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Let $S$ be a semigroup and $I,J$ be two ideals of $S$. For a semilattice we know that $IJ=I\cap J$. Now the question is there a semigroup with the property $IJ=I\cap J$. thanks for your attention

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  • $\begingroup$ Are you asking which semigroups have the property that the intersection of any two ideals is there product? $\endgroup$ – Benjamin Steinberg Jul 18 '14 at 14:25
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I assume you are asking which semigroups have the property that the intersection of any two ideals is their product.

One condition that will do it is von Neumann regularity. $S$ is regular if, for all $s\in S$, there is $t\in S$ with $sts=s$.

In any semigroup $IJ\subseteq I\cap J$. If $S$ is regular and $s\in I\cap J$, then choosing $t$ with $sts=s$, we have $s\in I$ and $ts\in J$ and so $s=sts\in IJ$. Thus $IJ=I\cap J$.

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Steinberg's example is more related to semigroup theory. But why you do not use ring theory? For every two comaximal ideals in a ring with one, we have this property. Note that

  1. Every ideal of a ring, is an ideal of multiplicative semigroup of the ring

  2. Every monoid which is the multiplicative monoid of a ring can be used for creating your example.

example Let S be the multiplicative monoid of integers.

suppose I=2Z and J=3Z

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