Common factors in integral domain Z[sqrt(-13)]

Consider the integral domain $\displaystyle \mathbb{Z}\[\sqrt{-13}\] = \{ x + y \sqrt{-13} | x,y \in \mathbb{Z} \}$.

Show that $\displaystyle 2$ and $\displaystyle 3 + \sqrt{-13}$ have no common factors in $\displaystyle \mathbb{Z}\[\sqrt{-13}\]$ except for $\displaystyle 1$ and $\displaystyle -1$.

I can show that both $\displaystyle 2$ and $\displaystyle 3 + \sqrt{-13}$ are irreducible. However, is there another way to prove that $\displaystyle 1$ and $\displaystyle -1$ are the only common factors?

Re: Common factors in integral domain Z[sqrt(-13)]

Showing irreducibility is not enough, you must also prove that the only units in your domain are 1 and -1.

Re: Common factors in integral domain Z[sqrt(-13)]

But should showing irreducibility be the first step though? Or is there a different solution that does not involve proving irreducibility?

Re: Common factors in integral domain Z[sqrt(-13)]

Suppose $\displaystyle \alpha = a + \sqrt{-13} b$ is a common factor.

Then $\displaystyle 2 = \alpha x$ for some $\displaystyle x = c + \sqrt{-13}d$ and $\displaystyle N(2) = 2^2 = N(\alpha)N(x) = (a^2 + 13 b^2)(c^2 + 13d^2)$. It can be deduced that $\displaystyle a = \pm 1$ or $\displaystyle a = \pm 2$.

We also have $\displaystyle 3 + \sqrt{-13} = \alpha y$ for some $\displaystyle y = e + \sqrt{-13}f$ and $\displaystyle N(3 + \sqrt{-13}) = 22 = N(\alpha)N(y) = (a^2 + 13 b^2)(e^2 + 13f^2)$. It can be seen that $\displaystyle a = \pm 1$ is the only possible value.

Hence $\displaystyle \alpha = \pm 1$ is the only common factor.