Results 1 to 5 of 5

Math Help - intersection of ideals...

  1. #1
    Member
    Joined
    May 2008
    Posts
    86

    intersection of ideals...

    Hi... Thank you.
    I'm confused regarding the problem of finding an intersection of two specific ideals. How do I approach such a problem?
    I'm given R = Q[x] - the polynoms over Q.
    and these two ideals:
    I = I(x)R = (2x^3 + x^2 -6x -3)R; J= J(x)R = (2x^3-x^2-x)R.
    (I marked the specific polynoms as I(x) and J(x) for convinience).
    I unerstand what this means, and I understand that if p(x) is in I and J, then there exists r1(x) and r2(x) in R so that:
    p(x) = I(x)r1(x) = J(x)r2(x)
    However I have no idea how to simplify that and I feel I'm off the track...

    :-\
    Help?
    Thank you.
    Tomer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by aurora View Post
    Hi... Thank you.
    I'm confused regarding the problem of finding an intersection of two specific ideals. How do I approach such a problem?
    I'm given R = Q[x] - the polynoms over Q.
    and these two ideals:
    I = I(x)R = (2x^3 + x^2 -6x -3)R; J= J(x)R = (2x^3-x^2-x)R.
    (I marked the specific polynoms as I(x) and J(x) for convinience).
    I unerstand what this means, and I understand that if p(x) is in I and J, then there exists r1(x) and r2(x) in R so that:
    p(x) = I(x)r1(x) = J(x)r2(x)
    However I have no idea how to simplify that and I feel I'm off the track...

    :-\
    Help?
    Thank you.
    Tomer.
    first find the prime factorization of the generators of your ideal: I(x)=(x^2-3)(2x+1), \ \ J(x)=x(x-1)(2x+1). thus: I \cap J = \text{lcm}(I(x),J(x))\mathbb{Q}[x]=x(x-1)(2x+1)(x^2 - 3)\mathbb{Q}[x]. \ \Box


    Note: recall that \mathbb{Q}[x] is a UFD (a PID atually) and things in a UFD are very easy!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    86
    Thank you.
    A friend of mine had offered an explenation to why the intersection is what you offered.... but I would be glad to hear your explenation, becuase I'm not sure I understood it.
    I mean, you said the intersection is the lcm of both polynoms. I haven't studied anything like that. Are you basing this claim on a more general one? Could you quote it perhaps?
    Oh, and what's PID? I just started studying "fraction rings" (I think that's the translation... I''m not studying in English), and I think that's what you mean by UFD (but I could be wrong).
    If you could give a short explenation I would, as usual, be very glad

    Thank you!
    Tomer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by aurora View Post
    Thank you.
    A friend of mine had offered an explenation to why the intersection is what you offered.... but I would be glad to hear your explenation, becuase I'm not sure I understood it.
    I mean, you said the intersection is the lcm of both polynoms. I haven't studied anything like that. Are you basing this claim on a more general one? Could you quote it perhaps?
    Oh, and what's PID? I just started studying "fraction rings" (I think that's the translation... I''m not studying in English), and I think that's what you mean by UFD (but I could be wrong).
    If you could give a short explenation I would, as usual, be very glad

    Thank you!
    Tomer.
    see here for the definition of UFD. a PID (principal ideal domain) is an integral domain in which every ideal can be generated by one element only. ok, here's a simple explanation of what i did:

    following your notation, let K(x)=x(x-1)(2x+1)(x^2 - 3), and K=K(x)R. obviously K \subseteq I \cap J. for the other side, suppose f(x) \in I \cap J. then f(x)=I(x)g(x)=J(x)h(x), for some

    g(x), h(x) in R. thus: (x^2 - 3)g(x)=x(x-1)h(x). hence x^2 - 3 \mid x(x-1)h(x). but clearly \gcd(x^2 - 3, x(x-1))=1, thus x^2 - 3 \mid h(x). so let h(x)=(x^2 - 3)u(x). then we will have:

    f(x)=J(x)h(x)=x(x-1)(2x+1)(x^2 - 3)u(x) \in K. therefore I \cap J \subseteq K, and we're done!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2008
    Posts
    86
    Very nice.
    I'm tired of thanking you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Product ideals vs. products of ideals
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 17th 2011, 05:24 AM
  2. Prime Ideals, Maximal Ideals
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: March 7th 2011, 07:02 AM
  3. Ideals
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 8th 2010, 08:09 PM
  4. Intersection of Ideals
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 3rd 2010, 08:06 AM
  5. When are principal ideals prime ideals?
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 5th 2008, 12:18 PM

Search Tags


/mathhelpforum @mathhelpforum