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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.

  1. 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?

  2. 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 $$x_n=\min\left(\alpha_n-\min(\alpha_n,\beta_n),\beta_n-\min(\alpha_n,\beta_n)\right)=0$$

Sahiba Arora
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Buh
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1 Answers1

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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.

  • I feel like I don't understand something here still. They assume $d>1$ and then get a contradiction. But let's say $R$ is a field. Then this wouldn't work since for any 2 elements $a,b\in R$ we'd have $\gcd(a,b)=R- {0}$. – Buh Jan 10 '18 at 19:18
  • In a commutative ring, because $\mid$ is a pre-order, the $\gcd$ is only defined up to equivalence (called "association", denoted by $~$: $a\sim b\Leftrightarrow a\mid b\land b\mid a$). In a field, every two non-zero elements are associated, so there is no such thing as (nonzero) element $d$ such that $1\mid d$ but $d\not\mid 1$. Thus we have a contradiction before we've even started the proof. –  Jan 10 '18 at 19:23
  • Hmm, I see. Aight, thanks! – Buh Jan 10 '18 at 19:28
  • I myself find it awkward that they say $d \gt 1$ but there is no suitable symbol for "$a$ strictly divides $b$" that I know of (meaning $a\mid b\land a\not\sim b$) so maybe we can leave it to the authors of the answers over there to use $\lt$ and $\gt$. You are right that those proofs can all be reworked to only use divisibility, with maybe replacing equality with association here and there, and they would work in any integral domain (commutative ring with unity and with no zero divisors). (You need the "no zero divisors" condition to be able to conclude $gd\mid g\implies d\mid 1$.) –  Jan 10 '18 at 19:34
  • Well, you require $R$ to be factorial, otherwise the $\gcd$ might not exist. – Buh Jan 10 '18 at 19:36
  • There are rings where $\gcd$ exists but are not factorial... Check this out: https://en.wikipedia.org/wiki/B%C3%A9zout_domain. However, the statement can be understood as "if for some $a,b\ne 0$ all the $\gcd$'s on the LHS exist, then the LHS is equal to RHS, which is $1$" and it would still be a valid statement. –  Jan 10 '18 at 19:39
  • Nice! Thanks for that link. – Buh Jan 10 '18 at 19:41