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Tag 02Z1

Chapter 53: Algebraic Spaces > Section 53.14: Examples of algebraic spaces

Example 53.14.1. Let $k$ be a field of characteristic $\not = 2$. Let $U = \mathbf{A}^1_k$. Set $$ j : R = \Delta \amalg \Gamma \longrightarrow U \times_k U $$ where $\Delta = \{(x, x) \mid x \in \mathbf{A}^1_k\}$ and $\Gamma = \{(x, -x) \mid x \in \mathbf{A}^1_k, x \not = 0\}$. It is clear that $s, t : R \to U$ are étale, and hence $j$ is an étale equivalence relation. The quotient $X = U/R$ is an algebraic space by Theorem 53.10.5. Since $R$ is quasi-compact we see that $X$ is quasi-separated. On the other hand, $X$ is not locally separated because the morphism $j$ is not an immersion.

    The code snippet corresponding to this tag is a part of the file spaces.tex and is located in lines 2140–2155 (see updates for more information).

    \begin{example}
    \label{example-affine-line-involution}
    Let $k$ be a field of characteristic $\not = 2$. Let $U = \mathbf{A}^1_k$. Set
    $$
    j : R = \Delta \amalg \Gamma \longrightarrow U \times_k U
    $$
    where $\Delta = \{(x, x) \mid x \in \mathbf{A}^1_k\}$ and
    $\Gamma = \{(x, -x) \mid x \in \mathbf{A}^1_k, x \not = 0\}$.
    It is clear that $s, t : R \to U$ are \'etale, and hence
    $j$ is an \'etale equivalence relation.
    The quotient $X = U/R$ is an algebraic space by
    Theorem \ref{theorem-presentation}.
    Since $R$ is quasi-compact we see that $X$ is quasi-separated.
    On the other hand, $X$ is not locally separated because
    the morphism $j$ is not an immersion.
    \end{example}

    Comments (2)

    Comment #577 by Yogesh on May 19, 2014 a 3:52 pm UTC

    Since this is the first listed example of an algebraic space, can you add some detail (or links) on how it compares to the scheme $\mathbf{A}^1_k/\mu_2 = Spec (k[x^2])\simeq \mathbf{A}^1_k$, where $\mu_2=\{\pm 1 \}$ acts on $\mathbf{A}^1_k$ by $-1 \cdot x=-x$. I think there is a map from one to the other, but I'm not sure which way is goes.

    Comment #593 by Johan (site) on May 23, 2014 a 8:03 pm UTC

    There is a map $X \to \text{Spec}(k[x^2])$ which is bijective but not an isomorphism. This is discussed in Remark 05VS and Example 05Z6.

    There are also 7 comments on Section 53.14: Algebraic Spaces.

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