I have a question about this exercise: $$\gcd\left(\frac{a}{\gcd(a,b)}, \frac{b}{\gcd(a,b)}\right)=1$$ for some $a,b\in R^*$, $R$ being a factorial ring.
I feel like it is not necessary to establish an order "$<$" on the ring $R$. Is there a proof that doesn't rely on this? For my understanding it cannot be written like this anyway: It should rather say $\gcd(\dots)=R^\times$ where $R^\times$ denotes the unit group.
Let $g:=\gcd(a,b)$. Since $g\mid a$ and $g\mid b$ we have $\frac{a}{g},\frac{b}{g}\in R$. Now let's take a $d\in R$ such that $d\mid \frac{a}{g}$ and $d\mid\frac{b}{g}$, then $gd\mid a$ and $gd\mid b$. So, by definition of the $\gcd$ we have $gd\mid g$ which is only true if $d$ is a unit: There exists a $k$ such that $(gd)k=g$, so $g(dk)=g$ and thus $dk=1$. Does this work?
Since $R$ is factorial, can't I use the prime factorization suggested in an answer below? We'd have
$$\gcd\left(\frac{a}{g},\frac{b}{g}\right)=\prod_{n=1}^\infty p_n^{x_n}$$ where $$x_n=\min\left(\alpha_n-\min(\alpha_n,\beta_n),\beta_n-\min(\alpha_n,\beta_n)\right)=0$$
You are absolutely correct.
However, reading between the lines in the post you are linking to: in many cases they don't say that they use the order $\lt$ (or $\le$): they may well be using the (pre-)order $\mid$ ("divides").
In $\mathbb N$ and $\mathbb Z$, the $\gcd$ is the greatest common divisor not only in the sense of being the greatest common divisor as a number, but has a stronger property that it is a maximum element in the set of all the common divisors, with respect to the pre-order $\mid$. (In other words: (a) It is a common divisor and (b) Any other common divisor divides it.) Now, that notion translates to any commutative rings with unity, as you have noted yourself.
Thus, you and some of the people who have answered the linked post may already be on the same page.