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About a torsion module

  1. May 11, 2009 #1
    Notations:
    M denotes an abelian group under addtion
    R denotes a commutative ring with identity
    ann(?): Let M be an R-module and r∈R v∈M, then ann(v)={r∈R | rv=0}

    Terms:
    R-module: a module whose base ring is R
    torsion element: A nonzero element v∈M for which rv=0 for some nonzero r∈R
    torsion module: all elements of the module are torsion elements
    integral domain: a commutative ring R with identity with the property that for r,s∈R, rs≠0 if r≠0 and s≠0

    Question:
    Find a torsion module M for which ann(M)={0}.

    I wonder whether the base ring R of such M can only be an integral domain.
     
  2. jcsd
  3. May 11, 2009 #2
    On the contrary, if a commutative ring R has such a module M (and if M is finitely generated), then R cannot be an integral domain. Why? (Hint: In a domain, the intersection of nonzero ideals is nontrivial.)

    The original example I thought of was the ideal [tex] \langle x,y \rangle [/tex] in the [tex] k[x,y] [/tex]-module [tex]k[x,y]/\langle xy \rangle[/tex], but that doesn't quite work. I tried tweaking it a bit, but to no avail. Thus, the example I finally settled on uses a non-Noetherian ring R to circumvent the above argument that R cannot be an integral domain. Consider
    [tex] R = k [ x,\sqrt{x}, \sqrt[3]{x}, \ldots, \sqrt[n]{x}, \ldots ] [/tex].
    This is a non-Noetherian domain. The R-module [tex] R/\langle x \rangle [/tex] has the desired property (I think).
     
  4. May 20, 2009 #3
    Thanks! VKint
     
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