How can I use the fact that $6=2\cdot 3=(\sqrt{10}-2)(\sqrt{10}+2)$ to prove $Z[\sqrt{10}]$ is not a UFD?
They are different ways of factoring 6 into irreducibles?
2026-04-03 06:17:43.1775197063
How can I use the fact that $6=2\cdot 3=(\sqrt{10}-2)(\sqrt{10}+2)$ to prove $Z[\sqrt{10}]$ is not a UFD?
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Hints:
That's exactly the idea, but showing that these elements are irreducible must be proven. It must also be proven that for example $2$ is not equal to $u(\sqrt{10}-2)$ for some unit $u \in \Bbb{Z}[\sqrt{10}].$ So you should also determine the units of $\Bbb{Z}[\sqrt{10}]$.