## 63.14 Examples of algebraic spaces

In this section we construct some examples of algebraic spaces. Some of these were suggested by B. Conrad. Since we do not yet have a lot of theory at our disposal the discussion is a bit awkward in some places.

Example 63.14.1. 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 63.10.5. 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.

Example 63.14.2. 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 63.10.5. 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 64.16.2)

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 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,

$X$ 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.

See also Examples, Lemma 108.64.1, which shows that descent data need not be effective even for a projective morphism of schemes. That example gives a smooth separated algebraic space of dimension 3 over ${\mathbf C}$ which is not a scheme.

We will use the following lemma as a convenient way to construct algebraic spaces as quotients of schemes by free group actions.

Lemma 63.14.3. Let $U \to S$ be a morphism of $\mathit{Sch}_{fppf}$. Let $G$ be an abstract group. Let $G \to \text{Aut}_ S(U)$ be a group homomorphism. Assume

Then

\[ j : R = \coprod \nolimits _{g \in G} U \longrightarrow U \times _ S U, \quad (g, x) \longmapsto (g(x), x) \]

is an étale equivalence relation and hence

\[ F = U/R \]

is an algebraic space by Theorem 63.10.5.

**Proof.**
In the statement of the lemma the symbol $\text{Aut}_ S(U)$ denotes the group of automorphisms of $U$ over $S$. Assume $(*)$ holds. Let us show that

\[ j : R = \coprod \nolimits _{g \in G} U \longrightarrow U \times _ S U, \quad (g, x) \longmapsto (g(x), x) \]

is a monomorphism. This signifies that if $T$ is a nonempty scheme, and $h : T \to U$ is a $T$-valued point such that $g \circ h = g' \circ h$ then $g = g'$. Suppose $T \not= \emptyset $, $h : T \to U$ and $g \circ h = g' \circ h$. Let $t \in T$. Consider the composition $\mathop{\mathrm{Spec}}(\kappa (t)) \to \mathop{\mathrm{Spec}}(\kappa (h(t))) \to U$. Then we conclude that $g^{-1} \circ g'$ fixes $u = h(t)$ and acts as the identity on its residue field. Hence $g = g'$ by $(*)$.

Thus if $(*)$ holds we see that $j$ is a relation (see Groupoids, Definition 39.3.1). Moreover, it is an equivalence relation since on $T$-valued points for a connected scheme $T$ we see that $R(T) = G \times U(T) \to U(T) \times U(T)$ (recall that we always work over $S$). Moreover, the morphisms $s, t : R \to U$ are étale since $R$ is a disjoint product of copies of $U$. This proves that $j : R \to U \times _ S U$ is an étale equivalence relation.
$\square$

Given a scheme $U$ and an action of a group $G$ on $U$ we say the action of $G$ on $U$ is *free* if condition $(*)$ of Lemma 63.14.3 holds. This is equivalent to the notion of a free action of the constant group scheme $G_ S$ on $U$ as defined in Groupoids, Definition 39.10.2. The lemma can be interpreted as saying that quotients of schemes by free actions of groups exist in the category of algebraic spaces.

Definition 63.14.4. Notation $U \to S$, $G$, $R$ as in Lemma 63.14.3. If the action of $G$ on $U$ satisfies $(*)$ we say $G$ *acts freely* on the scheme $U$. In this case the algebraic space $U/R$ is denoted $U/G$ and is called the *quotient of $U$ by $G$*.

This notation is consistent with the notation $U/G$ introduced in Groupoids, Definition 39.20.1. We will later make sense of the quotient as an algebraic stack without any assumptions on the action whatsoever; when we do this we will use the notation $[U/G]$. Before we discuss the examples we prove some more lemmas to facilitate the discussion. Here is a lemma discussing the various separation conditions for this quotient when $G$ is finite.

Lemma 63.14.5. Notation and assumptions as in Lemma 63.14.3. Assume $G$ is finite. Then

if $U \to S$ is quasi-separated, then $U/G$ is quasi-separated over $S$, and

if $U \to S$ is separated, then $U/G$ is separated over $S$.

**Proof.**
In the proof of Lemma 63.13.1 we saw that it suffices to prove the corresponding properties for the morphism $j : R \to U \times _ S U$. If $U \to S$ is quasi-separated, then for every affine open $V \subset U$ which maps into an affine of $S$ the opens $g(V) \cap V$ are quasi-compact. It follows that $j$ is quasi-compact. If $U \to S$ is separated, the diagonal $\Delta _{U/S}$ is a closed immersion. Hence $j : R \to U \times _ S U$ is a finite coproduct of closed immersions with disjoint images. Hence $j$ is a closed immersion.
$\square$

Lemma 63.14.6. Notation and assumptions as in Lemma 63.14.3. If $\mathop{\mathrm{Spec}}(k) \to U/G$ is a morphism, then there exist

a finite Galois extension $k'/k$,

a finite subgroup $H \subset G$,

an isomorphism $H \to \text{Gal}(k'/k)$, and

an $H$-equivariant morphism $\mathop{\mathrm{Spec}}(k') \to U$.

Conversely, such data determine a morphism $\mathop{\mathrm{Spec}}(k) \to U/G$.

**Proof.**
Consider the fibre product $V = \mathop{\mathrm{Spec}}(k) \times _{U/G} U$. Here is a diagram

\[ \xymatrix{ V \ar[r] \ar[d] & U \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & U/G } \]

Then $V$ is a nonempty scheme étale over $\mathop{\mathrm{Spec}}(k)$ and hence is a disjoint union $V = \coprod _{i \in I} \mathop{\mathrm{Spec}}(k_ i)$ of spectra of fields $k_ i$ finite separable over $k$ (Morphisms, Lemma 29.36.7). We have

\begin{align*} V \times _{\mathop{\mathrm{Spec}}(k)} V & = (\mathop{\mathrm{Spec}}(k) \times _{U/G} U) \times _{\mathop{\mathrm{Spec}}(k)}(\mathop{\mathrm{Spec}}(k) \times _{U/G} U) \\ & = \mathop{\mathrm{Spec}}(k) \times _{U/G} U \times _{U/G} U \\ & = \mathop{\mathrm{Spec}}(k) \times _{U/G} U \times G \\ & = V \times G \end{align*}

The action of $G$ on $U$ induces an action of $a : G \times V \to V$. The displayed equality means that $G \times V \to V \times _{\mathop{\mathrm{Spec}}(k)} V$, $(g, v) \mapsto (a(g, v), v)$ is an isomorphism. In particular we see that for every $i$ we have an isomorphism $H_ i \times \mathop{\mathrm{Spec}}(k_ i) \to \mathop{\mathrm{Spec}}(k_ i \otimes _ k k_ i)$ where $H_ i \subset G$ is the subgroup of elements fixing $i \in I$. Thus $H_ i$ is finite and is the Galois group of $k_ i/k$. We omit the converse construction.
$\square$

It follows from this lemma for example that if $k'/k$ is a finite Galois extension, then $\mathop{\mathrm{Spec}}(k')/\text{Gal}(k'/k) \cong \mathop{\mathrm{Spec}}(k)$. What happens if the extension is infinite? Here is an example.

Example 63.14.7. Let $S = \mathop{\mathrm{Spec}}(\mathbf{Q})$. Let $U = \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$. Let $G = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with obvious action on $U$. Then by construction property $(*)$ of Lemma 63.14.3 holds and we obtain an algebraic space

\[ X = \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})/G \longrightarrow S = \mathop{\mathrm{Spec}}(\mathbf{Q}). \]

Of course this is totally ridiculous as an approximation of $S$! Namely, by the Artin-Schreier theorem, see [Theorem 17, page 316, JacobsonIII], the only finite subgroups of $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ are $\{ 1\} $ and the conjugates of the order two group $\text{Gal}(\overline{\mathbf{Q}}/\overline{\mathbf{Q}} \cap \mathbf{R})$. Hence, if $\mathop{\mathrm{Spec}}(k) \to X$ is a morphism with $k$ algebraic over $\mathbf{Q}$, then it follows from Lemma 63.14.6 and the theorem just mentioned that either $k$ is $\overline{\mathbf{Q}}$ or isomorphic to $\overline{\mathbf{Q}} \cap \mathbf{R}$.

What is wrong with the example above is that the Galois group comes equipped with a topology, and this should somehow be part of any construction of a quotient of $\mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$. The following example is much more reasonable in my opinion and may actually occur in “nature”.

Example 63.14.8. Let $k$ be a field of characteristic zero. Let $U = \mathbf{A}^1_ k$ and let $G = \mathbf{Z}$. As action we take $n(x) = x + n$, i.e., the action of $\mathbf{Z}$ on the affine line by translation. The only fixed point is the generic point and it is clearly the case that $\mathbf{Z}$ injects into the automorphism group of the field $k(x)$. (This is where we use the characteristic zero assumption.) Consider the morphism

\[ \gamma : \mathop{\mathrm{Spec}}(k(x)) \longrightarrow X = \mathbf{A}^1_ k/\mathbf{Z} \]

of the generic point of the affine line into the quotient. We claim that this morphism does not factor through any monomorphism $\mathop{\mathrm{Spec}}(L) \to X$ of the spectrum of a field to $X$. (Contrary to what happens for schemes, see Schemes, Section 26.13.) In fact, since $\mathbf{Z}$ does not have any nontrivial finite subgroups we see from Lemma 63.14.6 that for any such factorization $k(x) = L$. Finally, $\gamma $ is not a monomorphism since

\[ \mathop{\mathrm{Spec}}(k(x)) \times _{\gamma , X, \gamma } \mathop{\mathrm{Spec}}(k(x)) \cong \mathop{\mathrm{Spec}}(k(x)) \times \mathbf{Z}. \]

This example suggests that in order to define points of an algebraic space $X$ we should consider equivalence classes of morphisms from spectra of fields into $X$ and not the set of monomorphisms from spectra of fields.

We finish with a truly awful example.

Example 63.14.9. Let $k$ be a field. Let $A = \prod _{n \in \mathbf{N}} k$ be the infinite product. Set $U = \mathop{\mathrm{Spec}}(A)$ seen as a scheme over $S = \mathop{\mathrm{Spec}}(k)$. Note that the projection maps $\text{pr}_ n : A \to k$ define open and closed immersions $f_ n : S \to U$. Set

\[ R = U \amalg \coprod \nolimits _{(n, m) \in \mathbf{N}^2, \ n \not= m} S \]

with morphism $j$ equal to $\Delta _{U/S}$ on the component $U$ and $j = (f_ n, f_ m)$ on the component $S$ corresponding to $(n, m)$. It is clear from the remark above that $s, t$ are étale. It is also clear that $j$ is an equivalence relation. Hence we obtain an algebraic space

\[ X = U/R. \]

To see what this means we specialize to the case where the field $k$ is finite with $q$ elements. Let us first discuss the topological space $|U|$ associated to the scheme $U$ a little bit. All elements of $A$ satisfy $x^ q = x$. Hence every residue field of $A$ is isomorphic to $k$, and all points of $U$ are closed. But the topology on $U$ isn't the discrete topology. Let $u_ n \in |U|$ be the point corresponding to $f_ n$. As mentioned above the points $u_ n$ are the open points (and hence isolated). This implies there have to be other points since we know $U$ is quasi-compact, see Algebra, Lemma 10.17.10 (hence not equal to an infinite discrete set). Another way to see this is because the (proper) ideal

\[ I = \{ x = (x_ n) \in A \mid \text{all but a finite number of }x_ n\text{ are zero}\} \]

is contained in a maximal ideal. Note also that every element $x$ of $A$ is of the form $x = ue$ where $u$ is a unit and $e$ is an idempotent. Hence a basis for the topology of $A$ consists of open and closed subsets (see Algebra, Lemma 10.21.1.) So the topology on $|U|$ is totally disconnected, but nontrivial. Finally, note that $\{ u_ n\} $ is dense in $|U|$.

We will later define a topological space $|X|$ associated to $X$, see Properties of Spaces, Section 64.4. What can we say about $|X|$? It turns out that the map $|U| \to |X|$ is surjective and continuous. All the points $u_ n$ map to the same point $x_0$ of $|X|$, and none of the other points get identified. Since $\{ u_ n\} $ is dense in $|U|$ we conclude that the closure of $x_0$ in $|X|$ is $|X|$. In other words $|X|$ is irreducible and $x_0$ is a generic point of $|X|$. This seems bizarre since also $x_0$ is the image of a section $S \to X$ of the structure morphism $X \to S$ (and in the case of schemes this would imply it was a closed point, see Morphisms, Lemma 29.20.2).

Whatever you think is actually going on in this example, it certainly shows that some care has to be exercised when defining irreducible components, connectedness, etc of algebraic spaces.

## Comments (7)

Comment #1408 by yogesh more on

Comment #1416 by Johan on

Comment #1420 by yogesh more on

Comment #1434 by Johan on

Comment #1447 by yogesh more on

Comment #2476 by Jiayu Zhao on

Comment #2508 by Johan on