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Tag 0B80

Chapter 58: Limits of Algebraic Spaces > Section 58.17: Obtaining schemes

Lemma 58.17.3. Let $f: X \to S$ be a quasi-compact and quasi-separated morphism from an algebraic space to a scheme $S$. If for every $x \in |X|$ with image $s = f(x) \in S$ the algebraic space $X \times_S \mathop{\rm Spec}(\mathcal{O}_{S,s})$ is a scheme, then $X$ is a scheme.

Proof. Let $x \in |X|$. It suffices to find an open neighbourhood $U$ of $s = f(x)$ such that $X \times_S U$ is a scheme. As $X \times_S \mathop{\rm Spec}(\mathcal{O}_{S, s})$ is a scheme, then, since $\mathcal{O}_{S, s} = \mathop{\rm colim}\nolimits \mathcal{O}_S(U)$ where the colimit is over affine open neighbourhoods of $s$ in $S$ we see that $$ X \times_S \mathop{\rm Spec}(\mathcal{O}_{S, s}) = \mathop{\rm lim}\nolimits X \times_S U $$ By Lemma 58.5.11 we see that $X \times_S U$ is a scheme for some $U$. $\square$

    The code snippet corresponding to this tag is a part of the file spaces-limits.tex and is located in lines 3588–3594 (see updates for more information).

    \begin{lemma}
    \label{lemma-enough-local}
    Let $f: X \to S$ be a quasi-compact and quasi-separated morphism from an
    algebraic space to a scheme $S$. If for every $x \in |X|$ with image
    $s = f(x) \in S$ the algebraic space $X \times_S \Spec(\mathcal{O}_{S,s})$
    is a scheme, then $X$ is a scheme.
    \end{lemma}
    
    \begin{proof}
    Let $x \in |X|$. It suffices to find an open neighbourhood $U$ of
    $s = f(x)$ such that $X \times_S U$ is a scheme.
    As $X \times_S \Spec(\mathcal{O}_{S, s})$ is a scheme, then, since
    $\mathcal{O}_{S, s} = \colim \mathcal{O}_S(U)$ where the colimit is
    over affine open neighbourhoods of $s$ in $S$ we see that
    $$
    X \times_S \Spec(\mathcal{O}_{S, s}) = \lim X \times_S U
    $$
    By Lemma \ref{lemma-limit-is-scheme} we see that $X \times_S U$
    is a scheme for some $U$.
    \end{proof}

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