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About R-module

  1. May 11, 2009 #1
    R denotes a commutative ring with identity

    R-module: a module whose base ring is R

    Prove that if a nonzero commutative ring R with identity has the property that every finitely generated R-module is free then R is a field.

    Idon't know how to complete the proof. The only thing I know is that if every finitely generated R-module is free then R-module R is also free, since it's generated by {1}.

    Thanks for any help!
  2. jcsd
  3. May 11, 2009 #2
    Hint: If I is an ideal of R, then R/I is a cyclic (in particular, finitely generated) R-module. If R/I is free, what does this tell you about I?
  4. May 20, 2009 #3
    Thanks for reply! VKint

    What exactly does the term cyclic mean here? I am not familar with some terms.

    There may be some grammatical mistakes above, sorry for that, I've done my best to express what I want to say.
  5. May 20, 2009 #4
    By cyclic, I mean that the module [tex] R/I [/tex] can be generated by just one element (i.e., the coset [tex] 1 + I [/tex]) over [tex] R [/tex]. Thus, since [tex] R/I [/tex] is finitely generated, you can apply your assumption to the [tex] R [/tex]-module [tex] R/I [/tex] to conclude that [tex] R/I [/tex] must be free.

    One way of defining a field is as a ring with no nontrivial ideals. Thus, all you need to show is that [tex] R [/tex] can't have any nonzero ideals (besides itself, of course). Can you see why it would be contradictory for [tex] R/I [/tex] to be free if [tex] I [/tex] were not either [tex] 0 [/tex] or [tex] R [/tex]?
  6. Jun 19, 2009 #5
    VKint, I had some trouble in my work recently, so I didn't reply in time, I'm sorry.

    I don't kown whether the following is right, may you can help me again, thanks a lot!

    If v∈M\I and i∈I, then v and any i are relatively prime, if not, then there's a scalar s which divides both v and i, then s is contained in I, otherwise the scalar i divided by s doesn't belong to I.

    So, there exist scalars s1 and s2 for which s1v + s2i = 1,
    then s1v + s2i + I = 1 + I → s1v + I = 1 + I
    since s2i∈I, so s1v=1, that is to say s1 IS the multiplicative inverse of v.

    The left could be solved by induction.
    Last edited: Jun 20, 2009
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