Results 1 to 2 of 2

Math Help - k[x]-module

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    5

    k[x]-module

    Let K be a field, and (x) -> K[x] be an injective map where (x) is the principal ideal generated by x.

    (i) Prove that the quotient K[x]-module, K[x]/(x), is isomorphic to K as an abelian group and also as a K-vector space.

    Hence there is an exact sequence of K[x]-modules :

    0 -> (x) -> K[x] -> K -> 0

    (ii) Is this sequence split as a sequence of K[x]-modules ?

    (iii) Is this sequence split as a sequence of K-modules ?


    (i) is clear. But I can't make sense nor find a solution for (ii) and (iii).
    Which is why I require your help !
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Simply82358 View Post
    Let K be a field, and (x) -> K[x] be an injective map where (x) is the principal ideal generated by x.

    (i) Prove that the quotient K[x]-module, K[x]/(x), is isomorphic to K as an abelian group and also as a K-vector space.

    Hence there is an exact sequence of K[x]-modules :

    0 -> (x) -> K[x] -> K -> 0

    (ii) Is this sequence split as a sequence of K[x]-modules ?

    (iii) Is this sequence split as a sequence of K-modules ?


    (i) is clear. But I can't make sense nor find a solution for (ii) and (iii).
    Which is why I require your help !
    the answer to (ii) is negative and to (iii) is positive. to see this, let I=\langle x \rangle and \pi: K[x] \longrightarrow K[x]/I be the natural homomorphism. let \text{id} be the identity map over K[x]/I.

    suppose that \alpha: K[x]/I \longrightarrow K[x] is a K[x] homomorhism such that \pi \alpha = \text{id}. now, for any p(x) \in K[x] we have xp(x) \in I and thus 0=\alpha(xp(x)+I)=x \alpha(p(x)+I),

    which means \alpha(p(x)+I)=0, i.e. \alpha = 0 and thus \text{id}=\pi \alpha = 0, which is non-sense.

    for (ii) just define \beta : K[x]/I \longrightarrow K[x] by \beta(p(x)+I)=p(0). see that \beta is a well-defined K homomorphism and \pi \beta = \text{id}. (note that p(0)+I=p(x)+I.)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: November 30th 2011, 03:50 AM
  2. Module Help
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 5th 2011, 04:13 PM
  3. About some module over a DVR
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 13th 2011, 01:48 AM
  4. R-module
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 20th 2009, 11:14 AM
  5. Module 77
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 26th 2008, 02:30 PM

Search Tags


/mathhelpforum @mathhelpforum