A modification to yesterday's question and three responses to the question,
which I think are still valid for the modified question.........DMD
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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.
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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
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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
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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