2
$\begingroup$

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.

  • 1
    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

1 Answers 1

2

$I$ 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 $\Bbb Z[X]$.

  • 0
    OK, I see, thank you very much2012-10-29