$ \mathbf{Z}[x]$ is not PID.
we know $\mathbb Z$ is a Unique Factorization Domain, so $\mathbb Z[x]$ is UFD, but why isn't it PID (since I think $\mathbb Z$ is PID)?
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5Consider the ideal $(2,x)$. – Viktor Vaughn Dec 19 '14 at 20:58
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2Another one: http://math.stackexchange.com/questions/500254/is-mathbbzx-a-principal-ideal-domain – user26857 Dec 19 '14 at 21:05
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@SpamIAm but (2,x) is principle in Q[x] right? then what should be the element generating the ideal? – annimal Dec 19 '14 at 21:05
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@annimal Who said anything about $\mathbb{Q}[x]$? And no element can generate the ideal, so what do you mean by "should"? – Andrew Dudzik Dec 19 '14 at 21:09
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@Slade: I think he means $\langle 2, x \rangle$ is principal in $\mathbb{Q}[x]$, so what is the generator for that ideal? The confusion, I think, arises from the fact that $2$ is invertible in $\mathbb{Q}$, but not in $\mathbb{Z}$. (In fact, the ideal $\langle 2,x \rangle$ is the entire ring $\mathbb{Q}[x]$!) – Alex Wertheim Dec 19 '14 at 21:12
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$2$ is a unit in $\mathbb{Q}[x]$, hence generates the entire ring. The idea is to show that $(2,x)$ is not a principal ideal of $\mathbb{Z}[x]$ – Viktor Vaughn Dec 19 '14 at 21:12
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@SpamIAm sorry, I'm asking a new question, why (2,x) is principle in Q[x] ? – annimal Dec 19 '14 at 21:12
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@SpamIAm OK thanks – annimal Dec 19 '14 at 21:13
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@annimal read the earlier comments. $(2,x)$ is generated by $1$ in $Q[x]$. – Kevin Carlson Dec 19 '14 at 21:14
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If it were a PID, then every nonzero prime ideal would be maximal. But $\mathbb Z[X]/p \mathbb Z[X] = \mathbb (Z/p\mathbb Z)[X]$ is an integral domain which is not a field.
Bruno Joyal
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1Well, I've posted a link under this question more or less for that answer. – user26857 Dec 19 '14 at 21:18
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@user26857 See also the more general theorem, and see also here, both back in the glory days of the site. – Bill Dubuque Dec 19 '14 at 22:21
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See this $2011$ answer for a clearer way to present such arguments. – Bill Dubuque May 04 '23 at 06:05