# Simple question regarding proving some ring isn't a UFD

Suppose we have the domain D={a+b$\sqrt{10}$|a,b $\in \mathbb{Z}$}. I want to show this is not a UFD and I am given the hint to find two factorization of 6 using the norm N(a+b$\sqrt{10}$)=$a^2-10b^2$. I know the norm of 6 is 36. But to find the factors of 6 do I just proceed by brute force?

• In this case wouldn't that identity just be the norm? – user227630 Mar 31 '15 at 0:47

Hint 1 :Note that $N(xy)=N(x)N(y)$. So $N(x)$ divides $N(xy)$. Try to find a solution to $N(x)=6$.