Let $R$ be a UFD. Suppose that $a$ and $b$ are nonzero nonunit elements of $R$. If $d_1$ and $d_2$ are gcds of $a$ and $b$, show that $d_1$ and $d_2$ are associates
I have tried to solve this problem, but I am not quite sure what I have done. Is this the way? If so, are there any missing details? UFDs are being a bit complicated for me.
Since $R$ is a UFD, we can write $a=up_1^{m_1}\cdots p_k^{m_k}$ and $b=vp_1^{n_1}\cdots p_k^{n_k}$, where $p_i$'s are irreducible elements nonasocciates and $u,v$ are units in $R$ ($m_i, n_i \in \mathbb{Z}^+$). Then $d_1=p_1^{l_1}\cdots p_k^{l_k}$ and $d_2=p_1^{r_1}\cdots p_k^{r_k}$, where $l_i$ and $r_i$ are the smallest of $m_i$ and $n_i$. But note that $l_i=r_i$ for all $i$. So $d_1|d_2$ and $d_2|d_1$. But we know that in an integral domain two elements are associates iff they divides mutualy. Completing the "proof".
You don't need to appeal to prime compositions and it gets confusing and requires detail checking when you do. It is best to work with the definition of a gcd in a UFD in this case.
Any element dividing both $a$ and $b$ will divide a gcd by definition. Hence $d_1$ divides $d_2$ and vice-versa. We get $$d_1=ud_2=uvd_1.$$ UFDs are integral domains, hence we can cancel $d_1$. This implies $uv=1$ and thus the claim.