Results 1 to 6 of 6

Math Help - A prime ideal

  1. #1
    Newbie
    Joined
    Feb 2009
    From
    Mukilteo, WA
    Posts
    10
    Awards
    1

    A prime ideal

    Hi everyone, this is my first post. Hope you don't mind that I'm using it to ask a homework question.

    Here's the problem:
    Let I = \{f\in\mathbb{Z}\left[x\right] : f(0) = 0\}.

    1. Show I is a prime ideal.
    2. Show I is not maximal.

    For 1), I let f(x),g(x)\in\mathbb{Z}\left[x\right] and f(x)g(x)\in I; that is, f(0)g(0) = 0. Because \mathbb{Z}\left[x\right] is an integral domain, it has no zero divisors; thus, either f(0) or g(0) must be 0 and hence in I. Is this right?

    I wasn't sure about 2); can I take J = \{f\in\mathbb{Z}\left[x\right] : f(0) = 0 or f(\frac{1}{2}) = 0\}?

    Thanks, Sam
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Hi

    1) Yep

    2) 2x\in J\ \text{and}\ -2x+1\in J\ , but...

    Try to work with an ideal generated by two elements.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    From
    Mukilteo, WA
    Posts
    10
    Awards
    1
    Thanks for your help clic-clac. I'm still a little confused, though.

    The polynomials such that f(0) = 0 are all polynomials with constant term 0. So I need to find an ideal that contains these and a polynomial with a non-zero constant term. So could I take \{f(x)\in\mathbb{Z}\left[x\right] : f(0) = 0\} \cup (2)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2009
    From
    Mukilteo, WA
    Posts
    10
    Awards
    1
    Actually I think (2) would be the whole ring. Hmm... how can I find a generating element that includes all functions that vanish at 0 without getting the whole ring?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Your new answer isn't so far from a solution: the idea of "adding" 2, which wasn't in your first ideal, is good.

    But
    a reunion of ideals is, in general, not an ideal. The construction that transforms a reunion of ideals into an ideal is the ideal generated by a set.

    To simplify notations, do you agree that \{f(x)\in\mathbb{Z}[x];\ f(0)=0\} is the ideal (x) in \mathbb{Z}[x] ?

    Since (x)\cup(2) is not a ideal, we can consider <(x)\cup(2)>, which is also written (x,2),\ (x)+(2) or x\mathbb{Z}[x]+2\mathbb{Z}[x]

    (x,2) contains (x) and 2 so it strictly contains (x). The question now is (x,2)\neq\mathbb{Z}[x]\ ?

    So take a p(x)\in (x,2), i.e. \exists a,b\in\mathbb{Z}[x],\ p(x)=xa(x)+2b(x). Is p(x)=1 possible? (That will answer the problem, since for any ideal \mathfrak{A},\ 1\in\mathfrak{A}\Leftrightarrow \mathfrak{A}=\mathbb{Z}[x])
    Last edited by clic-clac; February 19th 2009 at 12:07 AM. Reason: cor
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2009
    From
    Mukilteo, WA
    Posts
    10
    Awards
    1
    I was about to say what about b(x) = 1/2, then I realized we're working in Z[x]. Thank you so much clic-clac!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. ideal,nil,nilpotent ideal in prime ring
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 24th 2011, 07:57 AM
  2. prove N is a maximal ideal iff N is a prime ideal
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 20th 2011, 09:02 AM
  3. Ideal a is irreducible <--> a=p^n, p is prime ideal
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: July 3rd 2010, 10:54 PM
  4. Maximal Ideal, Prime Ideal
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 28th 2008, 02:39 PM
  5. Prime ideal but not maximal ideal
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 14th 2007, 09:50 AM

Search Tags


/mathhelpforum @mathhelpforum