I'd like to show that if $R$ is a commutative ring with connected Spectrum then:

$ R$ is a domain iff $R_{P}$ is a domain for all $P \in Spec(R)$ (where $R_P$ denotes localization at ideal $P$)

The $\Rightarrow$ direction is obvious. Let's suppose that $x,y \in R $ and $ x \cdot y =0$ (both $x,y$ are nontrivial and we can assume that neither of them is nilpotent). Let $I_x,I_y$ denote annoihilators of $x,y$. If $V(I_x) \cap V(I_y)$ is not empty (i.e. there exists prime ideal containing both annihilators) I am able to achieve contradiction. Hence, let's suppose that $V(I_x) \cap V(I_y) = \emptyset $. Now my problem is I don't know whether there can exist prime ideal containing both $x,y$ - if there is no such ideal then $V(I_x),V(I_y)$ is an closed covering, which proves that Spec is not connected. How can I show that this is indeed the case?

  • $\begingroup$ This must be wrong without an extra-condition like $R$ has finitely many minimal prime ideals. $\endgroup$ – user26857 Oct 13 '17 at 22:24

I think you will not be able to prove this, as there is a counterexample: https://stacks.math.columbia.edu/tag/0568. However, if you were to add the condition that $A$ was noetherian it would work.


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.