1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Algebra question - rings and ideals

  1. Feb 24, 2008 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [SOLVED] Algebra question - rings and ideals

    1. The problem statement, all variables and given/known data
    Let R be a (nonzero) commutative ring with identity and I be an ideal of I. Denote (I) the ideal of R[x] generated by I. The book says that (I) is the set of polynomials with coefficients in I. Why is that?

    3. The attempt at a solution

    Call A the set of polynomials with coefficients in I.

    R is commutative and so is R[x], therefor (I) is simply given by

    (I) = {a*p(x): a is in I and p(x) in R[x]}

    So clearly (I) is a subset of A.

    But for the other inclusion, given b_0+...+b_nx^n in A, we need to find an element a in I and a set {a_0,...,a_n} in R such that a*a_i = b_i for all i=1,...,n.

    How is this achieved??
     
  2. jcsd
  3. Feb 24, 2008 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    If b_0+...+b_nx^n is in A, then b_0, ..., b_n are all in I, and thus in (I). The latter is an ideal, so b_kx^k is certainly in (I), and so is the sum of such things.

    Edit:
    I don't really see how this follows. I would instead prove that A is an ideal of R that contains I and hence contains (I) by definition.
     
    Last edited: Feb 24, 2008
  4. Feb 24, 2008 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Of course it does not follows! I was confused.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Algebra question - rings and ideals
Loading...