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Consider the ring $\mathbb{Z}[\sqrt{5}i]=\{m+n\sqrt{5}i:m,n\in\mathbb{Z}\}$. Show that $21$ has two distinct factorisations into irreducibles in $\mathbb{Z}[\sqrt{5}i]$, which is thus not a UFD. Identify an irreducible element which is not prime in $\mathbb{Z}[\sqrt{5}i]$.

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I really don't understand this topic very well, but I do know that in $\mathbb{Z}[\sqrt{5}i]$, $$21 = 7 \times 3 = (4 + \sqrt{5}i) (4 - \sqrt{5}i)$$ The standard multiplicative norm is $$v(m+n\sqrt{5}i) = m^{2} + 5n^{2}$$ I know a unit must have norm $1$, so only units are $1$ or $-1$, and that $v(3) = 9$, $v(7) = 49$, $v(4\pm \sqrt{5}i) = 21$. How should I proceed from here?

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Prove that the elements $$3,\quad 7,\quad 4+\sqrt{5}i,\quad 4-\sqrt{5}i$$ are irreducible in $\mathbb{Z}[\sqrt{5}i]$ by showing that there is no element $a+b\sqrt{5}i$ that has norm $3$ or $7$, i.e. there are no $a,b\in\mathbb{Z}$ such that either $a^2+5b^2=3$ or $a^2+5b^2=7$.

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  • $\begingroup$ How would I show this? cause the only way I think of showing this is to just try numbers in Z to show it never equal 3 or 7, but I dont really see that as proving generally? $\endgroup$ – italy Jun 6 '15 at 20:33
  • $\begingroup$ smallest values of of m^{2} +5n^{2} are 0, 1 ,4 ,5 ,6 ,9 20, ..... $\endgroup$ – italy Jun 6 '15 at 20:36
  • $\begingroup$ @italy: Yes, it just needs some simple case work (e.g., it's impossible to get $3$ or $7$ if $|a|\geq 3$ or $|b|\geq 2$). $\endgroup$ – Zev Chonoles Jun 6 '15 at 20:37

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