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Simple question of interpretation - D&F Ch 10 - Proposition 30

  1. May 19, 2014 #1
    I am reading Dummit and Foote, Section 10.5 : Exact Sequences - Projective, Injective and Flat Modules.

    I am studying Proposition 30 (D&F, page 389)

    I need some help in order to interpret one of the statements in Proposition 30.

    Proposition 29 and its proof (Ch 10, D&F page 388) reads as follows:

    attachment.php?attachmentid=69913&stc=1&d=1400547687.png


    Statement (3) of Proposition 30 begins as follows:

    "(3) If P is a quotient of the R-module N ... "

    I am uncertain regarding the exact meaning of this statement ... I suspect it means that there exists a sub-module module Q of N such that P = N/Q ...

    BUT ... firstly, this seems a vague thing to assert and secondly, I am most uncertain of this interpretation ...

    Can someone please clarify the matter for me?

    Peter
     

    Attached Files:

  2. jcsd
  3. May 20, 2014 #2
    Your guess is essentially correct although [itex] Q [/itex] (which I will call [itex] L [/itex] to agree with the exact sequence in condition (3) ) should be a submodule of [itex] M [/itex] rather than [itex] N [/itex] ([itex] N [/itex] doesn't appear at all in condition (3).) So it means there exists some submodule [itex] L\subseteq M [/itex] such that [itex] P=M/L [/itex].

    For lots of these special types of modules (projective, injective, flat etc.) the definition is some nice property that holds for all vector spaces but not for all modules. We would really like it to hold for modules also so that standard vector space proofs can be extended to modules, so we simply define a special class of modules as those which do have this property.

    In this case, you know that if [itex] M [/itex] is a vector space and [itex] P=M/L [/itex] for some subspace [itex] L [/itex], then you can take a basis [itex] \{ v_1,\cdots, v_m \} [/itex] of L, extend it to a basis [itex] \{v_1,\cdots, v_m,v_{m+1}, \cdots, v_n\} [/itex] of [itex] M [/itex] and then [itex] M=L\oplus \mathrm{Span}(v_{m+1},\cdots, v_{n} ) [/itex]. Further, it is easy to check that [itex] P\cong \mathrm{Span}(v_{m+1},\cdots, v_n\} [/itex] (more precisely you take the image of the [itex] v_j[/itex]'s in the quotient) so we get the isomorphism
    [tex] M\cong L\oplus V/L=L\oplus P. [/tex]
    This shows that [itex] P [/itex] is a direct summand of [itex] V[/itex] as stated in condition (3). The fact that the exact sequence splits is of course equivalent to the direct sum decomposition (this is part of the splitting lemma.)

    So, this way of defining a projective module just tells you that a projective module is a module that we force to have the nice property that we can always split it into a direct some of any quotient plus a complement. If you are trying to prove something about modules and run into a situation where you need to split a module into a direct sum to imitate a proof which works for vector spaces, you can do it when the module is projective and you start with a quotient (similarly if you start with a subspace and need to find a splitting, you can do it when the module is injective.)
     
  4. May 20, 2014 #3
    Thanks Terandol ... Your post was EXTREMELY helpful to me ...

    Still reflecting on what you have said ...

    Gives me a good perspective on what is going on overall in the bigger picture of things ...

    Thanks again,

    Peter
     
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