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 \input{preamble} % OK, start here. % \begin{document} \title{Quotients of Groupoids} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent This chapter is devoted to generalities concerning groupoids and their quotients (as far as they exist). There is a lot of literature on this subject, see for example \cite{GIT}, \cite{seshadri_quotients}, \cite{KollarQuotients}, \cite{K-M}, \cite{KollarFinite} and many more. \section{Conventions and notation} \label{section-conventions-notation} \noindent In this chapter the conventions and notation are those introduced in Groupoids in Spaces, Sections \ref{spaces-groupoids-section-conventions} and \ref{spaces-groupoids-section-notation}. \section{Invariant morphisms} \label{section-invariant} \begin{definition} \label{definition-invariant} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j = (t, s) : R \to U \times_B U$ be a pre-relation of algebraic spaces over $B$. We say a morphism $\phi : U \to X$ of algebraic spaces over $B$ is {\it $R$-invariant} if the diagram $$\xymatrix{ R \ar[r]_s \ar[d]_t & U \ar[d]^\phi \\ U \ar[r]^\phi & X }$$ is commutative. If $j : R \to U \times_B U$ comes from the action of a group algebraic space $G$ on $U$ over $B$ as in Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-groupoid-from-action}, then we say that $\phi$ is {\it $G$-invariant}. \end{definition} \noindent In other words, a morphism $U \to X$ is $R$-invariant if it equalizes $s$ and $t$. We can reformulate this in terms of associated quotient sheaves as follows. \begin{lemma} \label{lemma-invariant} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j = (t, s) : R \to U \times_B U$ be a pre-relation of algebraic spaces over $B$. A morphism of algebraic spaces $\phi : U \to X$ is $R$-invariant if and only if it factors as $U \to U/R \to X$. \end{lemma} \begin{proof} This is clear from the definition of the quotient sheaf in Groupoids in Spaces, Section \ref{spaces-groupoids-section-quotient-sheaves}. \end{proof} \begin{lemma} \label{lemma-pullback-invariant} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j = (t, s) : R \to U \times_B U$ be a pre-relation of algebraic spaces over $B$. Let $U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. Let $X' \to X$ be any morphism of algebraic spaces. \begin{enumerate} \item Setting $U' = X' \times_X U$, $R' = X' \times_X R$ we obtain a pre-relation $j' : R' \to U' \times_B U'$. \item The pre-relation $j' : R' \to U' \times_B U'$ is the restriction of $j : R \to U \times_B U$ via $U' \to U$, see Groupoids in Spaces, Definition \ref{spaces-groupoids-definition-restrict-relation}. \item If $j$ is a relation, then $j'$ is a relation. \item If $j$ is a pre-equivalence relation, then $j'$ is a pre-equivalence relation. \item If $j$ is an equivalence relation, then $j'$ is an equivalence relation. \item If $j$ comes from a groupoid in algebraic spaces $(U, R, s, t, c)$ over $B$, then $j'$ comes from the restriction of this groupoid to $U'$, see Groupoids in Spaces, Definition \ref{spaces-groupoids-definition-restrict-groupoid}. \item If $j$ comes from the action of a group algebraic space $G/B$ on $U$ as in Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-groupoid-from-action} then $j'$ comes from the induced action of $G$ on $U'$. \end{enumerate} \end{lemma} \begin{proof} Omitted. Hint: Functorial point of view combined with the picture: $$\xymatrix{ R' = X' \times_X R \ar[dd] \ar[rr] \ar[rd] & & X' \times_X U = U' \ar'[d][dd] \ar[rd] \\ & R \ar[dd] \ar[rr] & & U \ar[dd] \\ U' = X' \times_X U \ar'[r][rr] \ar[rd] & & X' \ar[rd] \\ & U \ar[rr] & & X }$$ \end{proof} \begin{definition} \label{definition-pullback} In the situation of Lemma \ref{lemma-pullback-invariant} we call $j' : R' \to U' \times_B U'$ the {\it pullback} of the pre-relation $j$ to $X'$. We say it is a {\it flat pullback} if $X' \to X$ is a flat morphism of algebraic spaces. \end{definition} \section{Categorical quotients} \label{section-categorical} \noindent This is the most basic kind of quotient one can consider. \begin{definition} \label{definition-categorical} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j = (t, s) : R \to U \times_B U$ be pre-relation in algebraic spaces over $B$. \begin{enumerate} \item We say a morphism $\phi : U \to X$ of algebraic spaces over $B$ is a {\it categorical quotient} if it is $R$-invariant, and for every $R$-invariant morphism $\psi : U \to Y$ of algebraic spaces over $B$ there exists a unique morphism $\chi : X \to Y$ such that $\psi = \phi \circ \chi$. \item Let $\mathcal{C}$ be a full subcategory of the category of algebraic spaces over $B$. Assume $U$, $R$ are objects of $\mathcal{C}$. In this situation we say a morphism $\phi : U \to X$ of algebraic spaces over $B$ is a {\it categorical quotient in $\mathcal{C}$} if $X \in \Ob(\mathcal{C})$, and $\phi$ is $R$-invariant, and for every $R$-invariant morphism $\psi : U \to Y$ with $Y \in \Ob(\mathcal{C})$ there exists a unique morphism $\chi : X \to Y$ such that $\psi = \phi \circ \chi$. \item If $B = S$ and $\mathcal{C}$ is the category of schemes over $S$, then we say $U \to X$ is a {\it categorical quotient in the category of schemes}, or simply a {\it categorical quotient in schemes}. \end{enumerate} \end{definition} \noindent We often single out a category $\mathcal{C}$ of algebraic spaces over $B$ by some separation axiom, see Example \ref{example-categories} for some standard cases. Note that if $\phi : U \to X$ is a categorical quotient if and only if $U \to X$ is a coequalizer for the morphisms $t, s : R \to U$ in the category. Hence we immediately deduce the following lemma. \begin{lemma} \label{lemma-categorical-unique} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation in algebraic spaces over $B$. If a categorical quotient in the category of algebraic spaces over $B$ exists, then it is unique up to unique isomorphism. Similarly for categorical quotients in full subcategories of $\textit{Spaces}/B$. \end{lemma} \begin{proof} See Categories, Section \ref{categories-section-coequalizers}. \end{proof} \begin{example} \label{example-categories} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Here are some standard examples of categories $\mathcal{C}$ that we often come up when applying Definition \ref{definition-categorical}: \begin{enumerate} \item $\mathcal{C}$ is the category of all algebraic spaces over $B$, \item $B$ is separated and $\mathcal{C}$ is the category of all separated algebraic spaces over $B$, \item $B$ is quasi-separated and $\mathcal{C}$ is the category of all quasi-separated algebraic spaces over $B$, \item $B$ is locally separated and $\mathcal{C}$ is the category of all locally separated algebraic spaces over $B$, \item $B$ is decent and $\mathcal{C}$ is the category of all decent algebraic spaces over $B$, and \item $S = B$ and $\mathcal{C}$ is the category of schemes over $S$. \end{enumerate} In this case, if $\phi : U \to X$ is a categorical quotient then we say $U \to X$ is (1) a {\it categorical quotient}, (2) a {\it categorical quotient in separated algebraic spaces}, (3) a {\it categorical quotient in quasi-separated algebraic spaces}, (4) a {\it categorical quotient in locally separated algebraic spaces}, (5) a {\it categorical quotient in decent algebraic spaces}, (6) a {\it categorical quotient in schemes}. \end{example} \begin{definition} \label{definition-universal-categorical} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $\mathcal{C}$ be a full subcategory of the category of algebraic spaces over $B$ closed under fibre products. Let $j = (t, s) : R \to U \times_B U$ be pre-relation in $\mathcal{C}$, and let $U \to X$ be an $R$-invariant morphism with $X \in \Ob(\mathcal{C})$. \begin{enumerate} \item We say $U \to X$ is a {\it universal categorical quotient} in $\mathcal{C}$ if for every morphism $X' \to X$ in $\mathcal{C}$ the morphism $U' = X' \times_X U \to X'$ is the categorical quotient in $\mathcal{C}$ of the pullback $j' : R' \to U'$ of $j$. \item We say $U \to X$ is a {\it uniform categorical quotient} in $\mathcal{C}$ if for every flat morphism $X' \to X$ in $\mathcal{C}$ the morphism $U' = X' \times_X U \to X'$ is the categorical quotient in $\mathcal{C}$ of the pullback $j' : R' \to U'$ of $j$. \end{enumerate} \end{definition} \begin{lemma} \label{lemma-categorical-reduced} In the situation of Definition \ref{definition-categorical}. If $\phi : U \to X$ is a categorical quotient and $U$ is reduced, then $X$ is reduced. The same holds for categorical quotients in a category of spaces $\mathcal{C}$ listed in Example \ref{example-categories}. \end{lemma} \begin{proof} Let $X_{red}$ be the reduction of the algebraic space $X$. Since $U$ is reduced the morphism $\phi : U \to X$ factors through $i : X_{red} \to X$ (insert future reference here). Denote this morphism by $\phi_{red} : U \to X_{red}$. Since $\phi \circ s = \phi \circ t$ we see that also $\phi_{red} \circ s = \phi_{red} \circ t$ (as $i : X_{red} \to X$ is a monomorphism). Hence by the universal property of $\phi$ there exists a morphism $\chi : X \to X_{red}$ such that $\phi_{red} = \phi \circ \chi$. By uniqueness we see that $i \circ \chi = \text{id}_X$ and $\chi \circ i = \text{id}_{X_{red}}$. Hence $i$ is an isomorphism and $X$ is reduced. \medskip\noindent To show that this argument works in a category $\mathcal{C}$ one just needs to show that the reduction of an object of $\mathcal{C}$ is an object of $\mathcal{C}$. We omit the verification that this holds for each of the standard examples. \end{proof} \section{Quotients as orbit spaces} \label{section-orbits} \noindent Let $j = (t, s) : R \to U \times_B U$ be a pre-relation. If $j$ is a pre-equivalence relation, then loosely speaking the orbits'' of $R$ on $U$ are the subsets $t(s^{-1}(\{u\}))$ of $U$. However, if $j$ is just a pre-relation, then we need to take the equivalence relation generated by $R$. \begin{definition} \label{definition-orbit} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation over $B$. If $u \in |U|$, then the {\it orbit}, or more precisely the {\it $R$-orbit} of $u$ is $$O_u = \left\{ u' \in |U|\ : \begin{matrix} \exists n \geq 1, \ \exists u_0, \ldots, u_n \in |U|\text{ such that} \\ \text{for all }i \in \{0, \ldots, n - 1\}\text{ either } \\ u_i = u_{i + 1}\text{ or } \\ \exists r \in |R|, \ s(r) = u_i, t(r) = u_{i + 1} \text{ or } \\ \exists r \in |R|, \ t(r) = u_i, s(r) = u_{i + 1} \end{matrix} \right\}$$ \end{definition} \noindent It is clear that these are the equivalence classes of an equivalence relation, i.e., we have $u' \in O_u$ if and only if $u \in O_{u'}$. The following lemma is a reformulation of Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-pre-equivalence-equivalence-relation-points}. \begin{lemma} \label{lemma-pre-equivalence-equivalence-relation-points} Let $B \to S$ as in Section \ref{section-conventions-notation}. Let $j : R \to U \times_B U$ be a pre-equivalence relation of algebraic spaces over $B$. Then $$O_u = \{u' \in |U| \text{ such that } \exists r \in |R|, \ s(r) = u, \ t(r) = u'\}.$$ \end{lemma} \begin{proof} By the aforementioned Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-pre-equivalence-equivalence-relation-points} we see that the orbits $O_u$ as defined in the lemma give a disjoint union decomposition of $|U|$. Thus we see they are equal to the orbits as defined in Definition \ref{definition-orbit}. \end{proof} \begin{lemma} \label{lemma-invariant-map-constant-on-orbit} In the situation of Definition \ref{definition-orbit}. Let $\phi : U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. Then $|\phi| : |U| \to |X|$ is constant on the orbits. \end{lemma} \begin{proof} To see this we just have to show that $\phi(u) = \phi(u')$ for all $u, u' \in |U|$ such that there exists an $r \in |R|$ such that $s(r) = u$ and $t(r) = u'$. And this is clear since $\phi$ equalizes $s$ and $t$. \end{proof} \noindent There are several problems with considering the orbits $O_u \subset |U|$ as a tool for singling out properties of quotient maps. One issue is the following. Suppose that $\Spec(k) \to B$ is a geometric point of $B$. Consider the canonical map $$U(k) \longrightarrow |U|.$$ Then it is usually not the case that the equivalence classes of the equivalence relation generated by $j(R(k)) \subset U(k) \times U(k)$ are the inverse images of the orbits $O_u \subset |U|$. A silly example is to take $S = B = \Spec(\mathbf{Z})$, $U = R = \Spec(k)$ with $s = t = \text{id}_k$. Then $|U| = |R|$ is a single point but $U(k)/R(k)$ is enormous. A more interesting example is to take $S = B = \Spec(\mathbf{Q})$, choose some of number fields $K \subset L$, and set $U = \Spec(L)$ and $R = \Spec(L \otimes_K L)$ with obvious maps $s, t : R \to U$. In this case $|U|$ still has just one point, but the quotient $$U(k)/R(k) = \Hom(K, k)$$ consists of more than one element. We conclude from both examples that if $U \to X$ is an $R$-invariant map and if we want it to separate orbits'' we get a much stronger and interesting notion by considering the induced maps $U(k) \to X(k)$ and ask that those maps separate orbits. \medskip\noindent There is an issue with this too. Namely, suppose that $S = B = \Spec(\mathbf{R})$, $U = \Spec(\mathbf{C})$, and $R = \Spec(\mathbf{C}) \amalg \Spec(K)$ for some field extension $\sigma : \mathbf{C} \to K$. Let the maps $s, t$ be given by the identity on the component $\Spec(\mathbf{C})$, but by $\sigma, \sigma \circ \tau$ on the second component where $\tau$ is complex conjugation. If $K$ is a nontrivial extension of $\mathbf{C}$, then the two points $1, \tau \in U(\mathbf{C})$ are not equivalent under $j(R(\mathbf{C}))$. But after choosing an extension $\mathbf{C} \subset \Omega$ of sufficiently large cardinality (for example larger than the cardinality of $K$) then the images of $1, \tau \in U(\mathbf{C})$ in $U(\Omega)$ do become equivalent! It seems intuitively clear that this happens either because $s, t : R \to U$ are not locally of finite type or because the cardinality of the field $k$ is not large enough. \medskip\noindent Keeping this in mind we make the following definition. \begin{definition} \label{definition-geometric-orbits} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation over $B$. Let $\Spec(k) \to B$ be a geometric point of $B$. \begin{enumerate} \item We say $\overline{u}, \overline{u}' \in U(k)$ are {\it weakly $R$-equivalent} if they are in the same equivalence class for the equivalence relation generated by the relation $j(R(k)) \subset U(k) \times U(k)$. \item We say $\overline{u}, \overline{u}' \in U(k)$ are {\it $R$-equivalent} if for some overfield $k \subset \Omega$ the images in $U(\Omega)$ are weakly $R$-equivalent. \item The {\it weak orbit}, or more precisely the {\it weak $R$-orbit} of $\overline{u} \in U(k)$ is set of all elements of $U(k)$ which are weakly $R$-equivalent to $\overline{u}$. \item The {\it orbit}, or more precisely the {\it $R$-orbit} of $\overline{u} \in U(k)$ is set of all elements of $U(k)$ which are $R$-equivalent to $\overline{u}$. \end{enumerate} \end{definition} \noindent It turns out that in good cases orbits and weak orbits agree, see Lemma \ref{lemma-geometric-orbits}. The following lemma illustrates the difference in the special case of a pre-equivalence relation. \begin{lemma} \label{lemma-weak-orbit-pre-equivalence} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $\Spec(k) \to B$ be a geometric point of $B$. Let $j : R \to U \times_B U$ be a pre-equivalence relation over $B$. In this case the weak orbit of $\overline{u} \in U(k)$ is simply $$\{ \overline{u}' \in U(k) \text{ such that } \exists \overline{r} \in R(k), \ s(\overline{r}) = \overline{u}, \ t(\overline{r}) = \overline{u}' \}$$ and the orbit of $\overline{u} \in U(k)$ is $$\{ \overline{u}' \in U(k) : \exists\text{ field extension }k \subset K, \ \exists\ r \in R(K), \ s(r) = \overline{u}, \ t(r) = \overline{u}'\}$$ \end{lemma} \begin{proof} This is true because by definition of a pre-equivalence relation the image $j(R(k)) \subset U(k) \times U(k)$ is an equivalence relation. \end{proof} \noindent Let us describe the recipe for turning any pre-relation into a pre-equivalence relation. We will use the morphisms \begin{equation} \label{equation-list} \begin{matrix} j_{diag} & : & U & \longrightarrow & U \times_B U, & u & \longmapsto & (u, u) \\ j_{flip} & : & R & \longrightarrow & U \times_B U, & r & \longmapsto & (s(r), t(r)) \\ j_{comp} & : & R \times_{s, U, t} R & \longrightarrow & U \times_B U, & (r, r') & \longmapsto & (t(r), s(r')) \end{matrix} \end{equation} We define $j_1 = (t_1, s_1) : R_1 \to U \times_B U$ to be the morphism $$j \amalg j_{diag} \amalg j_{flip} : R \amalg U \amalg R \longrightarrow U \times_B U$$ with notation as in Equation (\ref{equation-list}). For $n > 1$ we set $$j_n = (t_n, s_n) : R_n = R_1 \times_{s_1, U, t_{n - 1}} R_{n - 1} \longrightarrow U \times_B U$$ where $t_n$ comes from $t_1$ precomposed with projection onto $R_1$ and $s_n$ comes from $s_{n - 1}$ precomposed with projection onto $R_{n - 1}$. Finally, we denote $$j_\infty = (t_\infty, s_\infty) : R_\infty = \coprod\nolimits_{n \geq 1} R_n \longrightarrow U \times_B U.$$ \begin{lemma} \label{lemma-make-pre-equivalence} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation over $B$. Then $j_\infty : R_\infty \to U \times_B U$ is a pre-equivalence relation over $B$. Moreover \begin{enumerate} \item $\phi : U \to X$ is $R$-invariant if and only if it is $R_\infty$-invariant, \item the canonical map of quotient sheaves $U/R \to U/R_\infty$ (see Groupoids in Spaces, Section \ref{spaces-groupoids-section-quotient-sheaves}) is an isomorphism, \item weak $R$-orbits agree with weak $R_\infty$-orbits, \item $R$-orbits agree with $R_\infty$-orbits, \item if $s, t$ are locally of finite type, then $s_\infty$, $t_\infty$ are locally of finite type, \item add more here as needed. \end{enumerate} \end{lemma} \begin{proof} Omitted. Hint for (5): Any property of $s, t$ which is stable under composition and stable under base change, and Zariski local on the source will be inherited by $s_\infty, t_\infty$. \end{proof} \begin{lemma} \label{lemma-geometric-orbits} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation over $B$. Let $\Spec(k) \to B$ be a geometric point of $B$. \begin{enumerate} \item If $s, t : R \to U$ are locally of finite type then weak $R$-equivalence on $U(k)$ agrees with $R$-equivalence, and weak $R$-orbits agree with $R$-orbits on $U(k)$. \item If $k$ has sufficiently large cardinality then weak $R$-equivalence on $U(k)$ agrees with $R$-equivalence, and weak $R$-orbits agree with $R$-orbits on $U(k)$. \end{enumerate} \end{lemma} \begin{proof} We first prove (1). Assume $s, t$ locally of finite type. By Lemma \ref{lemma-make-pre-equivalence} we may assume that $R$ is a pre-equivalence relation. Let $k$ be an algebraically closed field over $B$. Suppose $\overline{u}, \overline{u}' \in U(k)$ are $R$-equivalent. Then for some extension field $k \subset \Omega$ there exists a point $\overline{r} \in R(\Omega)$ mapping to $(\overline{u}, \overline{u}') \in (U \times_B U)(\Omega)$, see Lemma \ref{lemma-weak-orbit-pre-equivalence}. Hence $$Z = R \times_{j, U \times_B U, (\overline{u}, \overline{u}')} \Spec(k)$$ is nonempty. As $s$ is locally of finite type we see that also $j$ is locally of finite type, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type}. This implies $Z$ is a nonempty algebraic space locally of finite type over the algebraically closed field $k$ (use Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-type}). Thus $Z$ has a $k$-valued point, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-surjective-geometric-points}. Hence we conclude there exists a $\overline{r} \in R(k)$ with $j(\overline{r}) = (\overline{u}, \overline{u}')$, and we conclude that $\overline{u}, \overline{u}'$ are $R$-equivalent as desired. \medskip\noindent The proof of part (2) is the same, except that it uses Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-large-enough} instead of Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-surjective-geometric-points}. This shows that the assertion holds as soon as $|k| > \lambda(R)$ with $\lambda(R)$ as introduced just above Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-surjective-geometric-points}. \end{proof} \noindent In the following definition we use the terminology $k$ is a field over $B$'' to mean that $\Spec(k)$ comes equipped with a morphism $\Spec(k) \to B$. \begin{definition} \label{definition-set-theoretically-invariant} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation over $B$. \begin{enumerate} \item We say $\phi : U \to X$ is {\it set-theoretically $R$-invariant} if and only if the map $U(k) \to X(k)$ equalizes the two maps $s, t : R(k) \to U(k)$ for every algebraically closed field $k$ over $B$. \item We say $\phi : U \to X$ {\it separates orbits}, or {\it separates $R$-orbits} if it is set-theoretically $R$-invariant and $\phi(\overline{u}) = \phi(\overline{u}')$ in $X(k)$ implies that $\overline{u}, \overline{u}' \in U(k)$ are in the same orbit for every algebraically closed field $k$ over $B$. \end{enumerate} \end{definition} \noindent In Example \ref{example-not-invariant} we show that being set-theoretically invariant is too weak'' a notion in the category of algebraic spaces. A more geometric reformulation of what it means to be set-theoretically invariant or to separate orbits is in Lemma \ref{lemma-separates-orbits}. \begin{lemma} \label{lemma-set-theoretic-invariant} In the situation of Definition \ref{definition-set-theoretically-invariant}. A morphism $\phi : U \to X$ is set-theoretically $R$-invariant if and only if for any algebraically closed field $k$ over $B$ the map $U(k) \to X(k)$ is constant on orbits. \end{lemma} \begin{proof} This is true because the condition is supposed to hold for all algebraically closed fields over $B$. \end{proof} \begin{lemma} \label{lemma-invariant-set-theoretically-invariant} In the situation of Definition \ref{definition-set-theoretically-invariant}. An invariant morphism is set-theoretically invariant. \end{lemma} \begin{proof} This is immediate from the definitions. \end{proof} \begin{lemma} \label{lemma-set-theoretically-invariant-invariant-when-reduced} In the situation of Definition \ref{definition-set-theoretically-invariant}. Let $\phi : U \to X$ be a morphism of algebraic spaces over $B$. Assume \begin{enumerate} \item $\phi$ is set-theoretically $R$-invariant, \item $R$ is reduced, and \item $X$ is locally separated over $B$. \end{enumerate} Then $\phi$ is $R$-invariant. \end{lemma} \begin{proof} Consider the equalizer $$Z = R \times_{(\phi, \phi) \circ j, X \times_B X, \Delta_{X/B}} X$$ algebraic space. Then $Z \to R$ is an immersion by assumption (3). By assumption (1) $|Z| \to |R|$ is surjective. This implies that $Z \to R$ is a bijective closed immersion (use Schemes, Lemma \ref{schemes-lemma-immersion-when-closed}) and by assumption (2) we conclude that $Z = R$. \end{proof} \begin{example} \label{example-not-invariant} There exist reduced quasi-separated algebraic spaces $X$, $Y$ and a pair of morphisms $a, b : Y \to X$ which agree on all $k$-valued points but are not equal. To get an example take $Y = \Spec(k[[x]])$ and $$X = \mathbf{A}^1_k \Big/ \big(\Delta \amalg \{(x, -x) \mid x \not = 0\}\big)$$ the algebraic space of Spaces, Example \ref{spaces-example-affine-line-involution}. The two morphisms $a, b : Y \to X$ come from the two maps $x \mapsto x$ and $x \mapsto -x$ from $Y$ to $\mathbf{A}^1_k = \Spec(k[x])$. On the generic point the two maps are the same because on the open part $x \not = 0$ of the space $X$ the functions $x$ and $-x$ are equal. On the closed point the maps are obviously the same. It is also true that $a \not = b$. This implies that Lemma \ref{lemma-set-theoretically-invariant-invariant-when-reduced} does not hold with assumption (3) replaced by the assumption that $X$ be quasi-separated. Namely, consider the diagram $$\xymatrix{ Y \ar[d]_{-1} \ar[r]_1 & Y \ar[d]^a \\ Y \ar[r]^a & X }$$ then the composition $a \circ (-1) = b$. Hence we can set $R = Y$, $U = Y$, $s = 1$, $t = -1$, $\phi = a$ to get an example of a set-theoretically invariant morphism which is not invariant. \end{example} \noindent The example above is instructive because the map $Y \to X$ even separates orbits. It shows that in the category of algebraic spaces there are simply too many set-theoretically invariant morphisms lying around. Next, let us define what it means for $R$ to be a set-theoretic equivalence relation, while remembering that we need to allow for field extensions to make this work correctly. \begin{definition} \label{definition-set-theoretic-equivalence} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation over $B$. \begin{enumerate} \item We say $j$ is a {\it set-theoretic pre-equivalence relation} if $$\overline{u} \sim_R \overline{u}' \Leftrightarrow \begin{matrix} \exists\text{ field extension }k \subset K, \ \exists\ r \in R(K), \\ s(r) = \overline{u}, \ t(r) = \overline{u}' \end{matrix}$$ defines an equivalence relation on $U(k)$ for all algebraically closed fields $k$ over $B$. \item We say $j$ is a {\it set-theoretic equivalence relation} if $j$ is universally injective and a set-theoretic pre-equivalence relation. \end{enumerate} \end{definition} \noindent Let us reformulate this in more geometric terms. \begin{lemma} \label{lemma-set-theoretic-pre-equivalence-geometric} In the situation of Definition \ref{definition-set-theoretic-equivalence}. The following are equivalent: \begin{enumerate} \item The morphism $j$ is a set-theoretic pre-equivalence relation. \item The subset $j(|R|) \subset |U \times_B U|$ contains the image of $|j'|$ for any of the morphisms $j'$ as in Equation (\ref{equation-list}). \item For every algebraically closed field $k$ over $B$ of sufficiently large cardinality the subset $j(R(k)) \subset U(k) \times U(k)$ is an equivalence relation. \end{enumerate} If $s, t$ are locally of finite type these are also equivalent to \begin{enumerate} \item[(4)] For every algebraically closed field $k$ over $B$ the subset $j(R(k)) \subset U(k) \times U(k)$ is an equivalence relation. \end{enumerate} \end{lemma} \begin{proof} Assume (2). Let $k$ be an algebraically closed field over $B$. We are going to show that $\sim_R$ is an equivalence relation. Suppose that $\overline{u}_i : \Spec(k) \to U$, $i = 1, 2$ are $k$-valued points of $U$. Suppose that $(\overline{u}_1, \overline{u}_2)$ is the image of a $K$-valued point $r \in R(K)$. Consider the solid commutative diagram $$\xymatrix{ \Spec(K') \ar@{..>}[r] \ar@{..>}[d] & \Spec(k) \ar[d]_{(\overline{u}_2, \overline{u}_1)} & \Spec(K) \ar[d] \ar[l] \\ R \ar[r]^-j & U \times_B U & R \ar[l]_-{j_{flip}} }$$ We also denote $r \in |R|$ the image of $r$. By assumption the image of $|j_{flip}|$ is contained in the image of $|j|$, in other words there exists a $r' \in |R|$ such that $|j|(r') = |j_{flip}|(r)$. But note that $(\overline{u}_2, \overline{u}_1)$ is in the equivalence class that defines $|j|(r')$ (by the commutativity of the solid part of the diagram). This means there exists a field extension $k \subset K'$ and a morphism $r' : \Spec(K) \to R$ (abusively denoted $r'$ as well) with $j \circ r' = (\overline{u}_2, \overline{u}_1) \circ i$ where $i : \Spec(K') \to \Spec(K)$ is the obvious map. In other words the dotted part of the diagram commutes. This proves that $\sim_R$ is a symmetric relation on $U(k)$. In the similar way, using that the image of $|j_{diag}|$ is contained in the image of $|j|$ we see that $\sim_R$ is reflexive (details omitted). \medskip\noindent To show that $\sim_R$ is transitive assume given $\overline{u}_i : \Spec(k) \to U$, $i = 1, 2, 3$ and field extensions $k \subset K_i$ and points $r_i : \Spec(K_i) \to R$, $i = 1, 2$ such that $j(r_1) = (\overline{u}_1, \overline{u}_2)$ and $j(r_1) = (\overline{u}_2, \overline{u}_3)$. Then we may choose a commutative diagram of fields $$\xymatrix{ K & K_2 \ar[l] \\ K_1 \ar[u] & k \ar[l] \ar[u] }$$ and we may think of $r_1, r_2 \in R(K)$. We consider the commutative solid diagram $$\xymatrix{ \Spec(K') \ar@{..>}[r] \ar@{..>}[d] & \Spec(k) \ar[d]_{(\overline{u}_1, \overline{u}_3)} & \Spec(K) \ar[d]^{(r_1, r_2)} \ar[l] \\ R \ar[r]^-j & U \times_B U & R \times_{s, U, t} R \ar[l]_-{j_{comp}} }$$ By exactly the same reasoning as in the first part of the proof, but this time using that $|j_{comp}|((r_1, r_2))$ is in the image of $|j|$, we conclude that a field $K'$ and dotted arrows exist making the diagram commute. This proves that $\sim_R$ is transitive and concludes the proof that (2) implies (1). \medskip\noindent Assume (1) and let $k$ be an algebraically closed field over $B$ whose cardinality is larger than $\lambda(R)$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-large-enough}. Suppose that $\overline{u} \sim_R \overline{u}'$ with $\overline{u}, \overline{u}' \in U(k)$. By assumption there exists a point in $|R|$ mapping to $(\overline{u}, \overline{u}') \in |U \times_B U|$. Hence by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-large-enough} we conclude there exists an $\overline{r} \in R(k)$ with $j(\overline{r}) = (\overline{u}, \overline{u}')$. In this way we see that (1) implies (3). \medskip\noindent Assume (3). Let us show that $\Im(|j_{comp}|) \subset \Im(|j|)$. Pick any point $c \in |R \times_{s, U, t} R|$. We may represent this by a morphism $\overline{c} : \Spec(k) \to R \times_{s, U, t} R$, with $k$ over $B$ having sufficiently large cardinality. By assumption we see that $j_{comp}(\overline{c}) \in U(k) \times U(k) = (U \times_B U)(k)$ is also the image $j(\overline{r})$ for some $\overline{r} \in R(k)$. Hence $j_{comp}(c) = j(r)$ in $|U \times_B U|$ as desired (with $r \in |R|$ the equivalence class of $\overline{r}$). The same argument shows also that $\Im(|j_{diag}|) \subset \Im(|j|)$ and $\Im(|j_{flip}|) \subset \Im(|j|)$ (details omitted). In this way we see that (3) implies (2). At this point we have shown that (1), (2) and (3) are all equivalent. \medskip\noindent It is clear that (4) implies (3) (without any assumptions on $s$, $t$). To finish the proof of the lemma we show that (1) implies (4) if $s, t$ are locally of finite type. Namely, let $k$ be an algebraically closed field over $B$. Suppose that $\overline{u} \sim_R \overline{u}'$ with $\overline{u}, \overline{u}' \in U(k)$. By assumption the algebraic space $Z = R \times_{j, U \times_B U, (\overline{u}, \overline{u}')} \Spec(k)$ is nonempty. On the other hand, since $j = (t, s)$ is locally of finite type the morphism $Z \to \Spec(k)$ is locally of finite type as well (use Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-permanence-finite-type} and \ref{spaces-morphisms-lemma-base-change-finite-type}). Hence $Z$ has a $k$ point by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-surjective-geometric-points} and we conclude that $(\overline{u}, \overline{u}') \in j(R(k))$ as desired. This finishes the proof of the lemma. \end{proof} \begin{lemma} \label{lemma-set-theoretic-equivalence-geometric} In the situation of Definition \ref{definition-set-theoretic-equivalence}. The following are equivalent: \begin{enumerate} \item The morphism $j$ is a set-theoretic equivalence relation. \item The morphism $j$ is universally injective and $j(|R|) \subset |U \times_B U|$ contains the image of $|j'|$ for any of the morphisms $j'$ as in Equation (\ref{equation-list}). \item For every algebraically closed field $k$ over $B$ of sufficiently large cardinality the map $j : R(k) \to U(k) \times U(k)$ is injective and its image is an equivalence relation. \end{enumerate} If $j$ is decent, or locally separated, or quasi-separated these are also equivalent to \begin{enumerate} \item[(4)] For every algebraically closed field $k$ over $B$ the map $j : R(k) \to U(k) \times U(k)$ is injective and its image is an equivalence relation. \end{enumerate} \end{lemma} \begin{proof} The implications (1) $\Rightarrow$ (2) and (2) $\Rightarrow$ (3) follow from Lemma \ref{lemma-set-theoretic-pre-equivalence-geometric} and the definitions. The same lemma shows that (3) implies $j$ is a set-theoretic pre-equivalence relation. But of course condition (3) also implies that $j$ is universally injective, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-universally-injective}, so that $j$ is indeed a set-theoretic equivalence relation. At this point we know that (1), (2), (3) are all equivalent. \medskip\noindent Condition (4) implies (3) without any further hypotheses on $j$. Assume $j$ is decent, or locally separated, or quasi-separated and the equivalent conditions (1), (2), (3) hold. By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-when-universally-injective-radicial} we see that $j$ is radicial. Let $k$ be any algebraically closed field over $B$. Let $\overline{u}, \overline{u}' \in U(k)$ with $\overline{u} \sim_R \overline{u}'$. We see that $R \times_{U \times_B U, (\overline{u}, \overline{u}')} \Spec(k)$ is nonempty. Hence, as $j$ is radicial, its reduction is the spectrum of a field purely inseparable over $k$. As $k = \overline{k}$ we see that it is the spectrum of $k$. Whence a point $\overline{r} \in R(k)$ with $t(\overline{r}) = \overline{u}$ and $s(\overline{r}) = \overline{u}'$ as desired. \end{proof} \begin{lemma} \label{lemma-set-theoretic-equivalence} Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation over $B$. \begin{enumerate} \item If $j$ is a pre-equivalence relation, then $j$ is a set-theoretic pre-equivalence relation. This holds in particular when $j$ comes from a groupoid in algebraic spaces, or from an action of a group algebraic space on $U$. \item If $j$ is an equivalence relation, then $j$ is a set-theoretic equivalence relation. \end{enumerate} \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-separates-orbits} Let $B \to S$ be as in Section \ref{section-conventions-notation}. Let $j : R \to U \times_B U$ be a pre-relation. Let $\phi : U \to X$ be a morphism of algebraic spaces over $B$. Consider the diagram $$\xymatrix{ (U \times_X U) \times_{(U \times_B U)} R \ar[d]^q \ar[r]_-p & R \ar[d]^j \\ U \times_X U \ar[r]^c & U \times_B U }$$ Then we have: \begin{enumerate} \item The morphism $\phi$ is set-theoretically invariant if and only if $p$ is surjective. \item If $j$ is a set-theoretic pre-equivalence relation then $\phi$ separates orbits if and only if $p$ and $q$ are surjective. \item If $p$ and $q$ are surjective, then $j$ is a set-theoretic pre-equivalence relation (and $\phi$ separates orbits). \item If $\phi$ is $R$-invariant and $j$ is a set-theoretic pre-equivalence relation, then $\phi$ separates orbits if and only if the induced morphism $R \to U \times_X U$ is surjective. \end{enumerate} \end{lemma} \begin{proof} Assume $\phi$ is set-theoretically invariant. This means that for any algebraically closed field $k$ over $B$ and any $\overline{r} \in R(k)$ we have $\phi(s(\overline{r})) = \phi(t(\overline{r}))$. Hence $((\phi(t(\overline{r})), \phi(s(\overline{r}))), \overline{r})$ defines a point in the fibre product mapping to $\overline{r}$ via $p$. This shows that $p$ is surjective. Conversely, assume $p$ is surjective. Pick $\overline{r} \in R(k)$. As $p$ is surjective, we can find a field extension $k \subset K$ and a $K$-valued point $\tilde r$ of the fibre product with $p(\tilde r) = \overline{r}$. Then $q(\tilde r) \in U \times_X U$ maps to $(t(\overline{r}), s(\overline{r}))$ in $U \times_B U$ and we conclude that $\phi(s(\overline{r})) = \phi(t(\overline{r}))$. This proves that $\phi$ is set-theoretically invariant. \medskip\noindent The proofs of (2), (3), and (4) are omitted. Hint: Assume $k$ is an algebraically closed field over $B$ of large cardinality. Consider the associated diagram of sets $$\xymatrix{ (U(k) \times_{X(k)} U(k)) \times_{U(k) \times U(k)} R(k) \ar[d]^q \ar[r]_-p & R(k) \ar[d]^j \\ U(k) \times_{X(k)} U(k) \ar[r]^c & U(k) \times U(k) }$$ By the lemmas above the equivalences posed in (2), (3), and (4) become set-theoretic questions related to the diagram we just displayed, using that surjectivity translates into surjectivity on $k$-valued points by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-large-enough}. \end{proof} \noindent Because we have seen above that the notion of a set-theoretically invariant morphism is a rather weak one in the category of algebraic spaces, we define an orbit space for a pre-relation as follows. \begin{definition} \label{definition-orbit-space} Let $B \to S$ as in Section \ref{section-conventions-notation}. Let $j : R \to U \times_B U$ be a pre-relation. We say $\phi : U \to X$ is an {\it orbit space for $R$} if \begin{enumerate} \item $\phi$ is $R$-invariant, \item $\phi$ separates $R$-orbits, and \item $\phi$ is surjective. \end{enumerate} \end{definition} \noindent The definition of separating $R$-orbits involves a discussion of points with values in algebraically closed fields. But as we've seen in many cases this just corresponds to the surjectivity of certain canonically associated morphisms of algebraic spaces. We summarize some of the discussion above in the following characterization of orbit spaces. \begin{lemma} \label{lemma-orbit-space} Let $B \to S$ as in Section \ref{section-conventions-notation}. Let $j : R \to U \times_B U$ be a set-theoretic pre-equivalence relation. A morphism $\phi : U \to X$ is an orbit space for $R$ if and only if \begin{enumerate} \item $\phi \circ s = \phi \circ t$, i.e., $\phi$ is invariant, \item the induced morphism $(t, s) : R \to U \times_X U$ is surjective, and \item the morphism $\phi : U \to X$ is surjective. \end{enumerate} This characterization applies for example if $j$ is a pre-equivalence relation, or comes from a groupoid in algebraic spaces over $B$, or comes from the action of a group algebraic space over $B$ on $U$. \end{lemma} \begin{proof} Follows immediately from Lemma \ref{lemma-separates-orbits} part (4). \end{proof} \noindent In the following lemma it is (probably) not good enough to assume just that the morphisms $s, t$ are locally of finite type. The reason is that it may happen that some map $\phi : U \to X$ is an orbit space, yet is not locally of finite type. In that case $U(k) \to X(k)$ may not be surjective for all algebraically closed fields $k$ over $B$. \begin{lemma} \label{lemma-orbit-space-locally-finite-type-over-base} Let $B \to S$ as in Section \ref{section-conventions-notation}. Let $j = (t, s) : R \to U \times_B U$ be a pre-relation. Assume $R, U$ are locally of finite type over $B$. Let $\phi : U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. Then $\phi$ is an orbit space for $R$ if and only if the natural map $$U(k)/\big(\text{equivalence relation generated by }j(R(k))\big) \longrightarrow X(k)$$ is bijective for all algebraically closed fields $k$ over $B$. \end{lemma} \begin{proof} Note that since $U$, $R$ are locally of finite type over $B$ all of the morphisms $s, t, j, \phi$ are locally of finite type, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type}. We will also use without further mention Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-surjective-geometric-points}. Assume $\phi$ is an orbit space. Let $k$ be any algebraically closed field over $B$. Let $\overline{x} \in X(k)$. Consider $U \times_{\phi, X, \overline{x}} \Spec(k)$. This is a nonempty algebraic space which is locally of finite type over $k$. Hence it has a $k$-valued point. This shows the displayed map of the lemma is surjective. Suppose that $\overline{u}, \overline{u}' \in U(k)$ map to the same element of $X(k)$. By Definition \ref{definition-set-theoretically-invariant} this means that $\overline{u}, \overline{u}'$ are in the same $R$-orbit. By Lemma \ref{lemma-geometric-orbits} this means that they are equivalent under the equivalence relation generated by $j(R(k))$. Thus the displayed morphism is injective. \medskip\noindent Conversely, assume the displayed map is bijective for all algebraically closed fields $k$ over $B$. This condition clearly implies that $\phi$ is surjective. We have already assumed that $\phi$ is $R$-invariant. Finally, the injectivity of all the displayed maps implies that $\phi$ separates orbits. Hence $\phi$ is an orbit space. \end{proof} \section{Coarse quotients} \label{section-coarse} \noindent We only add this here so that we can later say that coarse quotients correspond to coarse moduli spaces (or moduli schemes). \begin{definition} \label{definition-coarse} Let $S$ be a scheme and $B$ an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation. A morphism $\phi : U \to X$ of algebraic spaces over $B$ is called a {\it coarse quotient} if \begin{enumerate} \item $\phi$ is a categorical quotient, and \item $\phi$ is an orbit space. \end{enumerate} If $S = B$, $U$, $R$ are all schemes, then we say a morphism of schemes $\phi : U \to X$ is a {\it coarse quotient in schemes} if \begin{enumerate} \item $\phi$ is a categorical quotient in schemes, and \item $\phi$ is an orbit space. \end{enumerate} \end{definition} \noindent In many situations the algebraic spaces $R$ and $U$ are locally of finite type over $B$ and the orbit space condition simply means that $$U(k)/\big(\text{equivalence relation generated by }j(R(k))\big) \cong X(k)$$ for all algebraically closed fields $k$. See Lemma \ref{lemma-orbit-space-locally-finite-type-over-base}. If $j$ is also a (set-theoretic) pre-equivalence relation, then the condition is simply equivalent to $U(k)/j(R(k)) \to X(k)$ being bijective for all algebraically closed fields $k$. \section{Topological properties} \label{section-topological} \noindent Let $S$ be a scheme and $B$ an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation. We say a subset $T \subset |U|$ is {\it $R$-invariant} if $s^{-1}(T) = t^{-1}(T)$ as subsets of $|R|$. Note that if $T$ is closed, then it may not be the case that the corresponding reduced closed subspace of $U$ is $R$-invariant (as in Groupoids in Spaces, Definition \ref{spaces-groupoids-definition-invariant-open}) because the pullbacks $s^{-1}(T)$, $t^{-1}(T)$ may not be reduced. Here are some conditions that we can consider for an invariant morphism $\phi : U \to X$. \begin{definition} \label{definition-topological} Let $S$ be a scheme and $B$ an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation. Let $\phi : U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. \begin{enumerate} \item \label{item-submersive} The morphism $\phi$ is submersive. \item \label{item-invariant-closed} For any $R$-invariant closed subset $Z \subset |U|$ the image $\phi(Z)$ is closed in $|X|$. \item \label{item-intersect-invariant-closed} Condition (\ref{item-invariant-closed}) holds and for any pair of $R$-invariant closed subsets $Z_1, Z_2 \subset |U|$ we have $$\phi(Z_1 \cap Z_2) = \phi(Z_1) \cap \phi(Z_2)$$ \item The morphism $(t, s) : R \to U \times_X U$ is universally submersive. \label{item-strong} \end{enumerate} For each of these properties we can also require them to hold after any flat pullback, or after any pullback, see Definition \ref{definition-pullback}. In this case we say condition (\ref{item-submersive}), (\ref{item-invariant-closed}), (\ref{item-intersect-invariant-closed}), or (\ref{item-strong}) holds {\it uniformly} or {\it universally}. \end{definition} \section{Invariant functions} \label{section-functions} \noindent In some cases it is convenient to pin down the structure sheaf of a quotient by requiring any invariant function to be a local section of the structure sheaf of the quotient. \begin{definition} \label{definition-functions} Let $S$ be a scheme and $B$ an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation. Let $\phi : U \to X$ be an $R$-invariant morphism. Denote $\phi' = \phi \circ s = \phi \circ t : R \to X$. \begin{enumerate} \item We denote $(\phi_*\mathcal{O}_U)^R$ the $\mathcal{O}_X$-sub-algebra of $\phi_*\mathcal{O}_U$ which is the equalizer of the two maps $$\xymatrix{ \phi_*\mathcal{O}_U \ar@<1ex>[rr]^{\phi_*s^\sharp} \ar@<-1ex>[rr]_{\phi_*t^\sharp} & & \phi'_*\mathcal{O}_R }$$ on $X_\etale$. We sometimes call this the {\it sheaf of $R$-invariant functions on $X$}. \item We say {\it the functions on $X$ are the $R$-invariant functions on $U$} if the natural map $\mathcal{O}_X \to (\phi_*\mathcal{O}_U)^R$ is an isomorphism. \end{enumerate} \end{definition} \noindent Of course we can require this property holds after any (flat or any) pullback, leading to a (uniform or) universal notion. This condition is often thrown in with other conditions in order to obtain a (more) unique quotient. And of course a good deal of motivation for the whole subject comes from the following special case: $U = \Spec(A)$ is an affine scheme over a field $S = B = \Spec(k)$ and where $R = G \times U$, with $G$ an affine group scheme over $k$. In this case you have the option of taking for the quotient: $$X = \Spec(A^G)$$ so that at least the condition of the definition above is satisfied. Even though this is a nice thing you can do it is often not the right quotient; for example if $U = \text{GL}_{n, k}$ and $G$ is the group of upper triangular matrices, then the above gives $X = \Spec(k)$, whereas a much better quotient (namely the flag variety) exists. \section{Good quotients} \label{section-good} \noindent Especially when taking quotients by group actions the following definition is useful. \begin{definition} \label{definition-good} Let $S$ be a scheme and $B$ an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation. A morphism $\phi : U \to X$ of algebraic spaces over $B$ is called a {\it good quotient} if \begin{enumerate} \item $\phi$ is invariant, \item $\phi$ is affine, \item $\phi$ is surjective, \item condition (\ref{item-intersect-invariant-closed}) holds universally, and \item the functions on $X$ are the $R$-invariant functions on $U$. \end{enumerate} \end{definition} \noindent In \cite{seshadri_quotients} Seshadri gives almost the same definition, except that instead of (4) he simply requires the condition (\ref{item-intersect-invariant-closed}) to hold -- he does not require it to hold universally. \section{Geometric quotients} \label{section-geometric} \noindent This is Mumford's definition of a geometric quotient (at least the definition from the first edition of GIT; as far as we can tell later editions changed universally submersive'' to submersive''). \begin{definition} \label{definition-geometric} Let $S$ be a scheme and $B$ an algebraic space over $S$. Let $j : R \to U \times_B U$ be a pre-relation. A morphism $\phi : U \to X$ of algebraic spaces over $B$ is called a {\it geometric quotient} if \begin{enumerate} \item $\phi$ is an orbit space, \item condition (\ref{item-submersive}) holds universally, i.e., $\phi$ is universally submersive, and \item the functions on $X$ are the $R$-invariant functions on $U$. \end{enumerate} \end{definition} \input{chapters} \bibliography{my} \bibliographystyle{amsalpha} \end{document}