# Showing that $2+\sqrt{-5}$ is not divisible by $3$ in $\mathbb{Z}[\sqrt{-5}]$

I'm trying to show that $\mathbb{Z}[\sqrt{-5}]$ is not a principal ideal domain by showing that $3$, which I've already shown is irreducible in $\mathbb{Z}[\sqrt{-5}]$, is not prime there.

To that effect, My goal is to show that although $3$ divides $(2+\sqrt{-5})(2-\sqrt{-5})=3^{2}$, neither of the factors $2+\sqrt{-5}$ or $2-\sqrt{-5}$ is divisible by $3$.

There are a plethora of examples out there showing $\mathbb{Z}[\sqrt{-5}]$ is not a PID by using similar techniques, but when it comes time to show that something like $2+\sqrt{-5}$ is not divisible by $3$, the details are completely glossed over.

I realize that another method of approaching this problem is to show that it is not a UFD, and I have actually already completed the proof that way. I also wanted to approach it this way, but I don't know how to show that $2+\sqrt{-5}$ is not divisible by $3$. There is nothing in my class notes that is helpful, so if anyone could please show me in detail how to do this at long last, I would be eternally grateful.

• These two links has some examples of similar applications. (1, 2). – user236182 Apr 28 '17 at 23:05

## 1 Answer

If $2+\sqrt{-5}$ is divisible by $3$ in $\mathbb{Z}[\sqrt{-5}]$ then there are integers $a$ and $b$ such that $$3(a+b\sqrt{-5})=2+\sqrt{-5}$$ which implies that $3a=2$, which is impossible. Similarly $3$ cannot divide $2-\sqrt{-5}$.

• THANK YOU!! That was so incredibly helpful. – ALannister Apr 28 '17 at 23:03
• No problem, happy to help. – carmichael561 Apr 28 '17 at 23:15