$\newcommand{\spec}{\operatorname{Spec}}\newcommand{\O}{\mathcal{O}}$Liu claims the following exercise:
Let $S$ be a locally Noetherian scheme. Let $X,Y$ be $S$-schemes of finite type and $s\in S$.
Then all morphisms of $S$-schemes $\phi:X\times_S\spec\O_{S,s}\to Y\times_S\spec\O_{S,s}$ arise as the base change of some $f:X\times_SU\to Y\times_SU$ along $\spec\O_{S,s}\to U$ where $U$ is some open neighbourhood of $s$.
Ok, sure, I believe you. What I don't believe is how this applies to make the following reduction in his proof of proposition $4.2$:
Let $Y$ be a locally Noetherian scheme, $f:X\to Y$ a separated morphism of finite type such that $\O_Y\to f_\ast\O_X$ is an isomorphism. Then there is an open $V$ of $Y$ such that $f^{-1}(V)\to V$ is an isomorphism and $X_y$ has no isolated points for $y\in Y\setminus V$.
It suffices to show for every $y\in Y$ for whom $X_y$ has an isolated point $x$, there is an open subset $W\ni y$ where $f^{-1}(W)\to W$ is an isomorphism.
It's clear that we may suppose $Y$ is affine to prove this. What's not clear is how we may suppose $Y$ is a local affine scheme, $\spec A$ for a local ring $(A,\mathfrak{m})$. It is supposed to follow from the above exercise.
I guess I am meant to consider $S=Y$; I guess $\phi$ should be $f\times1:X\times_Y\spec\O_{Y,y}\to Y\times_Y\spec\O_{Y,y}\cong\spec\O_{Y,y}$. This satisfies the same property regarding existence of an isolated point. I know $\phi$ arises as the base change of some $g:X\times_YU\to U$ where $U\ni y$ is open…. and what's the point?
Suppose we've shown this for local affine schemes, then I know $f\times 1$ has the intended property. Take $V'\subseteq\spec\O_{Y,y}$ such that $(f\times1)^{-1}(V')\to V'$ is an isomorphism. If $f\times1$ really is the base change of some $g:X\times_YU\to U$ along $\spec\O_{Y,y}\to U$, then… I'm not really sure. I need to return to $X$ somehow, but $g$ need not factor through $X\times_YU\to X$.
What's the idea?
Best Answer
There is indeed a shortcut taken here.
The statement is as follows: if $X,Y$ are schemes of finite type over some locally Noetherian scheme $S$, and if $f: X \rightarrow Y$ is an isomorphism after base changing by $\operatorname{Spec}{\mathcal{O}_{S,s}} \rightarrow S$ for some $s \in S$, then $f$ is an isomorphism after base changing by some open immersion $U \rightarrow S$ with $s \in U$.
Let’s prove it.
Step 1: We assume that $f$ is a closed immersion and we prove the statement.
Proof: we can assume that $S$, then $Y$ (hence $X$) is affine. Let’s say that $A=\mathcal{O}(S)$, $B=\mathcal{O}(Y)$ and $\mathcal{O}(X)=B/I$.
The assumption is that there is a prime $\mathfrak{p} \subset A$ such that $B_{\mathfrak{p}} \rightarrow (B/I)_{\mathfrak{p}}$ is an isomorphism. In other words, $I_{\mathfrak{p}}=0$.
Since $I$ is finitely generated, there is some $a \in A \backslash \mathfrak{p}$ with $I_a=0$. Hence $f$ is an isomorphism above the open subset $D(a) \subset S$.
Step 2: let $X$ be a $S$-scheme of finite type, where $S$ is locally Noetherian. Suppose that $X\times_S \operatorname{Spec}{\mathcal{O}_{S,s}}$ is empty for some $s \in S$. Then $X \times_S U$ is empty for some open subscheme $U \subset S$ containing $s$.
Proof: reduce to the affine case.
Step 3: We assume that $f$ is an immersion and we prove the statement.
Proof: we can assume that $S$ is affine. Let $Y’$ be an open subscheme of $Y$ such that $f$ factors through a closed immersion $g: X \rightarrow Y’$. By Step 1, we see that $g$ is an isomorphism above some open subset of $S$ containing $s$.
The conclusion follows from Step 2 applied to $Y \backslash Y’$.
Step 4: “Local uniqueness of a local extension”. More precisely: Let $X,Y$ be schemes of finite type over a locally Noetherian scheme $S$ and $f,g: X \rightarrow Y$ be two $S$-morphisms that coincide after base changing by $\operatorname{Spec}{\mathcal{O}_{S,s}} \rightarrow S$ for some $s \in S$, then they in fact coincide after base changing by some open immersion $U \rightarrow S$ with $s \in U$.
Proof: we can shrink $S$ and assume that it is affine. Consider the scheme $Z$, which is the fibre product of $(f,g): X \rightarrow Y \times_S Y$ and the diagonal $Y \rightarrow Y \times_S Y$. It has a natural immersion into $X$, and you can check that for every $X$-scheme $T$, one has $f_{|T}=g_{|T}$ iff $T \rightarrow X$ factors through $Z$.
It follows that the natural map $X \times_S \operatorname{Spec}{\mathcal{O}_{S,s}} \rightarrow X$ factors through $Z$. Hence $Z \times_S \operatorname{Spec}{\mathcal{O}_{S,s}} \rightarrow X \times_S \operatorname{Spec}{\mathcal{O}_{S,s}}$ is an isomorphism, and we conclude using Step 3 and the universal property for $Z$.
Step 5: the original statement at the beginning of the answer.
Proof: by Qing Liu’s exercise, we can find an open subscheme $U \subset S$ containing $s$ and morphisms $f_1=f: X \times_S U \rightarrow Y \times_S U$, $g_1: Y \times_S U \rightarrow X\times_S U$ extending $f$ and $f^{-1}$ respectively (seen as morphisms of $\mathcal{O}_{S,s}$-schemes).
Now, $f_1 \circ g_1: Y \times_S U \rightarrow Y \times_S U$ is a morphism extending the identity of $Y \times_S \operatorname{Spec}{\mathcal{O}_{S,s}}$, so by Step 4 there is some open subset $s \in U’ \subset U$ such that $f_1 \circ g_1$ is the identity above $U’$.
Similarly, there is an open subscheme $s \in U’’ \subset U$ such that $g_1 \circ f_1: X \times_S U \rightarrow X \times_S U$ is the identity.
Therefore, above the open subscheme $s \in U’ \cap U’’$, $f$ is an isomorphism with inverse map $g_1$.