I'm a bit lost... it seems every non-trivial ideal in a commutative ring is a principal ideal. but is it true? if not, could you pls give a counter example?
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5Take $(x, y) \subset k[x, y]$. – Qiaochu Yuan Dec 09 '14 at 02:06
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3In addition to @Qiaochu's example, you could also take $(2,X)\subseteq \mathbb{Z}[X]$. – Amitesh Datta Dec 09 '14 at 02:09
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@QiaochuYuan i'm beginner of algebra, could you pls elaborate a bit what is $(x,y)$ and what is $k[x,y]$? – athos Dec 10 '14 at 03:27
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@AmiteshDatta i'm beginner of algebra, could you pls elaborate a bit what is $(2,X)$ and what is $Z[X]$? – athos Dec 10 '14 at 03:28
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2Dear @athos, $\mathbb{Z}[X]$ is the polynomial ring over $\mathbb{Z}$ in one variable $X$. Also, $(2,X)$ is the ideal generated by $2,X\in \mathbb{Z}[X]$, that is, the set of all elements of $\mathbb{Z}[X]$ of the form $2p + Xq$ for $p,q\in \mathbb{Z}[X]$. – Amitesh Datta Dec 10 '14 at 04:08
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Yes now I see it.... Principal ideal indicates one dimension, while a ring could have 2 dimensions... Thx! – athos Dec 10 '14 at 13:17
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By non-trivial, do you an ideal that is not the whole ring itself? I'm going to proceed assuming that's what you mean.
If every ideal in a given integral domain $R$ is a principal ideal, then $R$ is a principal ideal domain, and then it's also a unique factorization domain. In $\mathbb{Z}[\sqrt{-2}]$, for example, every ideal is a principal ideal.
But now consider $\mathbb{Z}[\sqrt{-5}]$, which is neither a UFD nor a PID. $\langle 3 \rangle$ is a principal ideal but not a prime ideal. The ideal $\langle 3, 1 + \sqrt{-5} \rangle$, which consists of all numbers of the form $3a + b\sqrt{-5}$ with $\{a, b\} \in \mathbb{Z}[\sqrt{-5}]$ is a prime ideal but not a principal ideal.
Robert Soupe
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Alright, I'm going to change that to say "integral domain" rather than "ring." I hope integral domains are more accessible to beginners. – Robert Soupe Dec 11 '14 at 01:33