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Free-module, basis, rank

  1. Jul 30, 2011 #1
    I am trying to understand the notions of rank of an R-Module, free-module, basis, etc.
    I would like to understand this line (expand on it, find some critical examples/counterexamples ,etc) that I am quoting from Dummit & Foote:

    "If the ring R=F is a field, then any maximal set of F-linearly independent elements is a basis for M (the module) . For a general integral domain, however, an R-Module M of rank n need not have a basis, i.e., need not be a free R-Module even if M is torsion free, so some care is necessary...."

    Before this came the definition: For an integral domain R, the rank of an R-module is the maximal number of R-linearly independent elements of M.

    Thank you
  2. jcsd
  3. Jul 30, 2011 #2
    Hi arthurhenry! :smile:

    Here are some counterexamples that you could ponder on:

    - [itex]\mathbb{Z}_2[/itex] is a Z-module of rank 0, but not free.
    - [itex]\mathbb{Q}[/itex] is a non-free Z-module
    - The k[X,Y]-module (X,Y) of k[X,Y] is of rank 2, is torsion-free, but is not free.
  4. Jul 30, 2011 #3
    Thank you Micromass, this is great...

    I think I understand and I am trying to clarify; so response warranted if correct:

    1) Z_2 is a Z-Module of rank 0, because not even one element of Z_2 is R-indepenedent. That is, take the element 1 in Z_2, there exists a non-zero integer m in Z (for example 2) such that m.1=0
    So the rank is zero.
    Also, the element 1 in Z_2 can be expressed in more than (not unique) way. For example, 3.1 is the same as 5.1, where 5 and 3 are in the ring Z. So it is not free. Page 354 Dummit & Foote says that not only the module elements need to be unique, but also the ring elements.

    3) Here I am assuming you are referring to the ideal generated by (X,Y)...
    If so, then, an ideal is a subring and is an abelian group, so the ring acts on it and we get a module (I understand)
    It is rank 2 because rX+sY is not zero for any nonzero elements r and s in k[X,Y]. It is torsion free (i understand this part).
    It is not free, though, because X.X+Y.Y=1.(X^2)+1.(Y^2), so the uniqueness fails I believe.

    3) Q again fails to be free because of uniqueness I believe.
  5. Jul 30, 2011 #4
    There is yet another comment on
    http://planetmath.org/encyclopedia/RankOfAModule.html [Broken]

    which says that two different basis sets for an R-Module may have different cardinality.
    This one I cannot produce an example for (I am trying two finite cases for example)
    Last edited by a moderator: May 5, 2017
  6. Jul 30, 2011 #5
    Looks good!!
    That Q is not free is maybe a bit more difficult to see, but it follows from a very important theorem: the fundamental theorem of finitely generated abelian groups. You'll see this in Dummit & Foote sooner or later (section 5.2 page 158).
    I just wanted to put Q here because it's such a beautiful example!

    See http://planetmath.org/?op=getobj&from=objects&id=10670 :smile:
    Last edited by a moderator: May 5, 2017
  7. Jul 30, 2011 #6
    Okay, perhaps I rushed in saying that I see why Q is not a free Z-Module.

    I am not able to see it. I am also not seeing how I can apply Fund. Theroem. of Fin. Gen. Abelian groups.
    Are you suggesting Micromass that "use it to get a contadiction"...I guess I am not sure if I see what is fin. generated.
    Thank you

    p.s. don't want to give up thinking, but I was not able to see the connection
  8. Jul 30, 2011 #7
    OK, you don't really need it, what was I saying? :frown:

    Anyway, if Q were free, then it would be isomorphic to ZI for a set I. But a cardinality-argument show that ZI is only countable if I is finite. So Q has to be isomorphic to a certain Zn, but this is impossible: take any finite set in Q, you can always find a number not generated by that set.
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