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Module equivalence

  1. Aug 25, 2010 #1
    Module "equivalence"

    There is a problem in a book I'm not quite understanding.
    Let M be an R-module and let I=Ann(M). Show that M can be regarder as an R/I-Module where scalar multiplication is given by the rule m(I+r)=mr

    I don't understand what they mean by "regarded as". Am I suppose to show there is an isomorphism? (I don't know how to do that between modules over different rings), or just a bijection? Should I create a one-to-one mapping from the scalars R to R/I instead?

    Any help is appreciated
     
  2. jcsd
  3. Aug 25, 2010 #2

    Hurkyl

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    Re: Module "equivalence"

    Really, they probably just want you to show that the underlying additive group of M, together with that scalar multiplication, is an R/I-module.

    Bonus points for explicitly showing that, if [itex]\pi[/itex] is the projection from R to R/I, that [itex]r \cdot m = \pi(r) \cdot m[/itex]. (where the two [itex]\cdot[/itex] symbols refer to the appropriate scalar products)



    One main point is that, in the expression [itex]r \cdot m[/itex] where r is in R and m is in M, it doesn't matter if we mean this product as the scalar product on M, or if we are using r to name an element of R/I, and the product is the scalar product on the R/I-module defined in the problem; the result of the arithmetic expression is the same.
     
  4. Aug 26, 2010 #3
    Re: Module "equivalence"

    Ok, thank you Hurkyl!
    Just out of curiosity, what exactly does this result mean, the fact that the set M can be regarded as both an R-module and an R/I-module? Is this mostly a consequence of the properties of the quotient ring R/I (for example of, like you pointed out, the canonical homomorphism from R to R/I)?
    I'm finding it hard to wrap my head around it because R and R/I are not isomorphic (possibly not even of the same order if finite) yet behave the same as scalars over M?
     
  5. Aug 26, 2010 #4

    Hurkyl

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    Re: Module "equivalence"

    The more opportunities you have to use this fact, the more it will make sense.

    You already know that all of the elements of I behave exactly the same as scalar multipliers on M. So what do you get when you impose that they are all congruent? :smile:



    Incidentally, there are two well-behaved constructions related to a ring homomorphism f:R --> S.
    • (The underlying group of) any S-module M can be given an R-module structure defined by [itex]r\cdot m = f(r)\cdot m[/itex].
    • By the above, S itself can be viewed as an R-module. For any R-module M, the (underlying group of) the tensor product [itex]S \otimes_R M[/itex] can be given an S-module structure defined by [itex]s \cdot (s' \otimes m) = (ss') \otimes m[/itex].
    (These constructions extend to transform S-module homomorphisms into R-module ones and R-module into S-module ones respectively)


    When S is the quotient ring R/I, the constructions are even more well-behaved. The underlying group of the second construction is just (isomorphic to) the quotient [itex]M / IM[/itex]. Furthermore, the two constructions essentially become one-sided inverses -- applying the first construction and then the second construction gives you a module naturally isomorphic to the one you started with.

    Roughly speaking, this means that you can view the entire category of R/I-modules as if it was just a subcategory of R-modules. Another way of saying that is that it's unusually simple to use the module theory of R to study the module theory of R/I. How do you identify which R modules are also R/I modules under this correspondence? They are the ones with [itex]I \subseteq Ann(M)[/itex].
     
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