Lemma 49.10.7. Let $f : Y \to X$ be a morphism of schemes. If $f$ satisfies the equivalent conditions of Lemma 49.10.1 then for every $y \in Y$ there exist $n, d$ and a commutative diagram

$\xymatrix{ Y \ar[d] & V \ar[d] \ar[l] \ar[r] & Y_{n, d} \ar[d] \\ X & U \ar[l] \ar[r] & X_{n, d} }$

where $U \subset X$ and $V \subset Y$ are open, where $Y_{n, d} \to X_{n, d}$ is as in Example 49.10.5, and where the square on the right hand side is cartesian.

Proof. By Lemma 49.10.1 we can choose $U$ and $V$ affine so that $U = \mathop{\mathrm{Spec}}(R)$ and $V = \mathop{\mathrm{Spec}}(S)$ with $S = R[y_1, \ldots , y_ n]/(g_1, \ldots , g_ n)$. With notation as in Example 49.10.5 if we pick $d$ large enough, then we can write each $g_ i$ as $g_ i = \sum _{E \in T} g_{i, E}y^ E$ with $g_{i, E} \in R$. Then the map $A \to R$ sending $a_{i, E}$ to $g_{i, E}$ and the map $B \to S$ sending $x_ i \to y_ i$ give a cocartesian diagram of rings

$\xymatrix{ S & B \ar[l] \\ R \ar[u] & A \ar[l] \ar[u] }$

which proves the lemma. $\square$

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