Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook