Let $A$ be an integral domain. Let $m′$ be an element of $A$ that divides all the element $c \in A$ which divides $a, b \in A$. We would like to show that there exists an invertible element $x \in A$ such that $m=m′x$.
This question is the fourth sub-question that I am trying to solve. The previous sub-questions were :
Show that d divides a iff the principal ideal of (a) is included in the principal ideal (d).
Let $a, b \in A$ where $(a)=(b)$. Show that there exists invertible elements $x, y \in A$ such that $a=bx$ and $b=ay$. Also show that $y$ is then the inverse of $x$ and that these elements $x$, $y$ are unique (for $a$ and $b$ given).
Let $m \in A$ be such that $(m)$ is the principal ideal of the intersection of the principal ideal $(a)$ and the principal ideal $(b)$. We want to show that for all elements $c$ which divides $a$ and divides $b$ it can also be divisible by $m$.
I'm not totally certain that my solution is correct, so if people could give me a hand in checking my work, I'd really appreciate it!
We know that $m′|c$ and that $a|c$ and $b|c$ then if we define the principal ideal $(m)$ as the intersection of the principal ideal $(a)$ and the principal ideal $(b)$ then we have that $m|c$. We can then write $c=my$ with $y\in A$ and since we know that $m′|c$ we have that $c=m′x$ with $x\in A$. We then have that $my=m′x$. Since $1\in A$ then $y$ can be equal to $1$. We then get $m=m′x$.
I am not sure if I have proven the existence of an invertible element $x$.
Take $A=\mathbb{Z}$, $a=2, b=3$. Then $m^\prime=1$ satisfies the given condition i.e., if there is an element $c$ such that $c\mid 2$ and $c\mid 3$ then $m^\prime \mid c$. Now $m=6$. Thus there does not exist any invertible element $x$ such that $m^\prime x=m$. So I think that the question needs to be repaired.