Give an arbitrary commutative ring (not necessarily noetherian) $A$ with $\mathfrak{m}$ a maximal ideal. Is the completion $\hat{A}$ w.r.t. the ideal $\mathfrak{m}$ a local ring? If so, is the maximal ideal of $\hat{A}$ generated by $\mathfrak{m}$, i.e., is it equal to $\mathfrak{m}\hat{A}$?

I know that this is true when $A$ is noetherian.


closed as off-topic by YCor, Chris Godsil, Steven Sam, Stefan Kohl, S. Carnahan Jul 4 '14 at 23:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Chris Godsil, Steven Sam, S. Carnahan
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Yes. You can find those things in Bourbaki's Algèbre Commutative. $\endgroup$ – Vinteuil Jul 3 '14 at 12:19

Yes for the first question : see Bourbaki Commutative Algebra III, § 2, Proposition 19. No for the second question : see loc. cit., exercise 12.


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