1

Let $A$ is a commutative ring contains $1$, $B$ is a finitely generated A-module.

Here is my conjecture:

If every prime ideals $p$ of $A$ such that those submodules $pB$ is finitely generated then every ideals $I$ of $A$ , we have submodules $IB$ is finitely generated.

This problem pops in to my head very naturally. But I seem to be stuck in it.

I need some proofs for it if it's true, otherwise, a counterexample.

Thank in advance.

Anh_Rose 1210
  • 634
  • 1
    Have you tried to construct a counterexample with $B=A$? – Crostul Sep 30 '16 at 22:25
  • no. I'm not please say more specific. – Anh_Rose 1210 Sep 30 '16 at 22:32
  • Related to the above comment: http://math.stackexchange.com/questions/1130223/if-r-is-a-commutative-ring-in-which-all-the-prime-ideals-are-finitely-generated – A.G Oct 01 '16 at 13:59
  • Cohen's theorem for modules has been proved by Jothilingam as far as I remember, which say that If $M$ is a finitely generated $R$ module, and if $pM$ are finitely generated $R$ modules for every $p\in Spec(R)$, then $M$ is Noetherian. –  Oct 02 '16 at 06:23

0 Answers0