0
$\begingroup$

Let $f : X\to Y$ be a morphism between schemes over $\text{Spec}(\mathbf{Z}_p)$.

Assume:

  • $f$ is integral
  • both $X$ and $Y$ are universally closed and separated over $\mathbf{Z}_p$
  • $f$ mod $p^n$ is an isomorphism for every $n \ge 0$
  • $f_*\mathcal{O}_X = \mathcal{O}_Y$

Is $f$ an isomorphism?

Example If $X$ and $Y$ were of finite type, they would be proper and so $f$ would be finite. The flat locus of $f$ in $X$ is open and nonempty, and then it is $X$. So $f$ is finite flat of degree one, then an isomorphism. (We haven't even used the full assumptions (3) and (4)). The point of this question is: what if $X$ and $Y$ are not of finite type, not even locally?

$\endgroup$

1 Answer 1

3
$\begingroup$

Sure, by $(1)$ and $(4)$. Any integral morphism is affine by definition. If $f$ is an affine morphism with $f_* \mathcal{O}_X = \mathcal{O}_Y$, then $f$ is clearly an isomorphism.

$\endgroup$

You must log in to answer this question.