I'm asked to prove the following statement:
Let $R$ be a unique factorization domain, let $a,b$ be coprime elements. Show that for any ideal $J$ we have $$(ab)+J=((a)+J )\cap ((b) + J).$$
When $J=0$ it's trivial, but when $J \ne 0$, I can't work it out, although it seems easy.
This statement arises naturally when dealing with primary decomposition.
Update: Thanks Andraw Hubery for pointing out that it is not in general true for UFD.(Although valid for PID)
It's true for a principal ideal domain, since for $J=(x)$ this then becomes $$\mathrm{gcd}(ab,x)=\mathrm{gcd}(a,x)\cdot\mathrm{gcd}(b,x).$$
It's not true in general for unique factorisation domains. For example, let $R=K[x,y]$ be the polynomial ring over some field $K$, and take $a=x$, $b=y$ and $J=(x+y)$. Then $$(xy)+J=(xy,x+y) \quad \textrm{and}\quad (x)+J=(x,y)=(y)+J,$$ and clearly $(xy,x+y)\neq(x,y)$.