Results 1 to 5 of 5

Math Help - abstract algebra hw problem

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    3

    abstract algebra hw problem

    Hello,

    I can't seem to figure out a part of this question:

    Let F be a field and let I = { a(sub n)x^n + a(sub n-1)x^(n-1) + ... + a (sub 0), where the a(sub i) are elements of F and a(sub n) + ... + a (sub 0) = 0}. Show that I is an ideal of F[x].

    Here's what I have so far: So I defined an f(x) and g(x) that were elements of I and showed that f(x) - g(x) is in I. I'm trying to use the ideal test so I need to show now that, for an r(x) that's an element of F[x], r(x)f(x) and f(x)r(x) are in I. I can't figure out how to do this.

    Any help is much appreciated, thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Feb 2009
    From
    Loma Linda, CA
    Posts
    10
    Hi rush,

    Let  r \in F  \ { } f(x)=a_nx^n +a_{n-1}x^{n-1}+...+a_0 \in I

    then the right ideal  rf =\ { } r(a_nx^n +a_{n-1}x^{n-1}+...+a_0) distributing will give us

    (ra_n)x^n +(ra_{n-1})x^{n-1}+...+(ra_0)

    Is this in the ideal? note each of the coefficients ra_i \in F so all we need is for the coefficients to sum to 0
      ra_n +ra_{n-1}+...+ra_0 =r(a_n +a_{n-1}+...+a_0) =r(0)=0

    this places  rf \in I

    The other half I leave up to you
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    3
    doesn't the r really have to be an element in F[x] and not just one in F?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2009
    From
    Loma Linda, CA
    Posts
    10
    Sorry i misunderstood the definition of your F[x]. Now i understand that it is the set of linear combinations of x with elements in the field F. Then,
    regardless the element  r(x) \in F[x] will be of the form

     r(x)= b_n x^n +b_{n-1} x^{n-1}+...+b_o | b_i \in F

    but applying the same technique as before will be a little messy so an alternative you can try is to use the fact that

    a_n +a_{n-1}+...+a_0 =0 \iff f(1)=0 Since you are just summing the coefficients of f(x). Also note this is a characteristic of all elements in the ideal I.
    Using this result we can multiply the two functions and evaluate their product at 1

     h(x)=r(x)f(x)= (b_n x^n +b_{n-1} x^{n-1}+...+b_o)(a_nx^n +a_{n-1}x^{n-1}+...+a_0)

      h(1)=r(1)f(1)=(b_n (1)^n +b_{n-1} (1)^{n-1}+...+b_o)(a_n(1)^n +a_{n-1}(1)^{n-1}+...+a_0)

    (b_n +b_{n-1}+...+b_o)(a_n +a_{n-1}+...+a_0)=(b_n +b_{n-1}+...+b_o)(0)=0

    h(x)=r(x)f(x) \in I

    Hope this helps
    Last edited by Nokio720; March 15th 2009 at 07:15 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    3
    yes it does, thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Difficult Abstract Algebra Problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 4th 2010, 05:14 AM
  2. abstract algebra problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 17th 2010, 04:30 PM
  3. Abstract Algebra: a problem about ideals
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 8th 2010, 10:58 PM
  4. One more very confusing abstract algebra problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 18th 2010, 06:31 PM
  5. Abstract algebra homework problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 30th 2008, 08:52 AM

Search Tags


/mathhelpforum @mathhelpforum