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Let $I,J$ two ideals in a ring $R$. The product of ideals $IJ$ is included in $I \cap J$. For example we have equality in $\mathbb{Z}$ if generators have no common nontrival factors, in a ring $R$ when $I+J=(1)$, or when $R/IJ$ has no nonzero nilpotent elements. My question is not about equality, instead it is about strict inclusion.

Under what conditions $IJ \subsetneq I \cap J$ ?

If the question appears a little too general, then my primary aim is to see what happens under the hypothesis that $R$ is a Dedekind domain.

Oo3
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2 Answers2

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Hint: in a Dedekind (or Prüfer) domain $\:(I+J)\: (I\cap J)\: =\: IJ\ \ $ (gcd $*$ lcm law)

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Math Gems
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  • Good, however this does not tell much about the fact that $IJ$ is not equal to $I \cap J$. For example, if you consider a prime ideal in a Dedekind domain, it is $II \neq I$, and using the above identity this is difficult to see because it does not explicitly tell that $II$ is strict in $I \cap I=I$. – Oo3 Feb 16 '12 at 18:26
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    @Oo3 Generalizing PIDs, in a Dedekind domain: contains = divides. Thus $:(I+J): (I\cap J): =: IJ:$ implies $:I\cap J \supsetneq IJ$ $\iff$ $I\cap J\ |\ IJ$ properly $\iff$ its cofactor $I+J \ne 1$, i.e. by the gcd $*$ lcm law, the lcm is proper $\iff$ the gcd is proper. – Math Gems Feb 17 '12 at 04:06
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There is actually a complete answer, when we assume that the ring $R$ is commutative (which you do when you state the condition $I + J = R$). I stole it from David Speyer's answer here, where he gives a short proof too. Here is the statement:

Let $I,J \leq R$ for a commutative ring $R$. Then $IJ = I \cap J$ if and only if $\mathrm{Tor}^R_1(R/I,R/J) = 0$.

So the inclusion $IJ \subsetneq I \cap J$ is strict if and only if $\mathrm{Tor}^R_1(R/I,R/J) \neq 0$.

user26857
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Bogdan
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