# Irreducibles and factorization in $\mathbb{Z}[\sqrt{5}i]$

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 $\mathbb{Z}[\sqrt{5}i]$.

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?

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 $a^2+5b^2=7$.
• @italy: Yes, it just needs some simple case work (e.g., it's impossible to get $3$ or $7$ if $|a|\geq 3$ or $|b|\geq 2$). – Zev Chonoles Jun 6 '15 at 20:37