Lemma 76.41.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ separated of finite type, and $Y$ separated and quasi-compact. Then there exists a commutative diagram
where $X' \to X$ is proper surjective morphism and the morphism $X' \to \mathbf{P}^ n_ Y$ is an immersion.
Proof. By Limits of Spaces, Proposition 70.11.7 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 70.8.1. By Limits of Spaces, Lemma 70.5.9 we may assume that $Y_ i$ is separated for all $i$. By Limits of Spaces, Lemma 70.7.1 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 70.6.9 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 76.40.5. $\square$
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