Isomorphism of schemes over a DVR is determined by the isomorphism over its generic fiber?

Let $R$ be a discrete valuation ring. Let $X\to \operatorname{Spec} R$ and $Y\to \operatorname{Spec} R$ be separated (not necessarily proper) $R$-schemes which are flat over $\operatorname{Spec} R$.

Let $f: X\to Y$ a morphism of schemes over $\operatorname{Spec} R$, which induces an isomorphism on the generic fibers $f_{\eta}: X_{\kappa(\eta)}\to Y_{\kappa(\eta)}$, where $\eta$ is the generic point of $\operatorname{Spec} R$.

Question: Is it true that the original map $f: X\to Y$ is an isomorphism?

Remark: The answer is NO in this generality. In addition to the example in the comment by Notone, here is another example. Take $f=i: X=Y_{\kappa(\eta)}\to Y$ is the inclusion of the generic fiber of $Y$ in $Y$, then $f$ induces isomorphism on the generic fiber but $f$ itself is not an isomorphism.

Let me put more refined question:

In addition to above, assume that both $X$ and $Y$ are smooth over $\operatorname{Spec} R$ and $f: X\to Y$ is Nisnevich. Then is $f$ an isomorphism?

In case it is true under these assumptions, is it possible to remove the smoothness assumptions?

Your comments are most welcome.

Solution 1:

I’m answering a weakening of the refined question. Let $f: X \rightarrow Y$ be a quasi-compact surjective étale map (so I think Nisnevich and quasi-compact is sufficient) of flat separated $R$-schemes that is an isomorphism on the generic fiber. Then $f$ is an isomorphism.

(Without quasi-compactness and surjectivity the relative diagonal is still an isomorphism, but I’m not sure what the exact scheme-theoretical consequence is – is it that $f$ is an immersion and therefore an open immersion?)

Indeed, $f$ is a fpqc cover, so it’s enough to show that the base change of $f$ by $f$ (aka the first projection $\pi: X \times_Y X \rightarrow X$) is an isomorphism.

Now, let $s: X \rightarrow X \times_Y X$ be the diagonal, it’s an open immersion (as $f$ is étale) and closed immersion ($X,Y$ are separated $R$-schemes so $X$ is separated over $Y$). Now, $X \times_Y X$ is a flat $R$-scheme so its generic fiber is dense, ie $s$ has dense closed image, so $s$ is a surjective open immersion hence an isomorphism. As $\pi \circ s$ is the identity, then $\pi$ is an isomorphism and we are done.