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Earlier I asked a question which showed that $\mathbb{Z}[\sqrt{-n}]$ for $n$ a square free integer greater than 3 is not a UFD.

Since PID implies UFD, this also means $\mathbb{Z}[\sqrt{-n}]$ is not a PID. Is there an example of an ideal in this ring that is not principal?

My earlier question is here Why is $\mathbb{Z}[\sqrt{-n}]$ not a UFD?

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up vote 12 down vote accepted

If $R$ is a domain and $p$ is an irreducible that is not a prime, then any witness to the fact that $p$ is not a prime yields a witness to the fact that $R$ is not a PID: suppose that $p|ab$, but $p$ does not divide either $a$ or $b$. Then $(p,a)$ is not principal: if $(p,a)=(x)$, then $x|p$; since $p$ is irreducible, either $x$ is a unit, or $x$ is an associate of $p$.

If $x$ is an associate of $p$, then since $(p,a)=(x)$ we have that $x|a$, hence $p|a$, a contradiction.

If $x$ is a unit, then $(p,a)=(x)=(1)$. Therefore, there exist $\alpha,\beta\in R$ such that $1=\alpha p + \beta a$ (since every element of $(p,a)$ is of the form $rp+sa$ for some $r,s\in R$), hence multiplying through by $b$ gives $b=\alpha b p + \beta ab$, and since $p|ab$, it follows that $p|b$. This is another contradiction.

Thus, $(p,a)$ cannot be principal. Of course, neither can $(p,b)$.

Hence, if $n$ is even and $n\gt 2$, then $(2,\sqrt{-n})$ is not principal in $\mathbb{Z}[\sqrt{-n}]$. If $n$ is odd and $n\geq 3$, then $\text{}$$(2,1+\sqrt{-n})$ is not principal.

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If $x$ is a unit, then how does $p|ab$ imply $p|b$ exactly? – Danielle Intal Oct 9 '11 at 3:29
@DanielleIntal: I'll add it to the body, but it is exactly the same argument as used to prove Euclid's Lemma in the integers. – Arturo Magidin Oct 9 '11 at 3:30
Ok, thanks for adding it to the body. – Danielle Intal Oct 9 '11 at 3:32
By a "domain", I assume you mean what others might refer to as an "integral domain"? – goblin Sep 24 '15 at 13:43
@goblin The most widely used convention is that integral domains are commutative, and domains are not. On the other hand, the most widely used convention for 'prime' is only defined in a commutative ring, and 'irreducible' is only defined in an integral domain. – Travis Bemrose Nov 10 '15 at 23:54

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