morphism which is open but not universally open - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:47:29Z http://mathoverflow.net/feeds/question/42775 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42775/morphism-which-is-open-but-not-universally-open morphism which is open but not universally open unknown (google) 2010-10-19T12:20:59Z 2010-10-19T12:20:59Z <p>In someone's note, I have seen such an example, but I can't show that it is not universally open. Here is the example:</p> <p>Let $k$ be a field and $A = k[T]_{(T)}$, the discrete valuation ring obtained from the polynomial ring $k[T]$ localized at the prime ideal $(T)$. Let $\hat{A}$ be the completion with respect to $(T)$, which is just the power series ring $k[[ \ T \ ]]$.</p> <p>Now the natural map $A \rightarrow \hat{A}$ gives a open morphism $i : Spec(\hat{A}) \rightarrow Spec(A)$. Consider the base change $j : Spec(\hat{A}) \rightarrow Spec(A)$, we obtain a morphism $i^{'} : Spec( \hat{A} \otimes_{A} \hat{A} ) \rightarrow Spec( \hat{A} )$. Then the author said this is not an open morphism.</p> <p>There is a unique maximal ideal, called $m$ in $\hat{A} \otimes_{A} \hat{A}$, whose pullback in $\hat{A}$ under $i^{'}$ is the maximal ideal in $\hat{A}$. In order to show $i^{'}$ is not open, I need to show $m$ is also a minimal prime ideal. But I don't even know if $\hat{A} \otimes_{A} \hat{A}$ is an integral domain or not? </p> <p>There is another example in EGA, but I still want to know if the above example is right and how to see it.</p>