arXiv subjects

Rüdiger Göbel

Publications and source records attributed to Rüdiger Göbel.

22 records · Page 2Linked to original sources

arXiv2000

Philip Hall's Problem On Non-Abelian Splitters

Philip Hall raised around 1965 the following question which is stated in the Kourovka Notebook: Is there a non-trivial group which is isomorphic with every proper extension of itself by itself? We will decompose the problem into two parts: We want to find non-commutative splitters, that are groups G not= 1 with Ext(G,G)=1 . The class of splitters fortunately is quite large so that extra properties can be added to G. We can consider groups G with the following properties: There is a complete group L with cartesian product L^omega cong G, Hom(L^omega,S_omega)=0 (S_omega the infinite symmetric group acting on omega) and End(L,L)=Inn(L) cup {0}. We will show that these properties ensure that G is a splitter and hence obviously a Hall-group in the above sense. Then we will apply a recent result from our joint paper math.GR/0009089 which also shows that such groups exist, in fact there is a class of Hall-groups which is not a set.

math.GR↗

arXiv2000

Reflexive subgroups of the Baer-Specker group and Martin's axiom

In two recent papers (math.LO/0003164 and math.LO/0003165) we answered a question raised in the book by Eklof and Mekler (p. 455, Problem 12) under the set theoretical hypothesis of diamondsuit_{aleph_1} which holds in many models of set theory, respectively of the special continuum hypothesis (CH). The objects are reflexive modules over countable principal ideal domains R, which are not fields. Following H.Bass, an R-module G is reflexive if the evaluation map s:G ---> G^{**} is an isomorphism. Here G^*=Hom(G,R) denotes the dual module of G. We proved the existence of reflexive R-modules G of infinite rank with G not cong G+R, which provide (even essentially indecomposable) counter examples to the question mentioned above. Is CH a necessary condition to find `nasty' reflexive modules? In the last part of this paper we will show (assuming the existence of supercompact cardinals) that large reflexive modules always have large summands. So at least being essentially indecomposable needs an additional set theoretic assumption. However the assumption need not be CH as shown in the first part of this paper. We will use Martin's axiom to find reflexive modules with the above decomposition which are submodules of the Baer-Specker module R^omega.

math.LO↗

arXiv2000

On our paper `Almost Free Splitter', a correction

Let R be a subring of Q and recall from math.LO/9910161 that an R-module G is a splitter if Ext_R(G,G)=0. We correct the statement of Main Theorem 1.5 in math.LO/9910161. Assuming CH any aleph_1$-free splitter of cardinality aleph_1 is free over its nucleus as shown in math.LO/9910161. Generally these modules are very close to being free as explained below. This change follows from math.LO/9910161 and is due to an incomplete proof (noticed thanks to Paul Eklof) in the first section of math.LO/9910161. Assuming the negation of CH, in Shelah [Sh:F417] (work in progress) it will be shown that under Martin's axiom these splitters are free indeed. However there are models of set theory having non-free aleph_1-free splitter of cardinality aleph_1.

math.LO↗

arXiv1994

Subgroups of the Baer-Specker Group with Few Endomorphisms but Large Dual

The Baer-Specker group is the product of countably many copies of the additive group Z of integers. Assuming the continuum hypothesis, we construct a pure subgroup G of the Baer-Specker group with the following properties. Every endomorphism of G differs from a scalar multiplication by an endomorphism of finite rank. Yet G has uncountably many homomorphisms to Z.

math.LO↗