Unique factorization domain and principal ideal My task is to show that if P is unique factorization domain then for any $a,b \in P$ ideal $(a) \cap(b)$ is principal.
So I was thinking about using the fact that $(a) \cap (b) = (\operatorname{lcm}(a,b))$ (I'm not sure it applies there) and I don't know this would be it? 
 A: Your stated fact should hold for the domain $\mathbb{Z}$. How would you define the LCM for two elements in general? After that, mimic the proof of the fact in the case of $\mathbb{Z}$ should be fine.
A: Your hypothesis is true, actually, in a commutative ring with identity, $a, b\in R$ has the lcm-property if and only if $(a)\cap (b)$ is principal. 
Proof: If $(a)\cap (b)$ is principal, take $c\in R$ s.t. $(a)\cap (b)=(c)$, hence $a\mid c$ and $b\mid c$, ($c$ is a common multiple). In order to be $c$ the least common multiple of $a$ and $b$, another multiple of $a$ and $b$ must be a multiple of $c$. If $r\in R$ is a multiple of $a$ and $b$ then $r\in (a)\cap (b)=(c)$ so $r$ is a multiple of $c$. That means that the lcm of $a$ and $b$ exists.
On the other hand if the lcm exists, name it $c$. Convince yourself that $(c)\subseteq (a)\cap (b)$ as a result of $c$ to be a multiple of $a$ and $b$. On the other direction, if $u\in (a)\cap (b)$ then $a\mid u$ and $b\mid u$, and $c\mid u$ because $c$ is the least common multiple. So $(a)\cap (b)\subseteq (c)$.
Thus you don't need that P is a UFD, if the lcm exists, your statement holds.
