Results 1 to 2 of 2

Math Help - Action over quotient

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    48

    Action over quotient

    Let A a conmutative ring, M an A-module and J= \{ a \in A: am=0 \forall m \in M \}

    Prove that J is a subring of A and that  M is an R/J-module with action (s+J)m=sm.

    thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by roporte View Post
    Let A a conmutative ring, M an A-module and J= \{ a \in A: am=0 \forall m \in M \}

    Prove that J is a subring of A and that  M is an R/J-module with action (s+J)m=sm.

    thanks!
    J is actually more than just a subring ... it's an ideal of A! prove it! (J is called the annihilator of M in A) to show that M has a structure of an R/J module, we only need to show that the defined

    action is basically well-defined, because other properties of modules will be directly transfered from R-module structure of M. so suppose s_1 + J=s_2 + J. then s_1 -s_2 \in J. thus by the definition

    of J, we have: (s_1-s_2)m=0, \ \forall m \in M. hence: (s_1+J)m=s_1m=s_2m=(s_2 + J)m. Q.E.D.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. least action principle
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 17th 2011, 09:32 AM
  2. [SOLVED] Transitive action, blocks, primitive action, maximal subgroups.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 6th 2011, 07:39 PM
  3. Group action
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: April 15th 2010, 12:49 PM
  4. kernel of an action
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 26th 2008, 09:34 PM
  5. need a little limit help action
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 30th 2007, 09:34 PM

Search Tags


/mathhelpforum @mathhelpforum