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.
Maximal principal ideal which is not maximal
1
$\begingroup$
abstract-algebra
ring-theory
ideals
-
1$I=\langle X\rangle \subsetneq \langle 2,X\rangle$, so it is not maximal. – 2012-10-28
-
2Why post in the imperative? Are you assigning homework to us? – 2012-10-28
-
0No, I'm sorry, I did not intend that. – 2012-10-28
-
0but why $
$ is a maximal ideal principal of $\mathbb{Z}[X]$ – 2012-10-28 -
0I've never heard the phrase "maximal ideal principal". Could you define it? – 2012-10-28
-
0I 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
-
0I mean in all principal ideal of $\mathbb{Z}[X]$ then $
$ is maximal. OK? – 2012-10-28 -
0I 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 -
1Could 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