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A module problem

  1. May 5, 2007 #1

    Hurkyl

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    Let R be a commutative ring, with subring S.

    Let M be an R-module.

    Does there exist a S-module N such that [itex]N \otimes_S R \cong M[/itex] as R-modules? Preferably with N a sub-S-module of M?

    Even better, can we choose such modules N so that if we have an R-module homomorphism
    f:M --> M'​
    it yields an S-module homomorphism
    g:N --> N'​
    so that

    Code (Text):
            N (x) R ---> M
            |            |
    g (x) R |            | f
            V            V
            N' (x) R --> M'
     
    (with the horizontal arrows the aforementioned isomorphisms)
     
    Last edited: May 5, 2007
  2. jcsd
  3. May 5, 2007 #2

    matt grime

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    No. Is the simple answer. Such a module, M, I would be tempted to call S-free.

    For a well understood (by me) counter example, I like to think of group algebras. The functor [itex]?\otimes_{kH}kG[/itex] is the induction functor, and it is certainly not essentially surjective - though it is true, that every module is a summand of a kH-free object for some H (called the vertex subgroup, and this is a lovely theory of Green's from the 50s.

    In general, you're probably asking for an equivalence of categories here, given by restriction from R to S.
     
  4. May 5, 2007 #3

    Hurkyl

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    I didn't think it'd be that easy, but I couldn't see how to go about constructing a counterexample.


    I'm hoping to find circumstances for which, if I had two maps R --> T with S their equalizer, that S-mod --> R-mod would resemble an equalizer of the induced functors R --> T. And if not in Cat, then in what sort of 2-category it does.
     
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