For what commutative rings with infinitely many maximal ideals we can say that the intersection of any combination of finitely many maximal ideals is not zero? Obviously it holds for Dedekind domains because a product of fin many is not zero and the product is contained in the intersection. I would like some weaker condition... Thanks

  • 3
    $\begingroup$ Equally obviously, you can drop the word "Dedekind".... $\endgroup$ – Steven Landsburg Mar 7 '14 at 13:52

If finitely many distinct maximal ideals $\mathfrak{m}_1, ..., \mathfrak{m}_r$ of a commutative ring $R$ have intersection $0$, then by the Chinese remainder theorem the ring is isomorphic to the finite product of fields $\prod_{i=1}^r R/\mathfrak{m}_i$; in particular, it has only finitely many maximal ideals. So the answer to your question is: "For all."


Every Noetherian ring which is not Artinian; In other words every Noetherian ring of positive dimension. (because THEOREM 7.30 of R.Y.Sharp book "Steps in Commutative Algebra, Second edition" says :
Let $G$ be a module over the ring $R$, and assume that $G$ is annihilated by the product of finitely many (not necessarily distinct) maximal ideals of $R$, Then $G$ is a Noetherian $R$-module if and only if $G$ is an Artinian $R$-module

  • $\begingroup$ This answer is a very particular case of the other one (which gives a complete characterization of the rings into the question): a noetherian ring which is not artinian is obviously not a finite product of fields. $\endgroup$ – user26857 Nov 29 '14 at 17:27
  • 2
    $\begingroup$ I answered is at Mar 7 at 13:34 and the other is at Mar 7 at 14:52. $\endgroup$ – user 1 Nov 29 '14 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.