Im trying to prove that for some $a,b$ (not both zero) in a unique factorization domain then the gcd($a,b$) is of the form {$gu: u$ is a unit} where $g$=gcd($a,b$).
I'm looking for a hint. What I know is:
gcd($a,b$)= $g=ma+nb$ for some $m,n$ in my domain.
$g|a$ and $g|b$.
$a = a_1a_2\cdot\cdot\cdot a_n$ and $b=b_1b_2\cdot\cdot\cdot b_m$
I'd appreciate any help. Thanks in advance.
Using the definition of gcd, if you take two of them, they divide each other, so are associate
Bézout's identity does not necessarily hold in a unique factorization domain.
In a general domain, a greatest common divisor of $a$ and $b$ is defined to be any element $d$ such that
In a unique factorization domain a gcd of two elements is proved to exist (via the middle school rule), but in any domain we can prove that if a gcd of $a$ and $b$ exists, then it is determined up to invertible elements; more precisely
if $d$ and $e$ are greatest common divisors of $a$ and $b$, then there exists an invertible element $u$ such that $e=du$.
Proof. By property 1 (applied to $e$), $e\mid a$ and $e\mid b$; by property 2 (applied to $d$), $e\mid d$. Similarly, $d\mid e$.
Hence $d=eu$ and $e=dv$, so $e=euv$ and either $e=0$ or $uv=1$. If $uv=1$, then $u$ is invertible; if $e=0$, we can take $u=1$.$\quad\square$
