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Let $R$ be UFD and denote. Show that if If c|a and c|b then $cgcd(a/c, b/c)=gcd(a, b)$.

My attempt.

Let $gcd(\frac{a}{c},\frac{b}{c})=d$. First need to show $(a,b) \subset (cd)$. This is clear since if $x \in (a,b)$ then; $$x=ar+bs=c(a/c)t+c(b/c)s$$

and since $d$ divides both $a/c$ and $b/c$ we are done.

Now need to show that if $(u)$ is another princ. ideal with $(a,b) \subset (u)$ then $(cd) \subset (u)$.

this is also immediate since if $y \in (cd)$ then for some $s,t \in R$ we have $(a/c)v + (b/c)w=d$ and ; $$y=cdr=c[(a/c)v + (b/c)w]r=avr+bwr$$. Which seems to show $(a,b)=(cd)$ but this can't possibly be true;

Muselive
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    You are wrongly assuming that UFDs are Bezout domains, i,e, thet the gcds can be written as a linear combination of their arguments. See here for further discussion and proofs of the GCD DIstributive Law. – Bill Dubuque Sep 26 '20 at 16:28

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The Bezout identity only necessarily works in Bezout domains, which includes all PIDs, but not all UFDs. So the error is when you write $d$ as a linear combination of $a/c$ and $b/c$.

For example, $\mathbb{Z}[X,Y]$ is a UFD in which $gcd(X,Y)=1$. But we can't write $1$ as a linear combination of $X$ and $Y$.

Edit: Here is a sketch for the correct proof of that direction. Let $r=gcd(a,b)$ (mod units). It suffices to show $r$ divides $cd$. Note that $c$ divides $r$ by assumption, so it suffices to show $r/c$ divides $d$. By definition of $d$, it suffices to show $r/c$ divides $a/c$ and $b/c$. But this follows from the general property that if $x$ divides $y$, and $z$ divides both $x$ and $y$, then $x/z$ divides $y/z$.

In fact, this result holds in any GCD domain.

halrankard2
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  • What do you mean by ''mod units''? – Muselive Sep 26 '20 at 20:37
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    In general, $gcd(a,b)$ is only unique up to multiplication by units in the ring. In other words, there many be many elements of $R$ that fit the definition of "greatest common divisor of $a$ and $b$". But if $c$ and $d$ are two greatest common divisors of $a$ and $b$, then $c=ud$ for some unit $u$. – halrankard2 Sep 27 '20 at 10:36
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    So really, what the above proof shows is that if $d$ is a gcd of $a/c$ and $b/c$, then $cd$ is a gcd of $a$ and $b$. – halrankard2 Sep 27 '20 at 10:40