Intersection of two principal ideals is an ideal and lowest common multiple (if it is a PI) I think the first part of the proof would go like this: any element in $(a) \cap (b)$ can be written as $ar_1 = br_2$, so multiplying by an element $r \in R$ yields $ar_1r\in aR$ or $br_2r \in bR$, so it stays in $(a)\cap (b)$. Does this look correct?
I'm not sure how to approach the common multiple problem...how do I show if $(m) = (a) \cap (b)$, then $m$ is a least common multiply of $a$ and $b$?
Thanks a lot!
 A: As mentioned in the comment by Tom Price, in the first part you still need to check closure of subtraction.
For the second part here is an outline. Let $(m)=(a)\cap (b)$. This means that $m = ak$ and $m=bl$ for some elements in our ring (why?), i.e. $a|m$ and $b|m$. Thus $m$ is a common multiple of $a$ and $b$. Now we must show that it is the smallest common multiple. Suppose there was a smaller common multiple $m'$ of $a$ and $b$. Now we can write $m'=ak'$ and $m'=bl'$, which means $m' \in (a) \cap (b)$. Thus $m'\in (m)$, but this leads to a contradiction (how?).
A: Generally, the intersection of two ideals is always an ideal. This follows easily by direct verification:


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*Let $ x $ be an arbitrary element of $ R $ (our ring) and let $ I $ and $ J $ be ideals in our ring, with $ m \in I \cap J $. Then, by definition of an ideal, $ xm $ is in both $ I $ and $ J $, and is therefore in $ I \cap J $. This establishes that $ I \cap J $ absorbs multiplication. (This easily generalizes to left and right ideals.)

*Let $ x $ and $ y $ be elements of $ I \cap J $, then $ x + y $ is in both $ I $ and $ J $ and therefore in $ I \cap J $. The same goes for additive inverses, and so $ I \cap J $ is an additive subgroup of $ R $.


As it satisfies both conditions, it is an ideal by definition.
If $ I $ and $ J $ are principal ideals (generated by $ i $ and $ j $ respectively), then the set $ I \cap J $ is the set comprised of elements which are divisible by both $ i $ and $ j $. In a GCD domain, we can define $ \textrm{lcm}(a, b) = \gcd \{ x : x = r_1 a = r_2 b : r_1, r_2 \in R \} $ so that the least common multiple exists, denote $ \textrm{lcm}(i, j) = l $. By definition, any element divisible by both $ i $ and $ j $ is divisible by $ l $, so we have $I \cap J \subseteq (l) $. Assume that there were an element $ x \in (l) $ such that $ x \notin I $ or $ x \notin J $, then $ x $ is not divisible by at least one of $ i $ and $ j $. Now, note that by our definition of the lcm, as $ a $ and $ b $ are common divisors of every element in the set we defined, they must divide the greatest common divisor of all the elements in the set (the lcm). This allows us to conclude that $ l $ is divisible by both $ i $ and $ j $, and so is any element of the principal ideal $ (l) $. Therefore, no such $ x $ can exist, showing that $ I \cap J = (i) \cap (j) = (l) $ in any GCD domain.
