# The Stacks Project

## Tag 03FN

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 $S = \mathop{\mathrm{Spec}}(k[x])$ and $U = \mathop{\mathrm{Spec}}(k'[x])$. Note that $$U \times_S U = \mathop{\mathrm{Spec}}((k' \otimes_k k')[x]) = \Delta(U) \amalg \Delta'(U)$$ 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 57.16.2)
3. the fibre $0_X$ of $X \to S$ over $0_S$ is isomorphic to $\mathop{\mathrm{Spec}}(k') = 0_U$,
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 $S' = \mathop{\mathrm{Spec}}(k'[x])$ 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 $\mathop{\mathrm{Spec}}(k)$, 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 $\{\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)\}$ 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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