Example 64.14.2. 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

where $\Delta ' = (1, \sigma ) : U \to U \times _ S U$. Take

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 64.10.5. Here are some properties of $X$ (some of which will not make sense until later):

$X \to S$ is an isomorphism over $S \setminus \{ 0_ S\} $,

the morphism $X \to S$ is étale (see Properties of Spaces, Definition 65.16.2)

the fibre $0_ X$ of $X \to S$ over $0_ S$ is isomorphic to $\mathop{\mathrm{Spec}}(k') = 0_ U$,

$X$ is not a scheme because if it were, 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,

$X$ is not separated, but it is locally separated and quasi-separated,

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 26.14.3), and

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.

## Comments (2)

Comment #7684 by Mark on

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