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?*



Integral morphism between universally closed and separated schemes
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. 


