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

• Jun 8th 2012, 08:14 AM
math2011
Common factors in integral domain Z[sqrt(-13)]
Consider the integral domain $\mathbb{Z}$\sqrt{-13}$ = \{ x + y \sqrt{-13} | x,y \in \mathbb{Z} \}$.

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

I can show that both $2$ and $3 + \sqrt{-13}$ are irreducible. However, is there another way to prove that $1$ and $-1$ are the only common factors?
• Jun 8th 2012, 02:18 PM
wsldam
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.
• Jun 8th 2012, 07:01 PM
math2011
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?
• Jun 8th 2012, 08:45 PM
math2011
Re: Common factors in integral domain Z[sqrt(-13)]
Suppose $\alpha = a + \sqrt{-13} b$ is a common factor.

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

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

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