morphism which is open but not universally open

2010-10-19T12:20:59Z

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:

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 $k[[ \ T \ ]]$.

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.

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?

There is another example in EGA, but I still want to know if the above example is right and how to see it.

morphism which is open but not universally open - MathOverflow

morphism which is open but not universally open