Lemma 37.33.11. (Noetherian approximation and combining properties.) Let $P$, $Q$ be properties of morphisms of schemes which are stable under base change. Let $f : X \to S$ be a morphism of finite presentation of affine schemes. Assume we can find cartesian diagrams

$\vcenter { \xymatrix{ X_1 \ar[d]_{f_1} & X \ar[l] \ar[d]^ f \\ S_1 & S \ar[l] } } \quad \text{and}\quad \vcenter { \xymatrix{ X_2 \ar[d]_{f_2} & X \ar[l] \ar[d]^ f \\ S_2 & S \ar[l] } }$

of affine schemes, with $S_1$, $S_2$ of finite type over $\mathbf{Z}$ and $f_1$, $f_2$ of finite type such that $f_1$ has property $P$ and $f_2$ has property $Q$. Then we can find a cartesian diagram

$\xymatrix{ X_0 \ar[d]_{f_0} & X \ar[l] \ar[d]^ f \\ S_0 & S \ar[l] }$

of affine schemes with $S_0$ of finite type over $\mathbf{Z}$ and $f_0$ of finite type such that $f_0$ has both property $P$ and property $Q$.

Proof. The given pair of diagrams correspond to cocartesian diagrams of rings

$\vcenter { \xymatrix{ B_1 \ar[r] & B \\ A_1 \ar[u] \ar[r] & A \ar[u] } } \quad \text{and}\quad \vcenter { \xymatrix{ B_2 \ar[r] & B \\ A_2 \ar[u] \ar[r] & A \ar[u] } }$

Let $A_0 \subset A$ be a finite type $\mathbf{Z}$-subalgebra of $A$ containing the image of both $A_1 \to A$ and $A_2 \to A$. Such a subalgebra exists because by assumption both $A_1$ and $A_2$ are of finite type over $\mathbf{Z}$. Note that the rings $B_{0, 1} = B_1 \otimes _{A_1} A_0$ and $B_{0, 2} = B_2 \otimes _{A_2} A_0$ are finite type $A_0$-algebras with the property that $B_{0, 1} \otimes _{A_0} A \cong B \cong B_{0, 2} \otimes _{A_0} A$ as $A$-algebras. As $A$ is the directed colimit of its finite type $A_0$-subalgebras, by Limits, Lemma 32.10.1 we may assume after enlarging $A_0$ that there exists an isomorphism $B_{0, 1} \cong B_{0, 2}$ as $A_0$-algebras. Since properties $P$ and $Q$ are assumed stable under base change we conclude that setting $S_0 = \mathop{\mathrm{Spec}}(A_0)$ and

$X_0 = X_1 \times _{S_1} S_0 = \mathop{\mathrm{Spec}}(B_{0, 1}) \cong \mathop{\mathrm{Spec}}(B_{0, 2}) = X_2 \times _{S_2} S_0$

works. $\square$

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