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Showing every free module is Projective

  1. May 26, 2010 #1
    Take P a free module over an arbitrary ring. Show P is projective.

    Definition of Projective:
    If f: M-->N and g: P-->N are module homomorphisms with f surjective, then if P is projective there exists homomorphism h such that h: P-->M with f(h(x))=g(x) for all x.

    Obviously f has a right inverse, and so it is easy to find an h that works, but not a homomorphism h that works. I really have no idea about this and help would be appreciated.

    Second bit:

    Conversely, show that if P is a projective module over a Principal Ideal Domain, then P is free.
     
  2. jcsd
  3. May 26, 2010 #2

    Hurkyl

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    You have the definition of projective wrong:
    P is projective iff:
    For all homomorphisms f:M->N and g:P->N:
    If f is surjective, then:
    There exists a homomorphism h:P->M with fh = g​

    Anyways, isn't the set of all homomorphisms P->M easy to describe, when P is free?
     
  4. May 26, 2010 #3
    I can't see a difference between our definitions, but yours is certainly clearer.

    I don't know what that would be. I'd hazard a guess at matrices, but surely that relies on M being free as well?
     
  5. May 26, 2010 #4

    Hurkyl

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    I hate to just blurt out the answer to my question -- it's probably the most important property of free modules and I want to give you a chance to recall it, but anyways....

    If F is the free module on a set X of generators, then any function X-->M uniquely extends to module homomorphism F-->M. Conversely, any module homomorphism F-->M is completely determined by its values on X



    The difference between the definitions we gave is the order of quantifiers. e.g. by your definition, Z/2 would be a projective module over Z, as follows:

    Let M = Z/2 and N = Z/2.
    Let f:M->N and g:P->N be the identity homomorphisms.
    Note that f is surjective.
    Your definition says that I can now prove P projective, by letting h:P->M be the identity, because fh=g.

    However, the correct definition is univerally quantified over M,N,f, and g. I don't have to be able to find an h for just one particular choice of M,N,f,g -- I have to be able to find an h for every choice of M,N,f,g.
     
  6. May 26, 2010 #5
    About the spoiler; I didn't cover that in my course, I just checked my notes. But, it is fairly 'obvious' when you consider vector spaces, so I suppose I was just supposed to deduce it.

    And I see the difference now.

    Thanks for the help, I'll have another go at the problem.
     
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