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I've been trying to prove that $\mathbb{Z}[\sqrt{10}]$ is not factorial. I did this by defining the norm $N(a+b\sqrt{10})=a^2-10b^2$.

I was able to show for myself that $N(z)=\pm 2$ and $N(z)=\pm 5$ have no solutions, and my idea is to show that $2,5,\sqrt{10}$ are irreducible, and then $2\cdot 5=10=\sqrt{10}\sqrt{10}$ gives two factorizations of $10$ into nonassociate irreducibles.

The only hitch is I need to show $N(z)=\pm 1$ implies $z$ is a unit, from which it will follow that $2,5,\sqrt{10}$ are irreducibles since $N(2)=4$, $N(5)=25$, and $N(\sqrt{10})=-10$, but norms never take values $\pm 2,\pm 5$. I know that if $z$ is a unit, then $N(z)=\pm 1$, but I don't know how to show the converse. If there is a better solution, I'd be happy to see that instead. Thanks.

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Note that $N(a+b\sqrt{10}) = (a+b\sqrt{10})(a-b\sqrt{10})$. The factor on the right is an honest element of $\mathbf Z[\sqrt{10}]$, and if the product is $\pm1$... – Dylan Moreland Jul 17 '12 at 4:56
Oh, I should have noticed that factorization. Thanks! – Son Bi Jul 17 '12 at 5:00
up vote 3 down vote accepted

Since $N(z) = z\overline{z}$, where $\overline{a+b\sqrt{10}} = a-b\sqrt{10}$, and $z\in\mathbb{Z}[\sqrt{10}]$ implies $\overline{z}\in\mathbb{Z}[\sqrt{10}]$, it follows that if $N(z)=\pm 1$, then $z$ is a unit, with $z^{-1}=\pm\overline{z}$. Conversely, if $z$ is a unit, then multiplicativity of the norm map shows that $N(z)$ must be a unit in $\mathbb{Z}$.

Thus, $z\in\mathbb{Z}[\sqrt{10}]$ is a unit if and only if $N(z)=\pm 1$.

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Thank you! ${}{}$ – Son Bi Jul 17 '12 at 5:00

Hint $\rm\,\ z\:|\:z\,\bar z\:|\:1\:\Rightarrow\:z\:|\:1\:\Rightarrow\: z\,$ is a unit

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