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Thread: Principal ideal

  1. #1
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    Principal ideal

    Hey everyone,

    $\displaystyle I = \{ p \in R[X] : p(1) = p'(1) = 0 \} $
    I've got to show that $\displaystyle I$ is a principal ideal in $\displaystyle R[X]$.

    If I understand correctly, I've got to find some $\displaystyle q \in R[X]$ for which $\displaystyle qr(1) = qr'(1) = 0$ for any $\displaystyle r \in R[X]$. Is that right?

    .. because if it is, I have no clue how to find the answer
    I'm not asking for the whole solution to the problem, but a little hint would certainly be welcome.

    Thank you!
    Last edited by juef; Dec 2nd 2005 at 07:35 PM.
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  2. #2
    hpe
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    Quote Originally Posted by juef
    Hey everyone,

    I = {p in R[X] : p(1) = p'(1) = 0}
    I've got to show that I is a principal ideal in R[X].

    If I understand correctly, I've got to find some q in R[X] for which qr(1) = qr'(1) = 0 for any r in R[X]. Is that right?
    No, you have to find $\displaystyle q \in R[X]$ such that any $\displaystyle p \in I$ can be written as $\displaystyle p(X) = q(X)r(X)$ for some other $\displaystyle r(X)$. That is, you have to show that each $\displaystyle p \in I$ has one and the same common factor $\displaystyle q(X)$.

    Write $\displaystyle p \in I$ as a polynomial in (x-1), i.e.
    $\displaystyle p(x) = \sum_{k=0}^n a_k(x-1)^k$.
    Now check what the condition $\displaystyle p(1) =p'(1) = 0$ says about the coefficients $\displaystyle a_0,\, a_1, \dots, a_n$. This should give you an idea what q should be.
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  3. #3
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    Well, showing me the correct thing to prove is certainly something I should thank you for!

    Now I understand the principal ideal part. Prenny cool idea you got there too, writing $\displaystyle p$ as a polynomial in $\displaystyle x-1$... At first I didn't get why every $\displaystyle p \in I$ could be expressed that way, but now I do. I get everything.

    Thank you very much, you're really helping me there! Have a nice day!
    Last edited by juef; Dec 2nd 2005 at 08:05 PM.
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