Is $\mathbb{Z}[\sqrt{-7}]$ a UFD or not? I feel like there should be some easy example showing that it is not, but I can not think of one.
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1Common trick is to check $(1+ \sqrt{-7})(1-\sqrt{-7})$ – Future Jan 27 '16 at 03:04
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$2^3=(1+ \sqrt{-7})(1-\sqrt{-7})$
$2$ is irreducible since $2=ab \implies N(a)=2$, but there are no elements of norm $2$ since $x^2 + 7y^2 = 2$ has no solution because $y=0$ is the only choice for integral solutions to this equation.
$2$ is not prime since $2\mid (1+ \sqrt{-7})(1-\sqrt{-7})$ but $2\nmid 1 \pm \sqrt{-7}$.
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Yes it is a PID! But this is too deep for this question I guess. I don't know a simple way to show that avoiding the Finiteness Theorem. – Maffred Jan 27 '16 at 04:19
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You just have to believe the Minkowski bound, which seems to tell me that I don’t need to check any ideals at all for principalness. – Lubin Jan 27 '16 at 04:22
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