5
$\begingroup$

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.

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ The map $i'$ is open. Indeed, as you note, the only way it can fail to be open is if the diagonal closed point in the source is also open. But if such openness holds, this point would split off as a clopen subscheme (with suitable 1-point scheme structure) which is then necessarily flat over the base since $i'$ is flat. But that's absurd, since it would be a 1-point scheme lying over the closed point of the base and hence has uniformizer on the base pulling back to a nilpotent function, contradicting torsion-freeness over the base which is a consequence of flatness. $\endgroup$
    – BCnrd
    Commented Oct 19, 2010 at 13:44
  • $\begingroup$ Thank you, BCnrd. I would appretiate that anyone gives an example which is open but not universally open, an example different from the one in EGA IV.3, Remarque 14.3.9.i . $\endgroup$
    – user565739
    Commented Oct 19, 2010 at 14:36
  • $\begingroup$ What is wrong with that example? $\endgroup$
    – Laurent Moret-Bailly
    Commented Oct 19, 2010 at 15:02
  • $\begingroup$ To Moret-Bailly, you mean the example in EGA? I just took it a look . Just wanna know if there is another example which looks eaiser. $\endgroup$
    – user565739
    Commented Oct 19, 2010 at 15:25
  • 1
    $\begingroup$ This example is the "standard" one among morphisms of finite type between notherian schemes. In fact, the failure of an open morphism to be universally open is related to the non-normality of the base: see EGA IV.3 (14.4.9). $\endgroup$
    – Laurent Moret-Bailly
    Commented Oct 19, 2010 at 16:27

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .