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Prove that in $\mathbb{Z}[X]$ the ideal generated by $X$, i.e. $I=\langle X\rangle$, is a maximal principal ideal (that is, maximal among principal ideals), but is not a maximal ideal.

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$I=\langle X\rangle \subsetneq \langle 2,X\rangle$, so it is not maximal. –  Sigur Oct 28 '12 at 2:23
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Why post in the imperative? Are you assigning homework to us? –  kahen Oct 28 '12 at 2:27
    
No, I'm sorry, I did not intend that. –  Muniain Oct 28 '12 at 2:34
    
but why $<X>$ is a maximal ideal principal of $\mathbb{Z}[X]$ –  Muniain Oct 28 '12 at 2:42
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Could you please use \langle X\rangle instead of <X>. I'm sure you can see the spacing is all wrong and it makes it really hard to read. –  kahen Oct 28 '12 at 2:58

1 Answer 1

$I$ is not maximal because it's contained in $\langle 2,X\rangle$, as Sigur noticed, which is an ideal which stricly contains $I$ and is itself strict.

But it's maximal among principal ideals. Indeed, let $I'$ a principal ideal containing $I$, say generated by $P_0$. If $P\in I'\setminus I$, we have $P(0)\neq 0$ (otherwise $P\in I$). Write $P:=\underbrace{\sum_{j=1}^Na_jX^j}_{\in I\subset I'}+a_0$, then $a_0\in I'$. As $a_0=P_0Q_0$ for some $Q_0\in\Bbb Z[X]$, taking the degrees on both sided, $P_0$ is constant so $I'=\Bbb Z[X]$.

Conclusion: the only principal ideal containing $I$ is $\Bbb Z[X]$.

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So, $\langle X\rangle \subset \langle 1\rangle$, and $\langle 1\rangle$ indeed principal ideal. Are we convention a ideal of $\mathbb{Z}[X]$ have to different $\mathbb{Z}[X]$?. –  Muniain Oct 29 '12 at 8:42
    
Not in the definition of an ideal, but in the definition of a maximal ideal yes. –  Davide Giraudo Oct 29 '12 at 9:14
    
OK, I see, thank you very much –  Muniain Oct 29 '12 at 9:32

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