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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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1 Answer 1

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

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