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In a domain $R$ (commutatitive ring that has no zero divisors except $0$) the greatest common divisior of two elements $a,b\in R$ is a set of the form $dR^\times$ where $R^\times$ is the set of units of $R$ and $d\in R$ satisfies the following conditions:

(1) $d\mid a\land d\mid b$

(2) $\forall d'\in R:(d'\mid a\land d'\mid b\Rightarrow d'\mid d)$

Does the same definition also work if there are zero divisiors and what things about the GCD are still true or not true anymore?

Edit: To explain further what I mean: What general properties does a GCD have in the domain case (note that we don't assume a GCD domain, so a GCD might not always exist) and which of these properties are or are not also true in the general ring case?

Edit: To add an even more specific question: The ring of $p$-adic integers is a principal ideal domain and thus a GCD domain if $p$ is a prime. What happens if $p$ is not a prime? Does a GCD always exist?

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    $\begingroup$ Depends on what you mean by "work". Even in a domain it is quite possible for two elements to not have a greatest common divisor at all. $\endgroup$ Nov 17, 2016 at 10:16
  • $\begingroup$ What would be an example of a "thing about the GCD" you would want to be true? $\endgroup$ Nov 17, 2016 at 10:17
  • $\begingroup$ An example would be: If $gcd(a,b)$ and $gcd(a,c)$ both exist and are both equal to $R^\times$, then $gcd(a,bc)$ exists and is equal to $R^\times$. $\endgroup$
    – Mario
    Nov 17, 2016 at 10:20
  • $\begingroup$ @Mario: That's not true even in arbitrary domains...for instance, if $k$ is a field and $R=k[x,y,z,w]/(xy-zw)$, you get a counterexample by taking $a=x$, $b=z$, and $c=w$. $\endgroup$ Nov 17, 2016 at 10:26
  • $\begingroup$ This may be useful: en.wikipedia.org/wiki/… $\endgroup$
    – Crostul
    Nov 17, 2016 at 10:36

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The definition can fail to work, in the sense that the set of all $d$ satisfying (1) and (2) may not have the form $dR^\times$. For instance, let $R=\mathbb{Z}[x]/(5x)$, $a=x$, and $b=2x$. Notice that $b=2a$ and $a=3b$, so $a\mid b$ and $b\mid a$. It follows that $d=a$ and $d=b$ both satisfy conditions (1) and (2), and so should be elements of $\gcd(a,b)$. But there is no unit $u$ such that $b=au$ (it is not hard to prove that $\pm1$ are the only units in $R$).

However, the case of $n$-adic integers when $n$ is not prime is quite easy and everything works well. The reason is that if $p_1,\dots,p_m$ are the distinct prime factors of $n$, there is an isomorphism $$\mathbb{Z}_n\cong \mathbb{Z}_{p_1}\times\mathbb{Z}_{p_2}\times\dots\times\mathbb{Z}_{p_m}$$ (this is just a souped-up version of the Chinese remainder theorem). It follows that all questions of divisibility in $\mathbb{Z}_n$ can be answered by handling each coordinate of this product separately. So to find the GCD of two elements of $\mathbb{Z}_n$, you just find the GCDs of their components in each $\mathbb{Z}_{p_i}$. Similarly, an element of $\mathbb{Z}_n$ is a unit iff each of its components is a unit, so it follows that any two GCDs of a pair of elements differ by a unit factor (since this is true in each $\mathbb{Z}_{p_i}$).

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  • $\begingroup$ This is a great answer! Just one further question: Does this mean that in $\mathbb{Z}_n$ every two elements $a$ and $b$ have a GCD $d$ and the set of all GCDs of $a$ and $b$ is $d\mathbb{Z}_n^\times$? What happens if $n$ is not a prime and if $a$ and/or $b$ is a zero divisor? $\endgroup$
    – Mario
    Nov 17, 2016 at 11:14
  • $\begingroup$ Yes, that's correct. If $a$ is a zero divisor in $\mathbb{Z}_n$, that just means its coordinate in one of the $\mathbb{Z}_{p_i}$'s is $0$. So on that coordinate, you will be taking a GCD where one of the elements is $0$ (and so the GCD will just be all unit multiples of the other element). $\endgroup$ Nov 17, 2016 at 11:16

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