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Finitely generated modules as free modules

  1. Jul 13, 2011 #1
    I'm reading up on the classification of finitely generated modules over principal ideal domains. In doing so, I continuously come up on the statement "Let M be a finitely generated, free R-module."

    My question is, is this statement redundant? It seems to me that all finitely generated R-modules are necessarily free as R-modules. In particular, if M is an finitely generated R-module with minimal generating set [itex] A \subseteq M [/itex], then isn't the free R-module on A also M? Or am I missing a technical point?
  2. jcsd
  3. Jul 13, 2011 #2
    Hi Kreizhn! :smile:

    What you say is certainly true if R is a field (thus if we're working with vector spaces), as every module is free there. But R doesn't need to be a field.

    For example, let [itex]R=\mathbb{Z}[/itex] here. Then the free modules are all of the form [itex]\mathbb{Z}^n[/itex]. However, there are much more finitely generated modules. For example [itex]\mathbb{Z}_2[/itex] (the integers modulo 2) is certainly finitely generated, but it is not free.
  4. Jul 13, 2011 #3
    Ah yes, because in this instance, 1 generates [itex] \mathbb Z_2 [/itex] but the free module on the singleton would be [itex] \mathbb Z [/itex]?
  5. Jul 13, 2011 #4
    Indeed! :smile:
  6. Jul 13, 2011 #5
    I think I know where I made the mistake in my logic. M is finitely generated over R if there is a surjective homomorphism [itex] R^{\oplus A} \to M [/itex] for some finite subset [itex] A \subseteq M [/itex]. On the other hand, M is free if there exists a set B such that [itex] R^{\oplus B} \to M [/itex] is an isomorphism.

    So for this example, certainly [itex] \mathbb Z \to \mathbb Z /2\mathbb Z [/itex] is surjective, but there's no way this is could be an isomorphism.
  7. Jul 15, 2011 #6
    On this note, I would be interested in making something clear.

    When we think of free modules, does the word "free" essentially characterize the existence of a basis? Or is it possible for some modules to have a basis but not be free.

    In the case of our [itex] \mathbb Z_2 [/itex] example, I'm trying to think of whether {1} is a basis. Clearly {1} is a minimal generating set, but is it linearly independent? I feel that it isn't, since viewing [itex] \mathbb Z_2 [/itex] as a [itex] \mathbb Z [/itex] module, we have [itex] 2 \cdot 1 = 0 [/itex]. Is this correct?
  8. Jul 15, 2011 #7
    Hmm, I don't think we like using the word basis when not working in vector spaces. All free means is that there is an isomorphism [itex]\varphi:R^n\rightarrow M[/itex]. I guess you could see [itex]\varphi(1,0,...,0),...,\varphi(0,0,...,1)[/itex] as a basis of M. But we don't use that terminology (not sure why actually).

    Anyway, what you do have is that a module is finitely generated and free if there exists a finite set [itex]\{x_1,...,x_n\}[/itex] that generates the set and such that

    [tex]\sum{r_i x_i=0}~\Rightarrow~r_i=0[/tex]

    I guess we can call this linearly independent. But it's not standard terminology.
  9. Jul 15, 2011 #8
    I guess maybe the bases thing was a slight abuse of terminology, though I do have a definition of linear independence for general R-mods.

    Let [itex] i: I \to M [/itex] be a non-empty mapping from an index set I to an R-module M, and consider the free R-module on I denoted [itex] F^R(I) [/itex]. We know there is a canonical inclusion [itex] \iota: I \to F^R(I) [/itex], and so by the universal property of free R-modules, it follows that there exists an R-module homomorphism [itex] \phi: F^R(I) \to M [/itex], such that [itex] i = \phi \circ \iota [/itex]. If [itex] \phi [/itex] is injective, then [itex] i: I \to M [/itex] is linearly independent.

    I guess this answers my question though, since again there's no way that [itex] \mathbb Z \to \mathbb Z_2 [/itex] is injective. And I guess in particular, the kernel is the ideal [itex] 2 \mathbb Z [/itex].

    Okay, maybe a better question then. We know that all finite dimensional k-vector spaces (for k a field) are isomorphic. Does this hold in general for any given cardinals? Namely, let V and W be vector spaces whose module rank is both the cardinal [itex] \omega [/itex], is it necessary that [itex] V \cong_k W [/itex] as k-vector spaces?

    I use module rank because I'm not sure if "dimension" is appropriate in this context. I want to say that this is true, since if V and W are free k-modules of rank [itex] \omega [/itex] then there are isomorphisms [itex] k^{\oplus A} \to V, k^{\oplus B} \to W [/itex] where [itex] |A|=|B| = \omega [/itex]. But then there is a set bijection [itex] A \to B [/itex], so I figure this must make [itex] k^{\oplus A} \cong k^{\oplus B} [/itex] making V and W isomorphic as k-vector spaces.

    I've never seen this stated though, so I question whether or not I've done something wrong.
  10. Jul 15, 2011 #9
    Yes, this is true! Two vector spaces are equal if and only if they have the same dimension (=module rank). This holds for infinite cardinalities as well!!

    So if V and W both have dimension [itex]\aleph_0[/itex], then they are isomorphic!
  11. Jul 15, 2011 #10
    Excellent. Thanks so much.
  12. Jul 24, 2011 #11


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    Here is an important example that you might like to think about.

    A linear transformation of a finite dimensional vector space makes the vector space into a module over the ring of polynomials with coefficients in the base field. Since the ring of polynomials over a field is a principal ideal domain, the vector space is now a finitely generated module over a PID.

    The module structure is x.v = L(v) then extend by linearity to all of the polynomials.

    This module is not free. In fact it is a torsion module.
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