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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!

M47145
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Piepod
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  • Your proof of the first part is incomplete since you also need to show that the intersection is closed under addition, it's the same idea though. For the second part, what definition are you using for a "lowest common multiple" in an arbitrary ring? – Tom Price May 05 '16 at 20:29
  • Forgot to mention it's a Euclidean domain - m is a lcm of a and b if m|a and m|b, and if m'|a and m'|b, then m|m'. Thanks! – Piepod May 05 '16 at 21:56
  • @Piepod The definition you gave is for the greatest common divisor. Do you mean $gcd$ or $lcm$? – M47145 May 05 '16 at 22:21
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    An important subtlety is that while the existence of $ \textrm{lcm}(a, b) $ guarantees the existence of $ \gcd(a, b) $ (just pick $ ab / \textrm{lcm}(a, b) $), the converse is not true: $ \textrm{lcm}(a, b) $ may fail to exist even if a and b have a greatest common divisor. A stronger condition is necessary: every pair of elements in your ring must have a greatest common divisor, then the lcm can be defined as I did in my answer below. – Ege Erdil May 05 '16 at 23:07

2 Answers2

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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?).

M47145
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  • I am not sure whether we are allowed to assume that our domain is a principal ideal domain, as the result is true more generally (for arbitrary GCD domains.) – Ege Erdil May 05 '16 at 22:52
  • In the comments to the question OP suggests our domain is a Euclidean domain, thus it would be a PID. – M47145 May 05 '16 at 23:00
  • Ah, I see. Then your solution is shorter and possibly the solution that the OP was looking for :) – Ege Erdil May 05 '16 at 23:09
  • Ah I understand - but how do you get from (m)=(a)∩(b) to m=ak m=bl? I was originally thinking mj = ak = bl for some j,k,l but it doesn't seem multiplicative inverses necessarily exist to bring the j over. Are you setting j equal to the unit? – Piepod May 06 '16 at 00:24
  • $m\in (m) $ implies $m\in (a)\cap (b)$, which implies $m\in (a)$ and $m\in (b)$. Hence $m=ak$ and $m=bl$ for some $k,l$. – M47145 May 06 '16 at 00:30
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Generally, the intersection of two ideals is always an ideal. This follows easily by direct verification:

  • 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.

Ege Erdil
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