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In integral domains, prove or give a counterexample：

If $(a,b)\sim 1$, $(a,c)\sim1$, then $(a,bc)\sim1$.

That is to say, if $a,\,b$ are relatively prime and $a,\,c$ are relatively prime, then $a,\,bc$ are relatively prime.

Note that it needn't be a GCD domain. So usual proofs may not work here.

Any insights are much appreciated.

BTW："$(a,b)$" is the abbreviation for "$\gcd(a,b)$" and "$x\sim y$" denotes the statement that $x$ and $y$ are associates. In integral domains, a gcd is often not unique when it exists. So "$(a,b)$" here denotes an arbitrary gcd of a and b. For the same reason, we use "$\sim$ " instead of "$=$".

Sky subO
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    For a counterexample, my initial guess would be to take $a = 2$, $b = 1 - \sqrt{-5}$, $c = 1 + \sqrt{-5}$ in $\mathbb Z[\sqrt{-5}]$ – Rushy Sep 11 '21 at 20:49
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    What do you denote $(a,b)\sim 1$? – Bernard Sep 11 '21 at 21:13
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    Integral domains don’t even necessarily have a notion of GCD. (Presumably, that is what you mean by $(a,b)$) – Thomas Andrews Sep 11 '21 at 21:26
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    If by $(a,b)\sim 1$ you mean the ideal generated by $a,b$ is the whole ring $R$ (i.e., $(a)$ and $(b)$ are coprime ideals), then $R=aR+bR=aR+b(aR+cR)=aR+bcR$. – user10354138 Sep 12 '21 at 14:13
  • @Bernard 1 is a $\gcd$ of $a$ and $b$, in other words, $a$ and $b$ are relatively prime. – Sky subO Sep 12 '21 at 16:44
  • @user10354138 Sorry for the confusion. I've updated the post. It means $a$ and $b$ are coprime. – Sky subO Sep 12 '21 at 17:03
  • If you simply mean there exists $u,v,u',v'$ in the integral domain such that $ua+vb=1$ and $u'a+v'c=1$, you just have to multiply these relations. – Bernard Sep 12 '21 at 18:53
  • @SkysubO You still haven't define what you mean by coprime in this context. There are two inequivalent definitions of "$a,b$ coprime" you can use: (1) There are no common divisors other than units (2) the principal ideals are coprime ideals $aR+bR=R$. (2) implies (1) but not the other way round, e.g. $x+2$ and $x+4$ in $\mathbb{Z}[x]$. – user10354138 Sep 13 '21 at 01:39
  • @user10354138 Ah, I see. I mean $1$ is a $\gcd$ of $a$ and $b$, that is, (1) There are no common divisors other than units – Sky subO Sep 13 '21 at 08:08
  • @Bernard Unfortunately, we don't have the equation $ua+vb=1$. Let $d$ be a gcd of $a$ and $b$. Generally in integral domains, we only have $(a)+(b)\subseteq(d)$. So here we just know $(a)+(b)\subseteq(1)$. – Sky subO Sep 13 '21 at 08:44
  • @SkysubO: However, your ring is a gcd domain? – Bernard Sep 13 '21 at 09:31
  • @Bernard Just an integral domain. If it's a gcd domain, the statement is true. – Sky subO Sep 13 '21 at 15:01
  • What does ‘coprime’ mean in this case? – Bernard Sep 13 '21 at 15:08
  • ‘coprime’ means $1$ is a gcd of $a$ and $b$, in other words, $a$ and $b$ have no common divisors except units. – Sky subO Sep 13 '21 at 15:26
  • @Rushy Sounds a good idea! Can you prove that $\gcd(,2\ , 1-\sqrt{-5})\sim 1$ ? – Sky subO Sep 13 '21 at 16:26
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    They are irreducible, so either $(2, 1 - \sqrt{-5}) \sim 1$ or $2$ and $1 - \sqrt{5}$ are associate. But the only units in $\mathbb Z[\sqrt 5]$ are $1$ and $-1$, so they aren't associates. – Rushy Sep 13 '21 at 17:16
  • @Rushy I'm sure they are all irreducible and the fact that $2$ and $1−\sqrt{-5}$ are not associates. So my only concern or question is the statement "either...or...". – Sky subO Sep 13 '21 at 17:41
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    Assume that $2$ and $1 - \sqrt{-5}$ have a common divisor $d$, i.e. we have $2 = ad$ and $1 - \sqrt{-5}= bd$. Either $d$ is a unit, or we have that $d$ is not a unit, in which case both $a$ and $b$ must be units due to irreducibility of $2$ and $1 - \sqrt{-5}$. But then $ba^{-1}$ is a unit and $ba^{-1} \cdot 2 = ba^{-1}ad = bd = 1-\sqrt{-5}$, so $2$ and $1 - \sqrt{-5}$ are associates. – Rushy Sep 13 '21 at 17:58
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    And since the only associates of $2$ are $2$ and $-2$, we conclude that $d$ must be a unit and the claim follows. – Rushy Sep 13 '21 at 18:00
  • @Rushy Nice proof, your are great! Such a good counterexample is beyond my expectation. I thought before that the counterexample should be something where $\gcd(a,bc)$ doesn't exist. Please write it as a formal answer to let more people know. Thanks a million. – Sky subO Sep 13 '21 at 20:16

2 Answers2

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The claim is false. As a counterexample, consider $a = 2$, $b = 1 - \sqrt{-5}$ and $c = 1 + \sqrt{-5}$ in the ring $\mathbb Z[\sqrt{-5}]$. Note that $a, b,$ and $c$ are irreducible.

We show that $(2, 1 - \sqrt{-5}) \sim 1$. Assume that $2$ and $1 - \sqrt{-5}$ have some common divisor $d$. Then we can write $2 = xd$ and $1 - \sqrt{-5} = yd$ for some $x, y \in \mathbb Z[\sqrt{-5}]$. If $d$ is not a unit, then we have by irreducibility of $a$ and $b$ that $x$ and $y$ are units. But then $yx^{-1}$ is a unit and we find that $yx^{-1}\cdot 2 = yx^{-1}xd = yd = 1 - \sqrt{-5}$, i.e. $2$ and $1 - \sqrt{-5}$ are associate elements. This is a contradiction, since the only units in $\mathbb Z[\sqrt{-5}]$ are $1$ and $-1$, hence the only elements associate to $a = 2$ are $2$ and $-2$.

Thus we can conclude that $(2, 1 - \sqrt{-5}) \sim 1$, and by the same argument $(2, 1 - \sqrt{-5}) \sim 1$. However, $bc = (1 - \sqrt{-5})(1 + \sqrt{-5}) = 6$, so $(a, bc) = (2, 6)$. Clearly $2$ and $6$ have $2$ as a common divisor, so $(a, bc) \nsim 1$.

Rushy
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Let me provide a wrong proof here to show that how deceptive this claim is. Even textbook authors were deceived and accepted it as a lemma. Please see the picture below:

wrong proof

$(1)$ is the proposition sometimes called the GCD Distributive Law. Since there are better proofs for it, I've omitted the proof of $(1)$ in the translation：

Lemma $\quad$Let $R$ be an integral domain, $a,b,c\in R-\left\{0 \right\}$. Then

$(1)$ $\ \ c\,(a,b)\sim (ca,cb)$.

$(2)$ $\ \ (a,b)\sim 1$, $(a,c)\sim1$, then $(a,bc)\sim 1$.

Proof $\quad(1)$ ...

$(2)$ From $(a,b)\sim 1$ and $(1)$ we know that $(ac,bc)\sim c$, and obviously we have $(a,ac)\sim a$. So $(a,bc)\sim ((a,ac),bc)\sim$ $(a,(ac,bc))\sim (a,c)\sim 1$. QED

However, $(2)$ is false because of the answer by Rushy. What's wrong with the proof ?

Note that the GCD Distributive Law $(1)$ requires that $(ca,cb)$ exists in R (see this post). But the proof of $(2)$ employs $(1)$ without guaranteeing the existence of $(ac,bc)$. Therefore the statement "$(ac,bc)\sim c$" is incorrect. Consequently, the proof is wrong.

Sky subO
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