# How to compute $\gcd(a,b)$ if $N(a)=N(b)$?

Let $a,b$ be elements of an integral domain $R$. Let $N$ denote the norm. Let $x,y$ be other elements of the same integral domain $R$. I know that $\gcd(x,y)=\gcd(x,x-y)$ iff $N(x)>N(y)>0$. However if $N(a)=N(b)$ how do I compute $\gcd(a,b)$ ?

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What is the "norm" for a general domain? –  Math Gems Feb 1 '13 at 23:03
Maybe totally out of my depth, but if $N(a) = N(b)$ doesn't that mean that $a = u \cdot b$, where $u$ is a unit, and so $\gcd(a, b) = a$ (or $b$, or any other unit multiple thereof)? –  vonbrand Feb 3 '13 at 4:09
@vonbrand I think proving that $R$ is a UFD and $a=u*b$ where $u$ is a unit would be the answer to my question if a ring $R$ was given. –  mick Feb 4 '13 at 19:16
After consideration you only need to show R is a UFD. That is sufficient. (necc ?) So the question remains what to do for a non_UFD ring R. –  mick Feb 7 '13 at 23:04