$\Bbb Z[\sqrt{-5}]$ is not a PID

Question

I want to show:

In a PID $R$ two elements $a,b\in R$ always have a greatest common divisor. Therefore $\Bbb Z[\sqrt{-5}]$ is not a PID.

For the first part:

$I=\{ax+by:x,y\in R\}$ is an ideal, so since $R$ is a PID we have $I=(g)$ for some $g\in R$. In particular, $a=gr$ and $b=gs$ for some $r,s\in R$, which says that $g$ divides both $a$ and $b$.

If $d\in R$ is any other divisor of both $a$ and $b$, then $d$ divides any linear combination of $a$ and $b$. In particular, since $I=(g)$, $g$ is such a linear combination. This proves $d\mid g$, hence $g$ is a $\gcd$.

For the second part:

I have a hunch that $6$ and $2(1+\sqrt{-5})$ have no $\gcd$ since $6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$. How can I finish the argument?

Answer 1

Hint: use the fact that PID are unique factorization domains.

Let $N$ be the norm, $N(1+\sqrt{-5})=6$, suppose that $1 + \sqrt{-5}=ab, N(ab)=N(a)N(b)=6$, set $a=u+v\sqrt{-5}, N(a)=1$ implies $u^2+5v^2=1$ this implies $v=0, u^2=1$, $N(a)=2$ implies also $v=0, u^2=2$ impossible, you cannot have $N(a)=3$ with a similar argument, so $N(a)=1$ this implies that $a=1$ or $a=-1$ and $1+\sqrt{-5}$ is irreducible, if $N(a)=6=u^2+5v^2, u^2=1$, and $v^2=1$ this implies that $a=1+\sqrt{-5}$ or $a=-1-\sqrt{-5}$, thus $1+\sqrt{-5}$ is irreducible.

Show that 2 is irreducible by using the norm, thus since $6=2 \cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$, $Q(\sqrt{-5})$ is not a unique factorization domain, so it is not principal.