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!

Principal Ideals

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Let I be an ideal of the commutative ring R, and let J = {y in R such that y^2 in I}
    a) If R is the polynomial ring Q[x] and I is the principal ideal of R generated by x^4 + x^2, show that J is the principal ideal of R generated by x^3 + x
    b) If R is a Principal Ideal Domain, show that J is an ideal of R

    2. Relevant equations

    3. The attempt at a solution
    To solve for part a), I tried to write I = (x^4 + x^2) and (x^3 + x) explicitly and it turned out be long messy polynomial equations and i couldn't really show it rigorously that J = (x^3 + x) is true. For part b), one can take R to be Q[x] since Q[x] is a P.I.D. (since Q is a field), and by a), J must then be an ideal of R.

    Any help on part a) particularly and b) would be very much appreciated.
  2. jcsd
  3. Dec 3, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I used < > isntead of ( ) for denoting ideal generated by since I thought it might be confusing when I'm referring to a polynomial vs. the ideal generated by it

    If I = <x4 + x2>... that IS I... what kind of form does "explicitly" take? A generic polynomial is going to be one of the form f(x)(x4 + x2) where f is a polynomial.

    A good place to start would be to show <x3 + x> is a subset of J... show x3 + x is in J, then for any polynomial f(x), f(x)*(x3 + x) is contained in J also.

    Your answer to (b) is wrong. You can't assume R is a specific principal ideal domain, you have to show that the result holds for any principal ideal domain (for example, what if the domain was Z?)
  4. Dec 3, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

    Just think of what membership in each ideal implies about what irreducible factors the polynomials must have.
  5. Dec 5, 2008 #4
    i've solved part b) and one direction of a); precisely the part that <x^3 + x> is contained in J but I have some difficulty proving the other inclusion. This is what I have so far: take any polynomial f in J, so f^2 is in I => so (x^4 + x^2) divides f^2 => x^2 divides f^2 (or even x divides f^2) and (x^2 + 1) divides f^2 but I can't show how to deduce that (x^3 + x) divides f => J is contained in (x^3 + x)
  6. Dec 5, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper

    If x|f*g then x|f or x|g. Doesn't that follow from the irreducibility of x?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Principal Ideals
  1. Principal Ideal Domain (Replies: 3)