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Consider $\mathbb{Z[\sqrt{-5}]}=a+b\sqrt{-5}$ where $a,b \in \mathbb{Z}$. My understanding is that an integral domain is a PID if every ideal in the ring is principal.

For the above example, this I can see that it is an integral domain because it's a subring of $\mathbb{C}$, which contains 1. My issue comes with showing the 'principal' part, where I'm not quite sure where to start.

Trogdor
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  • No, it's not a subring of $\mathbb{Z}$. And it's not a PID, because it's not factorial. Is there a non invertible common divisor of $2$ and $1+\sqrt{-5}$? – egreg Apr 15 '15 at 22:54
  • What does containing $1$ have to do with being an integral domain? It is true that it is an integral domain, but $1\in \mathbb C$ is not the reason for that. – Thomas Andrews Apr 15 '15 at 23:07

2 Answers2

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In $\mathbb{Z}[\sqrt{-5}],$ $2$ and $3$ are irreducible, since $6 = 2 \cdot 3$ and $6 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$ Observe that $U(\mathbb{Z}[\sqrt{-5}]) = \{\pm1\}$ (group of units). Therefore, the factorization of $6$ are truly different $\Longrightarrow \mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain. Since all principal ideal domains are unique factorization domains, it follows that $\mathbb{Z}\left[\sqrt{-5} \right]$ is not a principal ideal domain either.

St Vincent
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Euclid's Lemma (2-line proof) $\ \ \gcd(a,b)=1,\ a\ |\ bc\ \Rightarrow\ a\mid c\ \ $ if $\ \gcd(ac,bc)\ $ exists.

If this fails, i.e. if $\,\gcd(a,b) = 1,\ a\mid bc\ $ but $\ a\nmid c\:$, then the $\,\gcd(ac,bc)\ $ fails to exist, so the ideal $\,(ac,bc)\,$ is not principal. Apply to: $ $ irreducible $\, 2\mid \omega\bar\omega = 6,\,$ but $\, 2\nmid \omega,\bar\omega,\,\ \omega = 1\!+\!\sqrt{-5}$.

Community
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Bill Dubuque
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