ (3 for all ordinals (3 with (3 < a, and for any descending chain Xl 2 X2 2 ... 2 Xn 2 ... of elements of P, there exists a positive integer no such that "'(Xn/Xn+l) < a for all n 2 no, i.e., for all n 2 no, "'(Xn/Xn+l) has previously been defined and it is an ordinal < a. If no ordinal a exists such that ",(P) = a, we say that P does not have Krull dimension. The dual Krull dimension of the poset P, denoted by ",O(P), is defined as being the Krull dimension ",(PO) of the opposite poset pO of P (if it exists). Recall that a module MR is called quotient finite dimensional (or qfd), if every factor module of M has finite uniform (or Goldie) dimension. For Open Problems 444 any module MR, we denote: C(M) := {X ~ M I X is cyclic}; F(M) := {X ~ M I X is finitely generated}. Recently, Albu and Rizvi in [1, Theorem 1.12] have shown that MR is Artinian {:} M is qfd and r;,(C(M)) ~ O. (Here, r;,( C(M)) means the Krull dimension of the poset C(M) of all cyclic submodules of M, ordered by inclusion.) It is natural to ask whether this result can be extended to an arbitrary Krull dimension a? Hence prove or disprove: Problem 1. For any right R-module M, r;,(MR) ~ a {:} M is qfd and r;,(C(M)) ~ a. Remark. We have shown that a slightly weaker version of this problem holds true in case one replaces C(M) by F(M), see [1, Theorem 1.17]: r;,(MR) ~ a {:} M is qfd and r;,(F(M)) ~ a. In [6, Lemma 6] (also [5, Theorem 6.3]), Huynh, Dung and Smith have shown the following interesting and useful result: The following are equivalent for a module MR and an ordinal a ~ O. (1) r;,(M) ~ a. (2) Every homomorphic image of M has an essential submodule E with r;,(E) ~ a. It would be very nice to obtain a dual of this result. Hence prove or disprove: Problem 2. The following are equivalent for a module MR and an ordinal a ~ O. (1) r;,°(M) ~ a. (2) Every submodule X of M has a small submodule S with r;,°(XjS) ~ a. Remarks. We suspect that this equivalence may hold true. Note that the equivalence holds true for a = 0 since r;,°(Xj S) ~ 0 for a small submodule S of X implies that X is finitely generated. In order to prove this result in general, we only need to show that condition (2) implies that M is qfd, and then use [1, Proposition 2.2]. A module M is called extending if every submodule of M is essential in a direct summand. We call a module FI-extending if every fully invariant Open Problems 445 submodule is essential in a direct summand. In a recent paper, Birken• meier, Muller and Rizvi [2) have shown that in contrast to the extending modules, arbitrary direct sums of FI-extending modules are always FI• extending. (The FI-extending property is also shown to carryover to matrix rings, unlike the extending property.) It is presently unknown if the FI-extending property is inherited by direct summands of arbitrary FI-extending modules. Problem 3. Let M be an FI-extending R-module. Is every direct sum• mand of M also FI-extending? If not, then find necessary and sufficient conditions for this to hold true. Remarks. Birkenmeier, C81ugareanu, Fuchs and Goeters in their inves• tigations have proved that for FI-extending abelian groups, the answer to this question is yes [3, Theorem 3.2). Recently Birkenmeier, Park and Rizvi [4) have shown that the answer is also affirmative if the FI-extending module is nonsingular. References [1) T. Albu and S. T. Rizvi, Chain conditions on quotients dimensional modules, Comm. Algebra, to appear. [2) G. F. Birkenmeier, B. J. MUller and S, T. Rizvi, Modules in which every fully invariant submodule is essential in a direct summand, preprint. [3) G. F. Birkenmeier, G. C81ugareanu, L. Fuchs and H. P. Goeters, The fully-invariant-extension property for abelian groups, Comm. Algebra, to appear. [4) G. F. Birkenmeier, J. K. Park and S. T. Rizvi, Modules with fully in• variant submodules essential in fully invariant summands, preprint. [5) D. V. Huynh, N. V. Dung and P. F. Smith, A characterization of rings with Krull dimension, J. Algebra 32 (1990), 104-112. Mingyi Wang A coalgebra C is called right (resp. left) conoetherian if every quotient comodule C / I of Cc (resp. cC) is finitely cogenerated. A coalgebra C is called right (resp. left) Co-Noetherian if every quotient module C / I of Cc (resp. cC) is quasi-finite. Problem 1. Is every right conoetherian coalagebra right Co-Noetherian? Does there exist a one-sided conoetherian or Co-Noetherian coalgebra? Open Problems 446 Problem 2. Let Te be a classical tilting comodule over a right conoethe• rian right semiperfect coalgebra and D the coendomorphism coalgebra of Te. Is D a left conoetherian left semiperfect coalgebra? Is DT a classical tilting comodule? References [1 J M. Wang and Z. X. Wu, Conoetherian coalgebras, Algebra Colloq. 5 (1998), 117-120. [2J M. Wang, Some co-hom functors and classical tilting comodules, SEA Bull. Math. 22 (1998), 455-468. [3J M. Wang, Tilting comodules over semi-perfect coalgebras, Algebra Colloq. 6 (1999), 461-472. Alexander Zimmermann Problem. Given a finite group G, is there a tilting complex in the de• rived category of the integral group ring Db(ZG) which is not a Morita bimodule?