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Thread: submodules of finitely generate modules

  1. #1
    Ant is offline
    Apr 2008

    submodules of finitely generate modules

    In my lecture the professor gave an example of a submodule of a finitely generated module not being free as:

    The submodule generated by [2] in Z/6Z. With the Z/6Z considered as a module over itself in the usual way.

    Because this submodule is of cardinality 3, it cannot be free. Why is this??

    Also, he stated that Z/2Z isn't free as a Z-module. Couldn't [1] form a basis?? Is it because it can't be linearly independent as x[1] for any even x yields 0. or Equivalently the uniqueness of expressions fails, ie. [1] = 1[1] = 3[1] = 5[1] etc?

    Does is therefore follow that Z/nZ for any n cannot be free as a Z-module?

    Thanks for anyhelp!
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  2. #2
    MHF Contributor

    Mar 2011

    Re: submodules of finitely generate modules

    to answer the second question first: yes, Z/2Z isn't free as a Z-module, because the set {[1]} isn't Z-linearly independent. so while {[1]} is a Z-spanning set, it's not a Z-basis. the ring you use for an abelian group to make it a module matters.

    the first question is the "deeper" one. for a commutative ring R, considered as an R-module, an R-submodule is an ideal of R.

    now an R-module M is said to be "free over the set S" if, for ANY other R-module N, and for any map (function) f:S-->N there is a unique R-module homomorphism (R-linear map) φ:M-->N with φ(s) = f(s) for all s in S.

    let's see what goes wrong with this in the case of M = Z/6Z, S = {[0],[2],[4]}.

    suppose we choose N = M, and the set-mapping f([0]) = [0], f([2]) = 2, f([4]) = [0].

    then, for our homomorphism φ, we must have φ([4]) = [0].

    since φ is to be R-linear, we must have φ([2]) = φ([4] + [4]) = φ([4]) + φ([4]) = [0] + [0] = [0], which is certainly NOT f([2]). so no such homomorphism φ even exists.

    put another way, {[2]} is not a Z/6Z-basis for ([2]), since we clearly have [3][2] = [0], with [3] ≠ [0] (so it's not Z/6Z-linearly independent).

    part of the problem here is that Z/6Z is not a PID (which has some nice ramifications for modules), even though Z/6Z is a cyclic ring (so all its ideals are principal). specifically, Z/6Z is not a domain (because 6 is not prime), so it has zero divisors. [2] is one of those zero-divisors, which is bad news for the submodule generated by [2].

    naively, this means the peculiarities of the ring R for an R-module, can play havoc with our intuition. the more "integer-like" the ring, the more an R-module behaves like we expect it to. PID's and (even better) euclidean domains make for very "nice modules", and at the far extreme, where R is a field, we get the very regular behavior of vector spaces (every F-module (vector space) IS free over any basis, which means we can often forget about which basis we're using and just remember the dimension).

    (note: when R is PID, such as is the case when R = Z, then the notions of free and torsion-free (no elements of finite (additive) order) coincide. so, yes, Z/nZ as a Z-module is not torsion-free, so it cannot be free over Z).
    Last edited by Deveno; Feb 12th 2013 at 01:37 PM.
    Thanks from Ant
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