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A partial answer:
It is easy to see that "for all nonzero elements $a$ and $b$, one has $\langle a\rangle\cap \langle b\rangle\neq 0$ iff for all nonzero ideals $I$ and $J$, has $I\cap J\neq 0.$" For Artinian local rings this is equivalent to saying that ring beis Gorenstein; see Bruns_Herzog Exercise 3.2.15.

A partial answer:
It is easy to see that "for all nonzero elements $a$ and $b$, $\langle a\rangle\cap \langle b\rangle\neq 0$ iff for all nonzero ideals $I$ and $J$, $I\cap J\neq 0.$" For Artinian local rings this is equivalent to ring be Gorenstein; see Bruns_Herzog Exercise 3.2.15

A partial answer:
It is easy to see that "for all nonzero elements $a$ and $b$, one has $\langle a\rangle\cap \langle b\rangle\neq 0$ iff for all nonzero ideals $I$ and $J$, has $I\cap J\neq 0.$" For Artinian local rings this is equivalent to saying that ring is Gorenstein; see Bruns_Herzog Exercise 3.2.15.

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user 1
  • 1.4k
  • 1
  • 13
  • 24

A partial answer:
It is easy to see that "for all nonzero elements $a$ and $b$, $\langle a\rangle\cap \langle b\rangle\neq 0$ iff for all nonzero ideals $I$ and $J$, $I\cap J\neq 0.$" For Artinian local rings this is equivalent to ring be Gorenstein; see Bruns_Herzog Exercise 3.2.15