Prove that if $a^2+5b^2 = p$ for some rational prime $p(a,b \neq 0)$ other than 5, then $a+\sqrt{-5}b$ is prime in $\mathbb{Z}[\sqrt{-5}]$.

Question

Above statement is taken from here. I started with let a,b,c $\in$ $\mathbb{Z}[\sqrt{-5}]$ such that a divides b.c and N(a) = p $\neq$ 5, then p divide N(c)(WLOG). So p divide ${c_{1}}^{2}+5{{c_{2}}}^2 $... what next ?

Answer 1

Since there have been two incorrect answers, here is a correct one. The idea is the same as the other answers, except that we use ideals instead of elements to avoid the pitfall that $\mathbb Z[\sqrt{-5}]$ is not a UFD.

Let $\mathfrak a = (a+\sqrt{-5}b)$ be the ideal generated by $a+\sqrt{-5}b$. Then $$N(\mathfrak a) = |N_{\mathbb Q(\sqrt{-5})/\mathbb Q}(a+\sqrt{-5}b)| = a^2+5b^2 = p.$$

If $\mathfrak a$ is not prime, then there are non-trivial ideals $\mathfrak{b,c}$ such that $\mathfrak{a=bc}$. But $p=N(\mathfrak a)=N(\mathfrak b)N(\mathfrak c)$, from which it follows that $N(\mathfrak b) = 1$ or $N(\mathfrak c)$. Hence, $\mathfrak b$ or $\mathfrak c$ is trivial, a condradiction.

Answer 2

Here is a variant of Mathmo's proof, which doesn't use as many properties of ideals. Consider the quotient ring $\mathbb Z[\sqrt{-5}]/(a+b\sqrt{-5})$. If $N(a+b\sqrt{-5})=a^2+5b^2=p$ is a prime, this ring has $p$ elements. Every ring with a prime number of elements is an integral domain, so in particular this quotient ring is.

Now suppose we have $x,y\in\mathbb Z[\sqrt{-5}]$ with $a+b\sqrt{-5}\mid xy$. Then $xy\equiv 0\pmod{a+b\sqrt{-5}}$, but since the quotient ring is an integral domain, either $x$ or $y$ is $0\pmod{a+b\sqrt{-5}}$, so one of them is divisible by $a+b\sqrt{-5}$. Hence $a+b\sqrt{-5}$ is prime.