Remark 76.43.4. We can ask if in Grothendieck's algebraization theorem (in the form of Lemma 76.43.3), we can get by with weaker separation axioms on the target. Let us be more precise. Let $A$, $I$, $S$, $S_ n$, $X$, $Y$, $X_ n$, $Y_ n$, and $g_ n$ be as in the statement of Lemma 76.43.3 and assume that

$X \to S$ is proper, and

$Y \to S$ is locally of finite type.

Does there exist a morphism of algebraic spaces $g : X \to Y$ over $S$ such that $g_ n$ is the base change of $g$ to $S_ n$? We don't know the answer in general; if you do please email stacks.project@gmail.com. If $Y \to S$ is separated, then the result holds by the lemma (there is an immediate reduction to the case where $X$ is finite type over $S$, by choosing a quasi-compact open containing the image of $g_1$). If we only assume $Y \to S$ is quasi-separated, then the result is true as well. First, as before we may assume $Y$ is quasi-compact as well as quasi-separated. Then we can use either [Bhatt-Algebraize] or from [Hall-Rydh-coherent] to algebraize $(g_ n)$. Namely, to apply the first reference, we use

where the last step uses a Grothendieck existence result for the derived category of the proper algebraic space $Y$ over $R$ (compare with Flatness on Spaces, Remark 77.13.7). The paper cited shows that this arrow determines a morphism $Y \to X$ as desired. To apply the second reference we use the same argument with coherent modules:

where the final equality is a consequence of Grothendieck's existence theorem (Theorem 76.42.11). The second reference tells us that this functor corresponds to a morphism $Y \to X$ over $R$. If we ever need this generalization we will precisely state and carefully prove the result here.

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