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non integral domain. For instance if we take the ring $\mathcal{Z}/9\mathcal{Z}=\{\bar0,\bar1,\bar2,\bar3,\bar4,\bar5,\bar6,\bar7,\bar8\}$. It's not an integral domain (indeed : $\bar3.\bar3=\bar0$ but $\bar3\neq \bar0$).

How can we determine $\gcd(\bar2,\bar7)$ ? Can we write a Bézout's relation ? I know that we can determine $(\mathcal{Z}/9\mathcal{Z})^{\times}$ using the congruences and Bézout which make the link between $\mathcal{Z}/9\mathcal{Z}$ and the ring $\mathcal{Z}$.

Moreover if we want to determine idempotent elements of this ring it can be a problem if we cannot use Bézout.

Thanks in advance !

Anonymous196
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Maman
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    The problem is with the g in $\gcd$. There is no ordering of your ring that makes sense. – Arthur Oct 25 '17 at 16:32
  • @Arthur it means that the notion of $\gcd$ has no sense here ? – Maman Oct 25 '17 at 16:36
  • I think it does. See for instance the fact that this article exists: https://en.wikipedia.org/wiki/GCD_domain. That's specifically for integral domains, but still, the fact that the concept has its own Wikipedia article should point to the fact that not every ring or every ID has it. – Arthur Oct 25 '17 at 16:38
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    @Arthur I have a hard time to reconcile your two comments. – quid Oct 25 '17 at 16:40
  • @quid Both my comments say that gcd doesn't always make sense. The main content in my second comment isn't the Wiki link, but the fact that the Wiki link exists. I could've made that clearer. – Arthur Oct 25 '17 at 16:42
  • @Arthur Now, does the notion of GCD make sense in the ring in the question. Yes or no. Your second comment quite directly says 'yes' your first seems to imply it is 'no.' This is confusing, at least for me. [Sorry for the double post, I messed up the ping at first.] – quid Oct 25 '17 at 16:45
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    @quid Ahh, I meant "I think it does" as agreeing with "it means the notion of $\gcd$ has no sense here ?" I see now how that is ambiguous, but I still believe that my response is correct (the question is "it means [...]?", and my answer is affirmative). At any rate, it's too late to edit. – Arthur Oct 25 '17 at 16:55
  • @Arthur thanks, that clarifies the situation. – quid Oct 25 '17 at 17:07

2 Answers2

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The first thing you have to understand is that even if $\gcd(a, b)$ exists, it is not unique, strictly speaking. Even in $\mathbb{Z}$, for instance both $3$ and $-3$ are gcd's of $6$ and $15$. But the change of $\pm$ is the only thing that can happen in $\mathbb{Z}$, because $\pm 1$ are the only units of $\mathbb{Z}$.

Now quasi's answer is correct, and you should understand why in that example, all units of the ring are gcd's for your example. However, being all units, of course they are all associated (meaning: If $d$ and $e$ are gcd's of $a$ and $b$, then there is a unit $u$ such that $d\cdot u=e$).

In a domain, that would always be the case: If gcd's exist, they are associated.

That is also true in your ring $\mathbb{Z}/9$, as you can check by going through all cases (and noticing that $6 = -3$ in that ring), and it will be true for any $\mathbb{Z}/n$. But it is not true in general in rings with zero-divisors. Examples for this are a bit more complicated. Some are given in answers to this question: Two principal ideals coincide if and only if their generators are associated, and here.

Torsten Schoeneberg
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By definition, in a commutative ring $R$, given two elements $a,b \in R$, an element $d \in R$ is a $\gcd$ of $a,b$, if

  • $d|a$ and $d|b\;$(i.e., $d$ is a common divisor of $a,b$).$\\[4pt]$
  • If $e \in R$ is such that $e|a$ and $e|b$, then $e|d$.

If all ideals of $R$ are principal (as is the case for $Z_9$), then any principal generator of the ideal $(a,b)$ is a $\gcd$ of $a,b$.

In particular, in the ring$Z_9$, the elements $2,7$ are units, so any unit is a $\gcd$ of $2,7$, For example, $1$ is a $\gcd$ of $2,7$.

quasi
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  • Thank you but the $\gcd$ should be $(a)+(b)$ ? Here we have $(\bar0), (\bar3)$ and $\mathcal{Z}/9\mathcal{Z}=(\bar1)$ as ideals – Maman Oct 25 '17 at 17:34
  • It seems to work like $\mathcal{Z}/p\mathcal{Z}$ with $p$ prime ? – Maman Oct 25 '17 at 17:39
  • If the ideal $(a,b)$ is a principal ideal, then any principal generator of the ideal $(a,b)$ is a $\gcd$ of $a,b$. – quasi Oct 25 '17 at 17:41
  • $(\bar{4},\bar6)$ $\subset$ $(\bar2)$ so the $\gcd(\bar4,\bar6)=(\bar2)$ – Maman Oct 25 '17 at 17:43
  • It's not "is in" that matters. If the ideal $(a,b)$ is principal, it's "is a principal generator" that matters. – quasi Oct 25 '17 at 17:46
  • We cannot use Bézout in the rings of the form $\mathcal{Z}/p^\alpha\mathcal{Z}$ ? – Maman Oct 25 '17 at 17:48
  • In the ring $Z_9$, since the element $4$ is a unit, it follows that the ideal $(4,6) = (1)$, hence any unit is a $\gcd$ of $4,6$ (i.e., any of the elements $1,2,4,5,7,8$ is a $\gcd$ of $4,6$. – quasi Oct 25 '17 at 17:51
  • If all ideals are principal, the Bezout property holds automatically. – quasi Oct 25 '17 at 17:52
  • If I take $(\bar3, \bar6)$ ? We have the fact that $(\bar3,\bar6)\subset (\bar3)$ – Maman Oct 25 '17 at 17:54
  • Find a principal generator. – quasi Oct 25 '17 at 17:55
  • Note: A $\gcd$ (assuming one exists) is an element, not an ideal. – quasi Oct 25 '17 at 17:56
  • In the ring $Z_9$, the elements $3,6$ are associates (each is a unit times the other), so $3$ is a $\gcd$ of $3,6$, but also $6$ is a $\gcd$ of $3,6$. Notationally, for the ring $Z_9$, it's legal to write $\gcd(3,6) = 3$, and it's also legal to write $\gcd(3,6)=6$. An alternative notation is to write $\gcd(3,6)={3,6}$, which simply means that the elements of the ring which are $\gcd's$ of $3,6$ are the elements of the set ${3,6}$. – quasi Oct 25 '17 at 18:00
  • Thank you but for instance $(\bar3, \bar6)$ is not a principal ideal but $(\bar3)$ is a principal generator ? – Maman Oct 25 '17 at 18:04
  • All ideals of $Z_9$ are principal ideals. – quasi Oct 25 '17 at 18:06
  • Also, a principal generator is an element, not an ideal. – quasi Oct 25 '17 at 18:06
  • Than you for being patient... It seems that it's the same procedures as $\mathcal{z}/p\mathcal{z}$... So strange – Maman Oct 25 '17 at 18:09
  • Strange perhaps, but also beneficial. It's nice to have $\gcd$s. Note: In general, for an arbitrary commutative ring with unity, not all pairs $a,b$ will necessarily have a $\gcd$. – quasi Oct 25 '17 at 18:16
  • Indeed if you take a non UFD it's the case $(\mathcal{Z}[i\sqrt{3}]$) ! – Maman Oct 25 '17 at 19:18