A modification to yesterday's question and three responses to the question, which I think are still valid for the modified question.........DMD __________________________________ Subject: Re: algebraic question, errata Date: Mon, 18 Aug 2003 14:57:17 -0400 From: Justin Smith On Mon, 2003-08-18 at 16:31, Don Davis wrote: > We cannot assume that the modules are finitely generated or free > (although they may be Z-free). I meant to say that they were Z-torsion free, but usually not Z-free. _________________________________________________ Subject: Re: algebraic question Date: Tue, 19 Aug 2003 07:52:53 -0400 From: "Vidhyanath Rao" > Suppose we have an infinite sequence > ... V_k \subseteq V_{k-1} \subseteq ... V_1 > of Z-modules such that each V_i is a direct summand of V_{i-1} > (and therefore, of V_1). > What can one say about V_{\infty} = intersection of all the V_i? > Is it a direct summand of V_1? Let V_1 = sum of denumerable number of copies of Z, with V_i the submodule of all elements whose the first i-1 coordinates are zero. Then V_i is the direct sum of V_{i+1} and the i-th copy of Z. The intersection of all V_i's is the trivial module. Nath Rao ____________________________________________________ Subject: Re: algebraic question Date: Tue, 19 Aug 2003 10:27:01 -0400 From: Tom Goodwillie I think not. Let P be the group of all infinite sequences (n_1,n_2,...) of integers. Let P_k be the subgroup defined by the condition n_i=0 for all i P for some V. Let V_k be the preimage of P_k in V. But I can't think why P (the product of countably many Z's) has no Z-basis. I think I used to know this. Can anyone enlighten me? Tom Goodwillie _________________________________________________ Subject: Re: algebraic question Date: Tue, 19 Aug 2003 12:22:48 -0400 (Eastern Daylight Time) From: "Nicholas J. Kuhn" The answer to Justin's question is no: V_infty need not be a direct summand of V_1. Here is a construction of an example... Z = integers. Let P be the infinite product Z x Z x ... It is a classic fact (with a sneaky proof) that P is NOT a free abelian group. Let V_1 be the free abelian group with basis P. Let V_infty be the kernel of the natural epi V_1 -->> P. Since P is not free, V_infty is not a summand of V_1. For n > 1, let V_n be the kernel of the composite V_1 -->> P -->> Z^{n-1}, where the second map is projection onto the first n-1 factors. It is easy to check: (i) V_n/V_n+1 = Z, so V_n+1 --> V_n splits, (ii) The intersection of the V_n's is V_infty, as defined above. Nick Kuhn