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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".

Hopmaths
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    More generally, prove that if $d$ is a common divisor of $a,b$ then $d$ is a divisor of any gcd. This shows that $d_1\mid d_2$ and $d_2\mid d_1.$ – Thomas Andrews Sep 01 '21 at 17:17
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    Note, you need units in the prime factorizations. of $d_1,d_2,$ too. Your argument works if you have already proven that a gcd's prime factorization has this property - that the exponent of $p$ is the minimum of the exponent of $p$ in $a$ and the exponent of $p$ in $b.$ – Thomas Andrews Sep 01 '21 at 17:20
  • @ThomasAndrews About your first comment, I think you are right the proof of this case for Euclidean Domains works like this. Regarding the second comment, in the book I use they do not specify a unit for the gcd written that way and the minimum exponent is already proven in the book. (The book is Abstract Algebra by Gregory Lee) – Hopmaths Sep 01 '21 at 20:13

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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.