Results 1 to 4 of 4

Math Help - How would you prove this?

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    296

    How would you prove this?

    If I and J are ideals in a ring R that is commutative with identity, and I=(a) and J=(b), prove that IJ=(ab)

    So if I=(a), then I is the set of all multiples of a, and likewise J is the set of all multiples of b. So IJ={i1j1+....+ikjk} with each i a multiple of a and each j a multiple of b, and so, each product a multiple of ab. So the sum of products that are multiples of ab is a multiple of ab?

    Is that correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by zhupolongjoe View Post
    If I and J are ideals in a ring R that is commutative with identity, and I=(a) and J=(b), prove that IJ=(ab)

    So if I=(a), then I is the set of all multiples of a, and likewise J is the set of all multiples of b. So IJ={i1j1+....+ikjk} with each i a multiple of a and each j a multiple of b, and so, each product a multiple of ab. So the sum of products that are multiples of ab is a multiple of ab?

    Is that correct?
    This shows that  IJ \subseteq (ab) .

    Now you need to show  (ab) \subseteq IJ . Can you see how this is done?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    296
    Hm....so you start with all multiples of ab, that is, all numbers of the form nab for some integer n. No, sorry, not really sure how to go this direction.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by zhupolongjoe View Post
    Hm....so you start with all multiples of ab, that is, all numbers of the form nab for some integer n. No, sorry, not really sure how to go this direction.
    For the other direction, you need to show  ab \in IJ . Since  IJ is an ideal, it follows that  (ab) \subseteq IJ .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove a(AB)=(aA)B=A(aB) ..
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 29th 2010, 05:14 AM
  2. Prove: f is one-to-one iff f is onto
    Posted in the Discrete Math Forum
    Replies: 12
    Last Post: June 25th 2010, 11:02 AM
  3. Replies: 2
    Last Post: August 28th 2009, 03:59 AM
  4. Please prove
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 7th 2009, 02:58 PM
  5. Prove this .
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: February 18th 2009, 05:09 AM

Search Tags


/mathhelpforum @mathhelpforum