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Tag 03FN

Chapter 56: Algebraic Spaces > Section 56.14: Examples of algebraic spaces

Example 56.14.2. Let $k$ be a field. Let $k'/k$ be a degree $2$ Galois extension with $\text{Gal}(k'/k) = \{1, \sigma\}$. Let $S = \mathop{\rm Spec}(k[x])$ and $U = \mathop{\rm Spec}(k'[x])$. Note that $$ U \times_S U = \mathop{\rm Spec}((k' \otimes_k k')[x]) = \Delta(U) \amalg \Delta'(U) $$ where $\Delta' = (1, \sigma) : U \to U \times_S U$. Take $$ R = \Delta(U) \amalg \Delta'(U \setminus \{0_U\}) $$ where $0_U \in U$ denotes the $k'$-rational point whose $x$-coordinate is zero. It is easy to see that $R$ is an étale equivalence relation on $U$ over $S$ and hence $X = U/R$ is an algebraic space by Theorem 56.10.5. Here are some properties of $X$ (some of which will not make sense until later):

  1. $X \to S$ is an isomorphism over $S \setminus \{0_S\}$,
  2. the morphism $X \to S$ is étale (see Properties of Spaces, Definition 57.15.2)
  3. the fibre $0_X$ of $X \to S$ over $0_S$ is isomorphic to $\mathop{\rm Spec}(k') = 0_U$,
  4. $X$ is not a scheme because if it where, then $\mathcal{O}_{X, 0_X}$ would be a local domain $(\mathcal{O}, \mathfrak m, \kappa)$ with fraction field $k(x)$, with $x \in \mathfrak m$ and residue field $\kappa = k'$ which is impossible,
  5. $X$ is not separated, but it is locally separated and quasi-separated,
  6. there exists a surjective, finite, étale morphism $S' \to S$ such that the base change $X' = S' \times_S X$ is a scheme (namely, if we base change to $S' = \mathop{\rm Spec}(k'[x])$ then $U$ splits into two copies of $S'$ and $X'$ becomes isomorphic to the affine line with $0$ doubled, see Schemes, Example 25.14.3), and
  7. if we think of $X$ as a finite type algebraic space over $\mathop{\rm Spec}(k)$, then similarly the base change $X_{k'}$ is a scheme but $X$ is not a scheme.

In particular, this gives an example of a descent datum for schemes relative to the covering $\{\mathop{\rm Spec}(k') \to \mathop{\rm Spec}(k)\}$ which is not effective.

    The code snippet corresponding to this tag is a part of the file spaces.tex and is located in lines 2157–2202 (see updates for more information).

    \begin{example}
    \label{example-non-representable-descent}
    Let $k$ be a field. Let $k'/k$ be a degree $2$ Galois extension
    with $\text{Gal}(k'/k) = \{1, \sigma\}$. Let $S = \Spec(k[x])$
    and $U = \Spec(k'[x])$. Note that
    $$
    U \times_S U =
    \Spec((k' \otimes_k k')[x]) =
    \Delta(U) \amalg \Delta'(U)
    $$
    where $\Delta' = (1, \sigma) : U \to U \times_S U$. Take
    $$
    R = \Delta(U) \amalg \Delta'(U \setminus \{0_U\})
    $$
    where $0_U \in U$ denotes the $k'$-rational point whose $x$-coordinate is zero.
    It is easy to see that $R$ is an \'etale equivalence relation on $U$ over $S$
    and hence $X = U/R$ is an algebraic space by
    Theorem \ref{theorem-presentation}. Here are some properties of $X$ (some
    of which will not make sense until later):
    \begin{enumerate}
    \item $X \to S$ is an isomorphism over $S \setminus \{0_S\}$,
    \item the morphism $X \to S$ is \'etale (see
    Properties of Spaces,
    Definition \ref{spaces-properties-definition-etale})
    \item the fibre $0_X$ of $X \to S$ over $0_S$ is isomorphic to
    $\Spec(k') = 0_U$,
    \item $X$ is not a scheme because if it where, then $\mathcal{O}_{X, 0_X}$
    would be a local domain $(\mathcal{O}, \mathfrak m, \kappa)$ with
    fraction field $k(x)$, with $x \in \mathfrak m$ and residue field
    $\kappa = k'$ which is impossible,
    \item $X$ is not separated, but it is
    locally separated and quasi-separated,
    \item there exists a surjective, finite, \'etale morphism $S' \to S$
    such that the base change $X' = S' \times_S X$ is a scheme (namely, if
    we base change to $S' = \Spec(k'[x])$ then $U$ splits into
    two copies of $S'$ and $X'$ becomes isomorphic to the affine line with
    $0$ doubled, see
    Schemes, Example \ref{schemes-example-affine-space-zero-doubled}), and
    \item if we think of $X$ as a finite type algebraic space over
    $\Spec(k)$, then similarly the base change $X_{k'}$ is a scheme
    but $X$ is not a scheme.
    \end{enumerate}
    In particular, this gives an example of a descent datum for schemes
    relative to the covering $\{\Spec(k') \to \Spec(k)\}$
    which is not effective.
    \end{example}

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