How would I prove there is no unique factorisation in $\mathbb Z[\sqrt{d}]$ for $d \leq-3$, where $d$ is a square-free integer?
I think it's something to do with the only invertible elements being $\pm 1$
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Maths student
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I believe this is wrong. – Will Jagy May 19 '15 at 19:16
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2Yes, wrong for $d=-4$. – Dietrich Burde May 19 '15 at 19:18
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oops I should say that d is square free – Maths student May 19 '15 at 19:18
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squarefree is still not enough, http://math.stackexchange.com/questions/1211145/for-which-values-of-d0-is-the-subring-of-quadratic-integers-of-mathbb-q – Will Jagy May 19 '15 at 19:19
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1@WillJagy The ring of integers is different from $Z[\sqrt{d}]$ for $d\equiv 1 \bmod 4$. – Dietrich Burde May 19 '15 at 19:21
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@DietrichBurde, alright. Given the apparent peculiarity here, I would like to know the source of the problem and what has been done in class prior to this exercise. Where was I: if you mean that squarefree IS good enough, I can accept that. – Will Jagy May 19 '15 at 19:27
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yeah I just realised I forgot to mention d is odd. sorry – Maths student May 19 '15 at 19:32
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If $d$ is even, then $2$ divides $\sqrt{d}^2=d$, but does not divide $\sqrt{d}$ for $d\le -3$ and $d$ squarefree, so $2$ is irreducible but not prime, which is impossible in a UFD. Similarly, if $d$ is odd, then $2$ divides $(1+\sqrt{d})(1-\sqrt{d})=1-d$ without dividing either of the factors, so again $2$ is a nonprime irreducible.
Dietrich Burde
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