Intersection of Ideals of coprime elements in a UFD

Question

Let $D$ be a UFD and let $a,b,c \in D$ be pairwise coprime, i.e. $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$. I would like to prove that then $(a,b) \cap (c) = (a,b) \cdot (c) = (ac,bc)$. Note that $(a,b)$ and $(c)$ are not necessarily comaximal ideals, so this does not follow immediately.

Here is my approach: $(ac,bc) = (a,b) \cdot (c) \subseteq (a,b) \cap (c)$ as this holds for all ideals. Now let $x \in (a,b) \cap (c)$. Since $x \in (a,b)$ there exist $d_1, d_2 \in D$ with $x=d_1a + d_2b$. As $x \in (c)$ we know that $c|x$, but how can I use that the elements being coprime of means that $c$ divides $d_1$ and $d_2$?

Answer 1

What you are trying to prove is false; consider $D=\Bbb{Z}[x,y]$ and $a=x$, $b=y$, $c=x+y$. Then \begin{eqnarray*} (a,b)\cap(c)&=&(x,y)\cap(x+y)=(x+y),\\ (ac,bc)&=&\big(x(x+y),y(x+y)\big), \end{eqnarray*} which are not the same; if $x+y\in\big(x(x+y),y(x+y)\big)$ then there exist $f,g\in\Bbb{Z}[x,y]$ such that $$x+y=x(x+y)f+y(x+y)g,$$ from which it follows that $1=xf+yg$, which is of course impossible.