Lemma 66.41.4. Let $S$ be a scheme. Let $f : X \to Y$ be a separated morphism of algebraic spaces over $S$. Suppose given a diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & Y }$

as in Definition 66.41.1 with $K \subset K'$ arbitrary. Then the dotted arrow exists making the diagram commute.

Proof. We have to show that we can find a morphism $\mathop{\mathrm{Spec}}(A) \to X$ fitting into the diagram.

Consider the base change $X_ A = \mathop{\mathrm{Spec}}(A) \times _ Y X$ of $X$. Then $X_ A \to \mathop{\mathrm{Spec}}(A)$ is a separated morphism of algebraic spaces (Lemma 66.4.4). Base changing all the morphisms of the diagram above we obtain

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X_ A \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar@{=}[r] & \mathop{\mathrm{Spec}}(A) }$

Thus we may replace $X$ by $X_ A$, assume that $Y = \mathop{\mathrm{Spec}}(A)$ and that we have a diagram as above. We may and do replace $X$ by a quasi-compact open subspace containing the image of $|\mathop{\mathrm{Spec}}(A')| \to |X|$.

The morphism $\mathop{\mathrm{Spec}}(A') \to X$ is quasi-compact by Lemma 66.8.9. Let $Z \subset X$ be the scheme theoretic image of $\mathop{\mathrm{Spec}}(A') \to X$. Then $Z$ is a reduced (Lemma 66.16.4), quasi-compact (as a closed subspace of $X$), separated (as a closed subspace of $X$) algebraic space over $A$. Consider the base change

$\mathop{\mathrm{Spec}}(K') = \mathop{\mathrm{Spec}}(A') \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(K) \to X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(K) = X_ K$

of the morphism $\mathop{\mathrm{Spec}}(A') \to X$ by the flat morphism of schemes $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(A)$. By Lemma 66.30.12 we see that the scheme theoretic image of this morphism is the base change $Z_ K$ of $Z$. On the other hand, by assumption (i.e., the commutative diagram above) this morphism factors through a morphism $\mathop{\mathrm{Spec}}(K) \to Z_ K$ which is a section to the structure morphism $Z_ K \to \mathop{\mathrm{Spec}}(K)$. As $Z_ K$ is separated, this section is a closed immersion (Lemma 66.4.7). We conclude that $Z_ K = \mathop{\mathrm{Spec}}(K)$.

Let $V \to Z$ be a surjective étale morphism with $V$ an affine scheme (Properties of Spaces, Lemma 65.6.3). Say $V = \mathop{\mathrm{Spec}}(B)$. Then $V \times _ Z \mathop{\mathrm{Spec}}(A') = \mathop{\mathrm{Spec}}(C)$ is affine as $Z$ is separated. Note that $B \to C$ is injective as $V$ is the scheme theoretic image of $V \times _ Z \mathop{\mathrm{Spec}}(A') \to V$ by Lemma 66.16.3. On the other hand, $A' \to C$ is étale as corresponds to the base change of $V \to Z$. Since $A'$ is a torsion free $A$-module, the flatness of $A' \to C$ implies $C$ is a torsion free $A$-module, hence $B$ is a torsion free $A$-module. Note that being torsion free as an $A$-module is equivalent to being flat (More on Algebra, Lemma 15.22.10). Next, we write

$V \times _ Z V = \mathop{\mathrm{Spec}}(B')$

Note that the two ring maps $B \to B'$ are étale as $V \to Z$ is étale. The canonical surjective map $B \otimes _ A B \to B'$ becomes an isomorphism after tensoring with $K$ over $A$ because $Z_ K = \mathop{\mathrm{Spec}}(K)$. However, $B \otimes _ A B$ is torsion free as an $A$-module by our remarks above. Thus $B' = B \otimes _ A B$. It follows that the base change of the ring map $A \to B$ by the faithfully flat ring map $A \to B$ is étale (note that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is surjective as $X \to \mathop{\mathrm{Spec}}(A)$ is surjective). Hence $A \to B$ is étale (Descent, Lemma 35.23.29), in other words, $V \to X$ is étale. Since we have $V \times _ Z V = V \times _{\mathop{\mathrm{Spec}}(A)} V$ we conclude that $Z = \mathop{\mathrm{Spec}}(A)$ as algebraic spaces (for example by Spaces, Lemma 64.9.1) and the proof is complete. $\square$

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