Lemma 57.18.4. Let $K$ be an algebraically closed field. Let $S$ be a finite type scheme over $K$. Let $X \to S$ and $Y \to S$ be finite type morphisms. There exists a countable set $I$ and for $i \in I$ a pair $(S_ i \to S, h_ i)$ with the following properties

$S_ i \to S$ is a morphism of finite type, set $X_ i = X \times _ S S_ i$ and $Y_ i = Y \times _ S S_ i$,

$h_ i : X_ i \to Y_ i$ is an isomorphism over $S_ i$, and

for any closed point $s \in S(K)$ if $X_ s \cong Y_ s$ over $K = \kappa (s)$ then $s$ is in the image of $S_ i \to S$ for some $i$.

**Proof.**
The field $K$ is the filtered union of its countable subfields. Dually, $\mathop{\mathrm{Spec}}(K)$ is the cofiltered limit of the spectra of the countable subfields of $K$. Hence Limits, Lemma 32.10.1 guarantees that we can find a countable subfield $k$ and morphisms $X_0 \to S_0$ and $Y_0 \to S_0$ of schemes of finite type over $k$ such that $X \to S$ and $Y \to S$ are the base changes of these.

By Lemma 57.18.1 there is a countable set $I$ and pairs $(S_{0, i} \to S_0, h_{0, i})$ such that

$S_{0, i} \to S_0$ is a morphism of finite type, set $X_{0, i} = X_0 \times _{S_0} S_{0, i}$ and $Y_{0, i} = Y_0 \times _{S_0} S_{0, i}$,

$h_{0, i} : X_{0, i} \to Y_{0, i}$ is an isomorphism over $S_{0, i}$.

such that every pair $(T \to S_0, h_ T)$ with $T \to S_0$ of finite type and $h_ T : X_0 \times _{S_0} T \to Y_0 \times _{S_0} T$ an isomorphism is isomorphic to one of these. Denote $(S_ i \to S, h_ i)$ the base change of $(S_{0, i} \to S_0, h_{0, i})$ by $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(k)$. We claim this works.

Let $s \in S(K)$ and let $h_ s : X_ s \to Y_ s$ be an isomorphism over $K = \kappa (s)$. We can write $K$ as the filtered union of its finitely generated $k$-subalgebras. Hence by Limits, Proposition 32.6.1 and Lemma 32.10.1 we can find such a finitely generated $k$-subalgebra $K \supset A \supset k$ such that

there is a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d]_ s \ar[r] & \mathop{\mathrm{Spec}}(A) \ar[d]^{s'} \\ S \ar[r] & S_0} \]

for some morphism $s' : \mathop{\mathrm{Spec}}(A) \to S_0$ over $k$,

$h_ s$ is the base change of an isomorphism $h_{s'} : X_0 \times _{S_0, s'} \mathop{\mathrm{Spec}}(A) \to X_0 \times _{S_0, s'} \mathop{\mathrm{Spec}}(A)$ over $A$.

Of course, then $(s' : \mathop{\mathrm{Spec}}(A) \to S_0, h_{s'})$ is isomorphic to the pair $(S_{0, i} \to S_0, h_{0, i})$ for some $i \in I$. This concludes the proof because the commutative diagram in (1) shows that $s$ is in the image of the base change of $s'$ to $\mathop{\mathrm{Spec}}(K)$.
$\square$

## Comments (0)