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Let $I = \{f(X) = \mathbb{C}[X] | f(0) = f(1) = f(−1) = 0\}$. Then $I$ is an ideal of $\mathbb{C}[X]$.

Deduce that $I = (X^3 −X)$ is the principal ideal generated by $X^3 −X$.

Can someone write $(X^3 −X)$ in set notation because I have no idea what it actually is.

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  • $\begingroup$ $(X^3-X)$ denotes the set of multiples of $X^3-X$ by another polynomial. It is an ideal because it is stable by sums and multiplication by elements of $\mathbf C[X]$. Explicitly: $\,(X^3-X)=\bigl\{P(X)(X^3-X)\mid P(X)\in \mathbf C[X]\bigr\}$. $\endgroup$
    – Bernard
    Apr 13 '15 at 18:09
  • $\begingroup$ I meant like if we have, $S = (m)= m\mathbb{Z} = \{mx | x ∈\mathbb{Z}\}$. What would it be in this case? $\endgroup$
    – snowman
    Apr 13 '15 at 18:11
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$$\langle X^3 - X\rangle = \{p(X)\dot\, (X^3 - X); p(X) \in \mathbb K[X]\}$$ is the ideal generated by $X^3 - X \in \mathbb K[X]$.

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  • $\begingroup$ where did K come from? Do u mean C? $\endgroup$
    – snowman
    Apr 13 '15 at 18:09
  • $\begingroup$ Any field $\mathbb K$. You may take $\mathbb K = \mathbb C $. $\endgroup$ Apr 13 '15 at 18:10
  • $\begingroup$ Why so much hatred? $\endgroup$ Apr 13 '15 at 18:42
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    $\begingroup$ clearly someone is spamming $\endgroup$
    – snowman
    Apr 13 '15 at 18:46

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