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Difference between R-algebra and R-module (R is a commutative ring)

  1. Oct 1, 2008 #1
    The definition of R-algebra and R-module looks somewhat similar when R is a commutative ring.

    For instance, R[x] is both an R-algebra and R-module.

    I'll appreciate if anyone shows an example which is an R-algebra but not an R-module, and vice versa.
    Thanks.
     
    Last edited: Oct 1, 2008
  2. jcsd
  3. Oct 1, 2008 #2

    morphism

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    Just look at how an R-algebra is defined: it's really an R-module with extra structure. So while every R-algebra is an R-module - with same action of R - the converse need not be true. I'll let you try to think of an example.
     
  4. Oct 1, 2008 #3
    The difference is quite significant. An R-module M is firstly an abelian group; it has not multiplicative structure. An R-algebra A is a ring which when you forget about the multiplicative structure is also an R-module.

    To show you an example of a module that is not an alegebra: pick your favourite vector space. That's just a module over a field, for example R^n. This is clearly not an algebra since you have no notion of multiplying vectors to get another vector.

    However, any algebra R-algebra A can easily be turned into an R-module, by simply forgetting the multiplicative structure in A. For example lets take the R-algebra R[x]; these are just polynomials with variable x and coefficients in R with scalar multiplication by R. Throw out the multiplicative structure leaves you with an R-module usually denoted by Rx (no brackets). Elements in here are of the form rx where r\in R. x^2 is not the module since you can't multiply x by itself. You can simply add elements rx+sx=(r+s)x, r,s\in R
     
  5. Oct 1, 2008 #4

    mathwonk

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    if R and S are rings, an R algebra structure on S is a ring map R-->S.

    If M is an abelian group, and R a ring, an R module structure on M is a ring map

    R->End(M).
     
  6. Oct 2, 2008 #5
    Thanks for all replies. Now, I am getting a grasp on the fundamental differences between them.

    I still don't understand and decrypt what mathwonk said though.
    Algebra structure and module structure correspond a ring map?

    Any references or additional remarks will be highly appreciated.
     
  7. Oct 2, 2008 #6

    morphism

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    You haven't given us what your definitions of R-algebra and R-module are. What mathwonk said could be taken as definitions.

    I'm going to assume that when you say a X is an R-algebra or an R-module, then you're thinking of R acting on X in an appropriate way. So if x is in X and r is in R, then we can look at the element "rx" that sits in the R-algebra/module X.

    Now if S is an R-algebra, then one can define a map f:R->S by setting f(r)=r1 (where 1 is the identity element of S). You can easily check that f is a ring homomorphism (and that the image of f lies in the center of S). In this way an R-algebra structure on S is a ring homomorphism from R into the center of S.

    And if M is an R-module, then for each r in R one can define a map f_r:M->M by setting f_r(m)=rm. This is easily seen to be a group endomorphism of the abelian group M, i.e. f_r lies in the ring End(M). Then if we let g:R->End(M) be defined by g(r)=f_r, we see that g is a ring homomorphism. In other words, R acts on M as endomorphisms of M.

    Like I said, all this can be used to define R-algebras and R-modules. Try to prove that these definitions are equivalent to what you have.

    As for references, any algebra book should talk about this stuff, see e.g. Dummit & Foote, Hungerford, Lang, ...
     
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