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

Question

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 "$=$".

Answer 1

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 $(a, bc) \nsim 1$.

Answer 2

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:

$(1)$ 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.