# Principal ideal

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• Dec 2nd 2005, 08:01 AM
juef
Principal ideal
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?

.. 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! :)
• Dec 2nd 2005, 01:57 PM
hpe
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 $q \in R[X]$ such that any $p \in I$ can be written as $p(X) = q(X)r(X)$ for some other $r(X)$. That is, you have to show that each $p \in I$ has one and the same common factor $q(X)$.

Write $p \in I$ as a polynomial in (x-1), i.e.
$p(x) = \sum_{k=0}^n a_k(x-1)^k$.
Now check what the condition $p(1) =p'(1) = 0$ says about the coefficients $a_0,\, a_1, \dots, a_n$. This should give you an idea what q should be.
• Dec 2nd 2005, 07:28 PM
juef
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 $p$ as a polynomial in $x-1$... At first I didn't get why every $p \in I$ 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!