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Thread: A prime ideal

  1. #1
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    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 $\displaystyle I = \{f\in\mathbb{Z}\left[x\right] : f(0) = 0\}$.

    1. Show $\displaystyle I$ is a prime ideal.
    2. Show $\displaystyle I$ is not maximal.

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

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

    Thanks, Sam
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  2. #2
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    Hi

    1) Yep

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

    Try to work with an ideal generated by two elements.
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  3. #3
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    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 $\displaystyle \{f(x)\in\mathbb{Z}\left[x\right] : f(0) = 0\} \cup (2)$?
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  4. #4
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    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?
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  5. #5
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    Your new answer isn't so far from a solution: the idea of "adding" $\displaystyle 2$, 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 $\displaystyle \{f(x)\in\mathbb{Z}[x];\ f(0)=0\}$ is the ideal $\displaystyle (x)$ in $\displaystyle \mathbb{Z}[x]$ ?

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

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

    So take a $\displaystyle p(x)\in (x,2),$ i.e. $\displaystyle \exists a,b\in\mathbb{Z}[x],\ p(x)=xa(x)+2b(x).$ Is $\displaystyle p(x)=1$ possible? (That will answer the problem, since for any ideal $\displaystyle \mathfrak{A},\ 1\in\mathfrak{A}\Leftrightarrow \mathfrak{A}=\mathbb{Z}[x]$)
    Last edited by clic-clac; Feb 19th 2009 at 12:07 AM. Reason: cor
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  6. #6
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    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!
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