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Suppose $I,J$ are ideals and $R$ is a PID. Why $IJ= I \cap J$ implies $I+J=R$?

Note. Original question asked the converse: why does $I+J=R$ imply $IJ=I\cap J$.

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non-trivial ideals of course –  user6495 Apr 4 '11 at 2:49
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@user6495: When you change a question so completely as to render already posted answers incorrect, it's best to "leave a trail" to let future readers know that the posters who replied did not misunderstand the question or answer incorrectly. –  Arturo Magidin Apr 4 '11 at 2:52
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2 Answers

up vote 6 down vote accepted

Answer to the original posed question.

You don't need $R$ to be a PID.

In any ring, $IJ\subseteq I$ and $IJ\subseteq J$ (in fact, this holds if $I$ is a right ideal and $J$ is a left ideal); so $IJ\subseteq I\cap J$.

For the reverse inclusion, if $R$ has a unity and $JI\subseteq IJ$, then $$(I\cap J) = (I\cap J)R = (I\cap J)(I+J) = (I\cap J)I + (I\cap J)J \subseteq IJ + IJ = IJ.$$

The conclusion thus holds, in particular, whenever $R$ is commutative with $1$.


Answer to the new question.

The question has now been changed to asking a converse: why is it the case that if $IJ=I\cap J$, then $I+J=R$ (in a PID)?

In a PID, this comes down to the fact that the least common multiple equals the product exactly when you are in the relatively prime case.

Let $I+J = (k)$, and $I=(a)$, $J=(b)$. Write $a=kx$ and $b=ky$. Then $kxy\in (a)\cap (b)$, so $(kxy)\subseteq I\cap J = IJ = (ab)=(k^2xy)$. Therefore, $k^2xy|kxy$, so $k$ is a unit, hence $I+J=R$.

Added. The converse no longer holds for arbitrary commutative rings with $1$. For example, taking $S$ to be a nonsimple ring. and let $L$ be a nontrivial ideal of $S$. Let $R=S\times S$, $I=\{(\ell,0)\mid \ell\in L\}$ and $J=\{(0,\ell)\mid \ell\in L\}$. Then $I\cap J = \{(0,0)\} = IJ$, but $I+J = L\times L\neq R$.

Of course, the example above is not a domain. There are many domains in which the result does hold (e.g., Dedekind domains). But here's an example: take $R$ to be the ring of all algebraic integers, and let $I=J$ be the ideal generated by all $n$th roots of $2$, $$I = J = (2^{1/n}\colon n\in\mathbb{N}).$$ Since $2^{1/n} = 2^{1/2n}2^{1/2n}$, then $IJ=I=I\cap J$, but $I+J = I \neq R$.

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my mistake, just corrected it. –  user6495 Apr 4 '11 at 2:50
    
thank you very much, it is clear now. –  user6495 Apr 4 '11 at 6:56
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It's $\rm\: (i,j) = 1\ \Rightarrow\ lcm(i,j)\ =\ i\ j\:.\ $ Proof: $\rm\ i,j\ |\ r\ \Rightarrow\ i\ |\ j\:(r/j)\ \Rightarrow\ i\ |\ r/j\ \Rightarrow\ i\ j\ |\ r\ $ by Euclid's Lemma.

Since you changed your question to the converse, it's probably simplest to use the general gcd * lcm law $\rm\ a\: b\ =\ gcd(a,b)\ lcm(a,b)\:,\:$ which immediately yields both directions of your question - see the proof below. For more on the universal definitions of gcd, lcm employed below see here and here.

THEOREM $\rm\;\; (a,b)\ =\ ab/[a,b] \;\;$ if $\rm\:[a,b] $ exists, $\:$ where $\rm\:[a,b]\:$ denotes $\rm\:lcm(a,b)\:.$

Proof: $\rm\quad\quad d\ |\ a,b \;\iff\; a,b\ |\ ab/d \;\iff\; [a,b]\ |\ ab/d \;\iff\; d\ |\ ab/[a,b]$

COROLLARY $\rm\ \ (a,b)\ = 1\ \iff\ [a,b] = ab$

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thanks! –  user6495 Apr 4 '11 at 6:56
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