Lemma 75.41.1. Let $S$ be a scheme. Let $Y$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $f : X \to Y$ be a separated morphism of finite type. Then there exists a commutative diagram

\[ \xymatrix{ X \ar[rd] & X' \ar[l] \ar[d] \ar[r] & \overline{X}' \ar[ld] \\ & Y } \]

where $X' \to X$ is proper surjective, $X' \to \overline{X}'$ is an open immersion, and $\overline{X}' \to Y$ is proper and representable morphism of algebraic spaces.

**Proof.**
By Limits of Spaces, Proposition 69.11.7 we can find a closed immersion $X \to X_1$ where $X_1$ is separated and of finite presentation over $Y$. Clearly, if we prove the assertion for $X_1 \to Y$, then the result follows for $X$. Hence we may assume that $X$ is of finite presentation over $Y$.

We may and do replace the base scheme $S$ by $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Write $Y = \mathop{\mathrm{lim}}\nolimits _ i Y_ i$ as a directed limit of quasi-separated algebraic spaces of finite type over $\mathop{\mathrm{Spec}}(\mathbf{Z})$, see Limits of Spaces, Proposition 69.8.1. By Limits of Spaces, Lemma 69.7.1 we can find an index $i \in I$ and a scheme $X_ i \to Y_ i$ of finite presentation so that $X = Y \times _{Y_ i} X_ i$. By Limits of Spaces, Lemma 69.6.9 we may assume that $X_ i \to Y_ i$ is separated. Clearly, if we prove the assertion for $X_ i$ over $Y_ i$, then the assertion holds for $X$. The case $X_ i \to Y_ i$ is treated by Lemma 75.40.3.
$\square$

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