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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.2012-10-28
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    Why post in the imperative? Are you assigning homework to us?2012-10-28
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    No, I'm sorry, I did not intend that.2012-10-28
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    but why $$ is a maximal ideal principal of $\mathbb{Z}[X]$2012-10-28
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    I've never heard the phrase "maximal ideal principal". Could you define it?2012-10-28
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    I think it's [romance language](https://en.wikipedia.org/wiki/Romance_language) word order for "maximal principal ideal", i.e. ideal which is maximal among principal ideals.2012-10-28
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    I mean in all principal ideal of $\mathbb{Z}[X]$ then $$ is maximal. OK?2012-10-28
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    I think if $$ is another principal ideal of $\mathbb{Z}[X]$ then $\subset $. This mean if $f(X)\in $ then exits $g(X)$ s.t $f(X)=g(X).h(X)$, then what is $k(X)$ s.t $f(X)=x.k(X)$?.2012-10-28
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    Could you please use `\langle X\rangle` instead of ``. I'm sure you can see the spacing is all wrong and it makes it *really* hard to read.2012-10-28

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