---
author:
- |
    János Kollár\
    [ ]{}\
    [with an appendix by]{} C. Raicu
bibliography:
- 'refs.bib'
title: Quotients by finite equivalence relations
---

Let $f:X\to Y$ be a finite morphism of schemes. Given $Y$, one can easily describe $X$ by the coherent sheaf of algebras $f_*{{\mathcal O}}_X$. Here our main interest is the converse. Given $X$, what kind of data do we need to construct $Y$? For this question, the surjectivity of $f$ is indispensable.

The fiber product $X\times_YX\subset X\times X$ defines an equivalence relation on $X$, and one might hope to reconstruct $Y$ as the quotient of $X$ by this equivalence relation. Our main interest is in the cases when $f$ is not flat. A typical example we have in mind is when $Y$ is not normal and $X$ is its normalization. In these cases, the fiber product $X\times_YX$ can be rather complicated. Even if $Y$ and $X$ are pure dimensional and CM, $X\times_YX$ can have irreducible components of different dimension and its connected components need not be pure dimensional. None of these difficulties appear if $f$ is flat [@MR0232781; @sga3] or if $Y$ is normal (\[quot.pure.dim.lem\]).

The aim of this note is to give many examples, review known results, pose questions and to prove a few theorems concerning finite equivalence relations.

Definition of equivalence relations
===================================

\[eq.rel.defn\] Let $X$ be an $S$-scheme and $\sigma:R\to X\times_SX$ a morphism (or $\sigma_1,\sigma_2:R\rightrightarrows X$ a pair of morphisms). We say that $R$ is an [*equivalence relation*]{} on $X$ if, for every scheme $T\to S$, we get a (set theoretic) equivalence relation $$\sigma(T):{\operatorname{Mor}}_S(T,R)\into
{\operatorname{Mor}}_S(T,X)\times {\operatorname{Mor}}_S(T,X).$$ Equivalently, the following conditions hold:

1.  $\sigma$ is a monomorphism (\[monom.defn\])

2.  (reflexive) $R$ contains the diagonal $\Delta_X$.

3.  (symmetric) There is an involution $\tau_R$ on $R$ such that $\tau_{X\times X}\circ\sigma\circ\tau_R=\sigma$, where $\tau_{X\times X}$ denotes the involution which interchanges the two factors of $X\times X$.

4.  (transitive) For $1\leq i<j\leq 3$ set $X_i:=X$ and let $R_{ij}:=R$ when it maps to $X_i\times_SX_j$. Then the coordinate projection of $R_{12}\times_{X_2}R_{23}$ to $X_1\times_SX_3$ factors through $R_{13}$: $$R_{12}\times_{X_2}R_{23}\to R_{13}\stackrel{\pi_{13}}{\longrightarrow}
    X_1\times_SX_3.$$

We say that $\sigma_1,\sigma_2:R\rightrightarrows X$ is a [*finite*]{} equivalence relation if the maps $\sigma_1,\sigma_2$ are finite. In this case, $\sigma:R\to X\times_SX$ is also finite, hence a closed embedding (\[monom.defn\]).

\[setth.eq.rel.defn\] Let $X$ and $R$ be reduced $S$-schemes. We say that a morphism $\sigma:R\to X\times_SX$ is a [*set theoretic equivalence relation*]{} on $X$ if, for every geometric point ${\operatorname{Spec}}K\to S$, we get an equivalence relation on $K$-points $$\sigma(K):{\operatorname{Mor}}_S({\operatorname{Spec}}K,R)\into
{\operatorname{Mor}}_S({\operatorname{Spec}}K,X)\times {\operatorname{Mor}}_S({\operatorname{Spec}}K,X).$$ Equivalently,

1.  $\sigma$ is geometrically injective.

2.  (reflexive) $R$ contains the diagonal $\Delta_X$.

3.  (symmetric) There is an involution $\tau_R$ on $ R$ such that $\tau_{X\times X}\circ\sigma\circ\tau_R=\sigma$ where $\tau_{X\times X}$ denotes the involution which interchanges the two factors of $X\times X$.

4.  (transitive) For $1\leq i<j\leq 3$ set $X_i:=X$ and let $R_{ij}:=R$ when it maps to $X_i\times_SX_j$. Then the coordinate projection of ${\operatorname{red}}\bigl(R_{12}\times_{X_2}R_{23}\bigr)$ to $X_1\times_SX_3$ factors through $R_{13}$: $${\operatorname{red}}\bigl(R_{12}\times_{X_2}R_{23}\bigr)\to R_{13}
    \stackrel{\pi_{13}}{\longrightarrow}
    X_1\times_SX_3.$$

Note that the fiber product need not be reduced, and taking the reduced structure above is essential, as shown by (\[nonred.noneq.exmp\]).

It is sometimes convenient to consider finite morphisms $p:R\to X\times_SX$ such that the injection $i:p(R)\into  X\times_SX$ is a set theoretic equivalence relation. Such a $p:R\to X\times_SX$ is called a [*set theoretic pre-equivalence relation.*]{}

\[nonred.noneq.exmp\] On $X:={{\mathbb C}}^2$ consider the ${{\mathbb Z}}/2$-action $(x,y)\mapsto (-x,-y)$. This can be given by a set theoretic equivalence relation $R\subset X_{x_1,y_1}\times X_{x_2,y_2}$ defined by the ideal $$(x_1-x_2,y_1-y_2)\cap (x_1+x_2,y_1+y_2)=
(x_1^2-x_2^2, y_1^2-y_2^2, x_1y_1-x_2y_2,  x_1y_2-x_2y_1)$$ in ${{\mathbb C}}[x_1,y_1,x_2,y_2]$. We claim that this is [*not*]{} an equivalence relation. The problem is with transitivity. The defining ideal of $R_{12}\times_{X_2}R_{23}$ in ${{\mathbb C}}[x_1,y_1,x_2,y_2,x_3,y_3]$ is $$(x_1^2-x_2^2, y_1^2-y_2^2, x_1y_1-x_2y_2,  x_1y_2-x_2y_1,
x_2^2-x_3^2, y_2^2-y_3^2, x_2y_2-x_3y_3,  x_2y_3-x_3y_2).$$ This contains $(x_1^2-x_3^2, y_1^2-y_3^2, x_1y_1-x_3y_3)$ but it does not contain $ x_1y_3-x_3y_1$. Thus there is no map $R_{12}\times_{X_2}R_{23}\to R_{13}$. Not, however, that the problem is easy to remedy. Let $R^*\subset X\times X$ be defined by the ideal $$(x_1^2-x_2^2, y_1^2-y_2^2, x_1y_1-x_2y_2)\subset {{\mathbb C}}[x_1,y_1,x_2,y_2].$$ We see that $R^*$ defines an equivalence relation. The difference between $R$ and $R^*$ is one embedded point at the origin.

\[cat.geom.quot\] Given two morphisms, $\sigma_1,\sigma_2:R\rightrightarrows X$, there is at most one scheme $q:X\to (X/R)^{cat}$ such that $q\circ \sigma_1=q\circ \sigma_2$ and $q$ is universal with this property. We call $(X/R)^{cat}$ the [*categorical quotient*]{} (or [*coequalizer*]{}) of $\sigma_1,\sigma_2:R\rightrightarrows X$.

The categorical quotient is easy to construct in the affine case. Given $\sigma_1,\sigma_2:R\rightrightarrows X$, the categorical quotient $(X/R)^{cat}$ is the spectrum of the $S$-algebra $$\ker\Bigl[{{\mathcal O}}_X\stackrel{\sigma_1^*-\sigma_2^*}{\longrightarrow} {{\mathcal O}}_R\Bigr].$$

Let $\sigma_1,\sigma_2:R\rightrightarrows X$ be a finite equivalence relation. We say that $q:X\to Y$ is a [*geometric quotient*]{} of $X$ by $R$ if

1.  $q:X\to Y$ is the categorical quotient $q:X\to (X/R)^{cat}$,

2.  $q:X\to Y$ is finite, and

3.  for every geometric point ${\operatorname{Spec}}K\to S$, the fibers of $q_K:X_K(K)\to Y_K(K)$ are the $\sigma\bigl(R_K(K)\bigr)$-equivalence classes of $X_K(K)$.

The geometric quotient is denoted by $X/R$.

The main example to keep in mind is the following, which easily follows from (\[quot.X/S.finite.lem\]) and the construction of $(X/R)^{cat}$ for affine schemes.

\[basic.quot.exmp\] Let $f:X\to Y$ be a finite and surjective morphism. Set $R:={\operatorname{red}}(X\times_YX)\subset X\times X$ and let $\sigma_i:R\to X$ denote the coordinate projections. Then the geometric quotient $X/R$ exists and $X/R\to Y$ is a finite and universal homeomorphism (\[univ.homeo.defn\]). Therefore, if $X$ is the normalization of $Y$, then $X/R$ is the weak normalization of $Y$. (See [@rc-book Sec.7.2] for basic results on semi-normal and weakly normal schemes.)

By taking the reduced structure of $X\times_YX$ above, we chose to focus on the set-theoretic properties of $Y$. However, as (\[sch.quot.exmp\]) shows, even if $X,Y$ and $X\times_YX$ are all reduced, $X/R\to Y$ need not be an isomorphism. Thus $X$ and $X\times_YX$ do not determine $Y$ uniquely.

In Section \[first.exmp.sec\] we give examples of finite, set theoretic equivalence relations $R\rightrightarrows X$ such that the categorical quotient $(X/R)^{cat}$ is non-Noetherian and there is no geometric quotient. This can happen even when $X$ is very nice, for instance a smooth variety over ${{\mathbb C}}$. Some elementary results about the existence of geometric quotients are discussed in Section \[basic.res.sec\].

An inductive plan to construct geometric quotients is outlined in Section \[induct.plan.sect\]. As an application, we prove in Section \[pos.char.sec\] the following:

\[quot.by.R.charp\] Let $S$ be a Noetherian ${{\mathbb F}}_p$-scheme and $X$ an algebraic space which is essentially of finite type over $S$. Let $R\rightrightarrows X$ be a finite, set theoretic equivalence relation. Then the geometric quotient $X/R$ exists.

There are many algebraic spaces which are not of finite type and such that the Frobenius map $F^q:X\to X^{(q)}$ is finite. By a result of Kunz (see [@mats-ca p.302]) such algebraic spaces are excellent. As the proof shows, (\[quot.by.R.charp\]) remains valid for algebraic spaces satisfying this property.

In the Appendix, C. Raicu constructs finite scheme theoretic equivalence relations $R$ on $X={{\mathbb A}}^2$ (in any characteristic) such that the geometric quotient $X/R$ exists yet $R$ is strictly smaller than the fiber product $X\times_{X/R}X$. Closely related examples are in [@venken; @philippe].

In characteristic zero, this leaves open the following:

\[sch.th.quot.quest\] Let $R\subset  X\times X$ be a scheme theoretic equivalence relation such that the coordinate projections $R\rightrightarrows X$ are finite.

Is there a geometric quotient $X/R$?

A special case of the quotient problem, called gluing or pinching, is discussed in Section \[glue.sec\]. This follows the works of [@artin70], [@ferrand] (which is based on an unpublished manuscript from ’70) and [@raoult].

First examples {#first.exmp.sec}
==============

The next examples show that in many cases, the categorical quotient of a very nice scheme $X$ can be non-Noetherian. We start with a nonreduced example and then we build it up to smooth ones.

\[first.exmp\] Let $k$ be a field and consider $g_i:k[x,\epsilon]\to k[x,\epsilon]$ where $$g_1\bigl(a(x)+\epsilon b(x)\bigr)=a(x)+\epsilon b(x)
{\quad\mbox{and}\quad}
g_2\bigl(a(x)+\epsilon b(x)\bigr)=a(x)+\epsilon \bigl(b(x)+a'(x)\bigr).$$

If ${\operatorname{char}}k=0$ then the coequalizer is the spectrum of $$\ker\Bigl[k[x,\epsilon]\stackrel{g_1^*-g_2^*}{\longrightarrow} 
k[x,\epsilon]\Bigr]=k+\epsilon k[x].$$ Note that $k+\epsilon k[x]$ is not Noetherian and its only prime ideal is $\epsilon k[x]$.

If ${\operatorname{char}}k=p$ then the coequalizer is the spectrum of the finitely generated $k$-algebra $$\ker\Bigl[k[x,\epsilon]\stackrel{g_1^*-g_2^*}{\longrightarrow} 
k[x,\epsilon]\Bigr]=k[x^p]+\epsilon k[x].$$

It is not surprising that set theoretic equivalence relations behave badly on nonreduced schemes. However, the above example is easy to realize on reduced and even on smooth schemes.

(cf. [@holmann p.342]) Let $p_i:Z\to Y_i$ be finite morphisms for $i=1,2$. We can construct out of them an equivalence relation on $Y_1\amalg Y_2$ where $R$ is the union of the diagonal with two copies of $Z$, one of which maps as $$(p_1,p_2):Z\to Y_1\times Y_2\subset \bigl( Y_1\amalg Y_2\bigr)
\times
\bigl( Y_1\amalg Y_2\bigr),$$ the other its symmetric pair. The categorical quotient $\bigl(\bigl( Y_1\amalg Y_2\bigr)/R\bigr)^{cat}$ is also the universal push-out of $Y_1\stackrel{p_1}{\leftarrow} Z
  \stackrel{p_2}{\to}Y_2$. If $Z$ and the $Y_i$ are affine over $S$, then it is the spectrum of the $S$-algebra $$\ker\Bigl[{{\mathcal O}}_{Y_1}+{{\mathcal O}}_{Y_2}\stackrel{p_1^*-p_2^*}{\longrightarrow} {{\mathcal O}}_Z\Bigr].$$

For the first example let $Y_1\cong Y_2:={\operatorname{Spec}}k[x,y^2,y^3]$ and $Z:={\operatorname{Spec}}k[u,v]$ with $p_i$ given by $$p_1^*: (x,y^2,y^3)\mapsto (u,v^2,v^3)
{\quad\mbox{and}\quad}
p_2^*:(x,y^2,y^3)\mapsto (u+v,v^2,v^3).$$ Since the $p_i^*$ are injective, the categorical quotient is the spectrum of the $k$-algebra $k[u,v^2,v^3]\cap k[u+v,v^2,v^3]$. Note that $$\begin{array}{lll}
k[u,v^2,v^3] & = & \bigl\{f_0(u)+
\sum_{i\geq 2} v^i f_i(u)\ :\ f_i\in k[u]\bigr\}{\quad\mbox{and}\quad}\\
k[u+v,v^2,v^3] & = & \bigl\{f_0(u)+vf'_0(u)+\sum_{i\geq 2} v^i f_i(u)
\ :\ f_i\in k[u]\bigr\}.
\end{array}$$ As in (\[first.exmp\]), if ${\operatorname{char}}k=0$ then the categorical quotient is the spectrum of the non-Noetherian algebra $k+\sum_{n\geq 2}v^nk[u]$. If ${\operatorname{char}}k=p$ then the geometric quotient is given by the finitely generated $k$-algebra $$k[u^p]+\sum_{n\geq 2}v^nk[u].$$

This example can be embedded into a set theoretic equivalence relation on a smooth variety.

Let $Y_1\cong Y_2:={{\mathbb A}}^3_{xyz}$, $Z:={{\mathbb A}}^2_{uv}$ and $$p_1^*: (x_1,y_1,z_1)\mapsto (u,v^2,v^3)
{\quad\mbox{and}\quad}
p_2^*(x_2,y_2,z_2)\mapsto (u+v,v^2,v^3).$$ By the previous computations, in characteristic zero the categorical quotient is given by $$k+(y_1,z_1)+(y_2, z_2)\subset k[x_1,y_1,z_1]+k[x_2,y_2,z_2],$$ where $(y_i,z_i)$ denotes the ideal $(y_i,z_i)\subset k[x_i,y_i,z_i]$. A minimal generating set is given by $$y_1x_1^m, z_1x_1^m, y_2x_2^m, z_2x_2^m\ :\ m=0,1,2, \dots$$

In positive characteristic the categorical quotient is given by $$k[x_1^p,x_2^p]+ (y_1,z_1)+(y_2, z_2)\subset k[x_1,y_1,z_1]+k[x_2,y_2,z_2].$$ A minimal generating set is given by $$x_1^p, x_2^p, y_1x_1^m, z_1x_1^m, y_2x_2^m, z_2x_2^m\ :\ m=0,1,\dots, p-1.$$

\[nagata.exmp\] The following example, based on [@Nagata69], shows that even for rings of invariants of finite group actions some finiteness assumption on $X$ is necessary in order to obtain geometric quotients.

Let $k$ be a field of characteristic $p>0$ and $K:= k(x_1,x_2,\dots)$, where the $x_i$ are algebraically independent over $k$. Let $$D:= \sum_ix_{i+1}\frac{\partial}{\partial x_i}{\quad\mbox{be a derivation of $K$.}\quad}$$ Let $F:=\{f\in K\vert D(f)=0\}$ be the subfield of constants. Set $$R=K+\epsilon K{\quad\mbox{where $\epsilon^2=0$ and}\quad} \sigma: f+\epsilon g\mapsto
f+\epsilon (g+D(f)).$$ $R$ is a local Artin ring. It is easy to check that $\sigma$ is an automorphism of $R$ of order $p$. The fixed ring is $R^{\sigma}=F+\epsilon K$. Its maximal ideal is $m:= (\epsilon K)$ and generating sets of $m$ correspond to $F$-vectorspace bases of $K$. Next we show that the $x_i$ are linearly independent over $F$ which implies that $R^{\sigma}$ is not Noetherian.

Assume that we have a relation $\sum_{i\leq n} f_ix_i=0$ We may assume that $f_n=1$ and $f_i\in F\cap k(x_1,\dots,x_r)$ for some $r$. Apply $D$ to get that $$0=\sum_{i\leq n} f_iD(x_i)=\sum_{i\leq n} f_ix_{i+1}.$$ Repeating $s$ times gives that $\sum_{i\leq n} f_ix_{i+s}=0$, or, equivalently $$x_{n+s}=-\sum_{i\leq n-1} f_ix_{i+s}.$$ This is impossible if $n+s>r$; a contradiction.

It is easy to see that $R$ is not a submodule of any finitely generated $R^{\sigma}$-module.

This example of [@naga] gives a 2-dimensional regular local ring $R$ and an automorphism of order 2 such that the ring of invariants is not Noetherian.

Let $k$ be a field of characteristic 2 and $K:= k(x_1, y_1, x_2, y_2, \dots)$, where the $x_i, y_i$ are algebraically independent over $k$. Let $R:=K[[u,v]]$ be the power series ring in 2 variables. Note that $R$ is a 2-dimensional regular local ring, but it is not essentially of finite type over $k$. Define a derivation of $K$ to $R$ by $$D_K:= \sum_i v(x_{i+1}u+y_{i+1}v)\frac{\partial}{\partial x_i} +
u(x_{i+1}u+y_{i+1}v)\frac{\partial}{\partial y_i}.$$ This extends to a derivation of $R$ to $R$ by setting $D_R|_K=D_K$ and $D_R(u)=D_R(v)=0$. Note that $D_R\circ D_R=0$, thus $\sigma: r\mapsto r+D_R(r)$ is an order 2 automorphism of $R$. We claim that the ring of invariants $R^{\sigma}$ is not Noetherian.

To see this, note first that $x_{i}u+y_{i}v\in R^{\sigma}$ for every $i$.

[*Claim.*]{} For every $n$, $x_{n+1}u+y_{n+1}v\not\in \bigl(x_{1}u+y_{1}v, \dots, x_{n}u+y_{n}v\bigr)
R^{\sigma}$.

Proof. Assume the contrary and write $$x_{n+1}u+y_{n+1}v=\sum_{i\leq n} r_i\bigl(x_{i}u+y_{i}v\bigr)
{\quad\mbox{where $r_i\in R^{\sigma}$.}\quad}$$ Working modulo $(u,v)^2$ and gathering the terms involving $u$, we get an equality $$x_{n+1}\equiv \sum_{i\leq n}  r_ix_{i} \mod R^{\sigma}\cap (u,v)R.$$ Applying $D_R$ and again gathering the terms involving $u$ we obtain that $x_{n+2}
\equiv\sum_{i\leq n}  r_ix_{i+1}$ modulo $R^{\sigma}\cap (u,v)R$. Repeating this $s$ times gives $$x_{n+s+1}=\sum_{i\leq n}  \bar r_ix_{i+s}{\quad\mbox{where $\bar r_i\in K$.}\quad}$$ Since the $\bar r_i$ involve only finitely many variables, we get a contradiction for large $s$. Thus $$(p_1)\subset (p_1,p_2)\subset (p_1,p_2,p_3)\subset \cdots$$ is an infinite increasing sequence of ideals in $R^{\sigma}$.

The next examples show that, if $S$ is a smooth projective surface, then a geometric quotient $S/R$ can be nonprojective (but proper) and if $X$ is a smooth proper 3-fold, $X/R$ can be an algebraic space which is not a scheme.

\[nonqpglue.exmp\] 1. Let $C,D$ be smooth projective curves and $S$ the blow up of $C\times D$ at a point $(c,d)$. Let $C_1\subset S$ be the birational transform of $C\times \{d\}$, $C_2:=C\times \{d'\}$ for some $d'\neq d$ and ${{\mathbb P}}^1\cong E\subset S$ the exceptional curve.

Fix an isomorphism $\sigma:C_1\cong C_2$. This generates an equivalence relation $R$ which is the identity on $S\setminus(C_1\cup C_2)$. As we see in (\[quot.by.R.lowdim\]), $S/R$ is a surface of finite type. Note however that the image of $E$ in $S/R$ is numerically equivalent to $0$, thus $S/R$ is not quasi projective. Indeed, let $M$ be any line bundle on $S/R$. Then $\pi^*M$ is a line bundle on $S$ such that $(C_1\cdot \pi^*M)=(C_2\cdot \pi^*M)$. Since $C_2$ is numerically equivalent to $C_1+E$, this implies that $(E\cdot \pi^*M)=0$.

2\. Take $S\cong {{\mathbb P}}^2$ and $Z:=(x(y^2-xz)=0)$. Fix an isomorphism of the line $(x=0)$ and the conic $(y^2-xz=0)$ which is the identity on their intersection. As before, this generates an equivalence relation $R$ which is the identity on their complement. By (\[quot.by.R.lowdim\]), ${{\mathbb P}}^2/R$ exists as a scheme but it is not projective.

Indeed, if $M$ is a line bundle on ${{\mathbb P}}^2/R$ then $\pi^*M$ is a line bundle on ${{\mathbb P}}^2$ whose degree on a line is the same as its degree on a conic. Thus $\pi^*H\cong {{\mathcal O}}_{{{\mathbb P}}^2}$ and so $H$ is not ample.

3\. Let $S=S_1\amalg S_2\cong {{\mathbb P}}^2\times \{1,2\}$ be 2 copies of ${{\mathbb P}}^2$. Let $E\subset {{\mathbb P}}^2$ be a smooth cubic. For a point $p\in E$, let $\sigma_p: E\times \{1\}\to E\times \{2\}$ be the identity composed with translation by $p\in E$. As before, this generates an equivalence relation $R$ which is the identity on their complement.

Let $M$ be a line bundle on $S/R$. Then $\pi^*M|_{S_i}\cong {{\mathcal O}}_{{{\mathbb P}}^2}(m_i)$ for some $m_i>0$, and we conclude that $${{\mathcal O}}_{{{\mathbb P}}^2}(m_1)|{E}\cong \tau_p^*\bigl({{\mathcal O}}_{{{\mathbb P}}^2}(m_2)|{E}\bigr).$$ This holds iff $m_1=m_2$ and $p\in E$ is a $3m_1$-torsion point. Thus the projectivity of $S/R$ depends very subtly on the gluing map $\sigma_p$.

Hironaka’s example in [@hartsh App.B.3.4.1] gives a smooth, proper variety $X$ and two curves ${{\mathbb P}}^1\cong C_1\cong  C_2\subset X$ such that $C_1+C_2$ is homologous to $0$. Let $g:C_1\cong C_2$ be an isomorphism and $R$ the corresponding equivalence relation.

We claim that there is a no quasi projective open subset $U\subset X$ which intersects both $C_1$ and $C_2$. Assume to the contrary that $U$ is such. Then there is an ample divisor $H_U\subset U$ which intersects both curves but does not contain either. Its closure $H\subset X$ is a Cartier divisor which intersects both curves but does not contain either. Thus $H\cdot (C_1+C_2)>0$, a contradiction.

This shows that if $p\in X/R$ is on the image of $C_i$ then $p$ does not have any affine open neighborhood since the preimage of an affine set by a finite morphism is again affine. Thus $X/R$ is not a scheme.

[@lip-lip]\[sch.quot.exmp\] Fix a field $k$ and let $a_1,\dots, a_n\in k$ be different elements. Set $A:=k[x,y]/\bigl( \prod_i (x-a_iy)\bigr)$. Then $Y:={\operatorname{Spec}}R$ is $n$ lines through the origin. Let $f:X\to Y$ its normalization. Thus $X=\amalg_i {\operatorname{Spec}}k[x,y]/(x-a_iy)$. Note that $$k[x,y]/(x-a_iy)\otimes_A k[x,y]/(x-a_jy)=
\left\{
\begin{array}{l}
 k[x,y]/(x-a_iy){\quad\mbox{if $a_i=a_j$, and}\quad}\\
 k \qquad \qquad\qquad\ \  {\quad\mbox{if $a_i\neq a_j$.}\quad}
\end{array}
\right.$$ Thus $X\times_YX$ is reduced. It is the union of the diagonal $\Delta_X$ and of $f^{-1}(0,0)\times f^{-1}(0,0)$. Thus $X/\bigl(X\times_YX\bigr)$ is a seminormal scheme which is isomorphic to the $n$ coordinate axes in ${{\mathbb A}}^n$. For $n\geq 3$, it is not isomorphic to $Y$.

One can also get similar examples where $Y$ is integral. Indeed, let $Y\subset {{\mathbb A}}^2$ be any plane curve whose only singularities are ordinary multiple points and let $f:X\to Y$ be its normalization. By the above computations, $X\times_YX$ is reduced and $X/\bigl(X\times_YX\bigr)$ is the seminormalization of $Y$.

If $Y$ is a reduced scheme with normalization $\bar Y\to Y$. Then, as we see in (\[quot.X/S.finite.lem\]), the geometric quotient $\bar Y/\bigl(\bar Y\times_Y\bar Y\bigr)$ exists. It coincides with the strict closure considered in [@lip-arf]. The curve case was introduced earlier in [@arf].

The related Lipschitz closure is studied in [@pham] and [@lip-lip].

Basic results {#basic.res.sec}
=============

In this section we prove some basic existence results for geometric quotients.

\[quot.X/S.finite.lem\] Let $S$ be a Noetherian scheme. Assume that $X$ is finite over $S$ and let $p_1,p_2:R\rightrightarrows X$ be a finite, set theoretic equivalence relation over $S$. Then the geometric quotient $X/R$ exists.

Proof. Since $X\to S$ is affine, the categorical quotient is the spectrum of the ${{\mathcal O}}_S$-algebra $$\ker\Bigl[{{\mathcal O}}_X\stackrel{p_1^*-p_2^*}{\longrightarrow} {{\mathcal O}}_R\Bigr].$$ This kernel is a submodule of the finite ${{\mathcal O}}_S$-algebra ${{\mathcal O}}_X$, hence itself a finite ${{\mathcal O}}_S$-algebra. The only question is about the geometric fibers of $X\to (X/R)^{cat}$. Pick any $s\in S$. Taking kernel commutes with flat base scheme extensions. Thus we may assume that $S$ is complete, local with closed point $s$ and algebraically closed residue field $k(s)$. We need to show that the reduced fiber of $(X/R)^{cat}\to S$ over $s$ is naturally isomorphic to ${\operatorname{red}}X_s/{\operatorname{red}}R_s$.

If $U\to S$ is any finite map then ${{\mathcal O}}_{{\operatorname{red}}U_s}$ is a sum of $m(U)$ copies of $k(s)$ for some $m(U)<\infty$. $U$ has $m(U)$ connected components $\{U_i:i=1,\dots, m(U)\}$ and each $U_i\to S$ is finite. Thus $U\to S$ uniquely factors as $$U\stackrel{g}{\to} \amalg_{m(U)}S\to S
{\quad\mbox{such that}\quad}
g_s:{\operatorname{red}}U_s\stackrel{\cong}{\longrightarrow} \amalg_{m(U)}{\operatorname{Spec}}k(s)$$ is an isomorphism, where $\amalg_mS$ denotes the disjoint union of $m$ copies of $S$.

Furthermore, if $\{R_j:j=1,\dots m(R)\}$ are the connected components of $R$, then $\sigma_1$ maps $R_j$ to some $U_i$ and we have a commutative diagram $$\begin{array}{ccc}
R_j&
\stackrel{\sigma_1}{\longrightarrow}
& U_i\\
\downarrow && \downarrow\\
S& =& S.
\end{array}$$ Applying this to $X\to S$ and $R\to S$, we obtain a commutative diagram $$\begin{array}{lcl}
{{\mathcal O}}_S^{m(X)}&
\stackrel{p_1^*(s)-p_2^*(s)}{\longrightarrow}
&{{\mathcal O}}_S^{m(R)}\\
\ \downarrow && \ \downarrow\\
{{\mathcal O}}_X&\stackrel{p_1^*-p_2^*}{\longrightarrow}& {{\mathcal O}}_R.
\end{array}$$ Thus, for $\tau:={p_1^*(s)-p_2^*(s)}$, we get a morphism $$\ker\Bigl[{{\mathcal O}}_S^{m(X)}
\stackrel{\tau}{\longrightarrow}
 {{\mathcal O}}_S^{m(R)}\Bigr]\to  {{\mathcal O}}_{(X/R)^{cat}}.$$ The kernel on the left is $m:=|X_s/R_s|$ copies of ${{\mathcal O}}_S$, hence we obtain a factorization $$(X/R)^{cat}\to \amalg_mS\to S
{\quad\mbox{such that}\quad}
{\operatorname{red}}(X/R)^{cat}_s\to \amalg_m{\operatorname{Spec}}k(s)$$ is an isomorphism.

For later reference, we record the following straightforward consequence.

\[quot.of.sub\] Let $R\rightrightarrows X$ be a finite, set theoretic equivalence relation such that $X/R$ exists. Let $Z\subset X$ be a closed $R$-invariant subscheme. Then $Z/R|_Z$ exists and $Z/R|_Z\to X/R$ is a finite and universal homeomorphism (\[univ.homeo.defn\]) onto its image.

\[subquot.neq.sub\] Even in nice situations, $Z/R|_Z\to X/R$ need not be a closed embedding, as shown by the following examples.

(\[subquot.neq.sub\].1) Set $X:={{\mathbb A}}^2_{xy}\amalg {{\mathbb A}}^2_{uv}$ and let $R$ be the equivalence relation that identifies the $x$-axis with the $u$-axis.

Let $Z=(y=x^2)\amalg (v=u^2)$. In $Z/R|_Z$ the two components intersect at a node, but the image of $Z$ in $X/R$ has a tacnode.

In this example the problem is clearly caused by ignoring the scheme structure of $R|_Z$. As the next example shows, similar phenomena happen even if $R|_Z$ is reduced. (\[subquot.neq.sub\].2) Set $Y:=(xyz=0)\subset {{\mathbb A}}^3$. Let $X$ be the normalization of $Y$ and $R:=X\times_YX$. Set $W:=(x+y+z=0)\subset  Y$ and let $Z\subset X$ be the preimage of $W$. As computed in (\[sch.quot.exmp\]), $R$ and $R|_Z$ are both reduced, $Z/R|_Z$ is the seminormalization of $W$ and $Z/R|_Z\to W$ is not an isomorphism.

\[bb.count.exmp\] The following counter example to (\[quot.X/S.finite.lem\]) is proposed in [@bial 6.2]. Consider the diagram $$\begin{array}{ccc}
 {\operatorname{Spec}}k[x,y] & \stackrel{p_1}{\longrightarrow} & {\operatorname{Spec}}k[x,y^2,y^3]\\
p_2\downarrow\hphantom{p_2} && \hphantom{q_2}\downarrow q_2\\
{\operatorname{Spec}}k[x+y,x+x^2, y^2,y^3]&  \stackrel{q_1}{\longrightarrow} &
{\operatorname{Spec}}k[x+x^2, xy^2, xy^3,  y^2,y^3]
\end{array}
\eqno{(\ref{bb.count.exmp}.1)}$$ It is easy to see that the $p_i$ are homeomorphisms but $q_2p_1=q_1p_2$ maps $(0,0)$ and $(-1,0)$ to the same point. If (\[bb.count.exmp\].1) were a universal push-out, one would get a counter example to (\[quot.X/S.finite.lem\]). However, it is not a universal push-out. Indeed, $$\begin{array}{rcl}
\tfrac13(x+y)^3+\tfrac12(x+y)^2&=&\ \ 
\bigl(\tfrac13 x^3+\tfrac12x^2\bigr)+ (x^2+x)y\ \ \ \ \ \ \ \ 
+xy^2+\tfrac12y^2+\tfrac13 y^3\\[1ex]
&=&
-\bigl(\tfrac23 x^3+\tfrac12x^2\bigr)+ (x^2+x)(x+y)+xy^2+\tfrac12
y^2+\tfrac13 y^3
\end{array}$$ shows that $\tfrac23 x^3+\tfrac12x^2$ is also in the intersection $k[x,y^2,y^3]\cap k[x+y,x+x^2, y^2,y^3]$.

Another case where $X/R$ is easy to obtain is the following.

\[quot.pure.dim.lem\] Let $p_1,p_2:R\rightrightarrows X$ be a finite, set theoretic equivalence relation where $X$ is normal, Noetherian and $X,R$ are both pure dimensional. Assume that one of the following holds:

1.  $X$ is defined over a field of characteristic 0,

2.  $X$ is essentially of finite type over $S$, or

3.  $X$ is defined over a field of characteristic $p>0$ and the Frobenius map $F^p:X\to X^{(p)}$ (\[frob.say\]) is finite.

Then the geometric quotient $X/R$ exists as an algebraic space. $X/R$ is normal, Noetherian and essentially of finite type over $S$ in case (2).

Proof. Let us first deal with the case when $R$ comes from a finite group acting on $X$. This case is a result of Deligne, discussed in [@Knutson71 IV.1.8].

For $x\in X$, let $G_x\subset G$ denote the stabilizer. Let $x\in U_x\subset X$ be a $G_x$-invariant affine open subset. By shrinking $U_x$ we may assume that $G_{x'}\subset G_x$ for every $x'\in U_x$.

In the affine case, quotients by finite groups are easy to get (\[inv.of.fin.gps\]); this is where the conditions (1–3) are used. Thus $U_x/G_x$ exists and it is easy to see that the $U_x/G_x$ give étale charts for $X/G$.

In the general case, it is enough to construct the quotient when $X$ is irreducible. Let $m$ be the separable degree of the projections $\sigma_i:R\to X$.

Consider the $m$-fold product $X\times\cdots\times X$ with coordinate projections $\pi_i$. Let $R_{ij}$ (resp. $\Delta_{ij}$) denote the preimage of $R$ (resp. of the diagonal) under $(\pi_i,\pi_j)$. A geometric point of $\cap_{ij} R_{ij}$ is a sequence of geometric points $(x_1,\dots, x_m)$ such that any 2 are $R$-equivalent and a geometric point of $\cap_{ij} R_{ij}\setminus \cup_{ij}\Delta_{ij}$ is a sequence $(x_1,\dots, x_m)$ that constitutes a whole $R$-equivalence class. Let $X'$ be the normalization of the closure of $\cap_{ij} R_{ij}\setminus \cup_{ij}\Delta_{ij}$. Note that every $\pi_{\ell}:\cap_{ij} R_{ij}\to X$ is finite, hence the projections $\pi'_{\ell}:X'\to X$ are finite.

The symmetric group $S_m$ acts on $X\times\cdots\times X$ by permuting the factors and this lifts to an $S_m$-action on $X'$. Over a dense open subset of $X$, the $S_m$-orbits on the geometric points of $X'$ are exactly the $R$-equivalence classes.

Let $X^*\subset X'/S_m\times X$ be the image of $X'$ under the diagonal map.

By construction, $X^*\to X$ is finite and one-to-one on geometric points over an open set. Since $X$ is normal, $X^*\cong X$ in characteristic 0 and $X^*\to X$ is purely inseparable in positive characteristic.

In characteristic 0, we thus have a morphism $X\to X'/S_m$ whose geometric fibers are exactly the $R$-equivalence classes. Thus $X'/S_m=X/R$.

Essentially the same works in positive characteristic, see Section \[pos.char.sec\] for details.

\[quot.norm.lem\] Let $p_1,p_2:R\rightrightarrows X$ be a finite, set theoretic equivalence relation such that $(X/R)^{cat}$ exists.

1.  If $X$ is normal and $X, R$ are pure dimensional then $(X/R)^{cat}$ is also normal.

2.  If $X$ is seminormal then $(X/R)^{cat}$ is also seminormal.

Proof. In the first case, let $Z\to (X/R)^{cat}$ be a finite morphism which is an isomorphism at all generic points of $(X/R)^{cat}$. Since $X$ is normal, $\pi:X\to (X/R)^{cat}$ lifts to $\pi_Z:X\to Z$. By assumption, $\pi_Z\circ p_1$ equals $\pi_Z\circ p_2$ at all generic points of $R$ and $R$ is reduced. Thus $\pi_Z\circ p_1=\pi_Z\circ p_2$. The universal property of categorical quotients gives $(X/R)^{cat}\to Z$, hence $Z=(X/R)^{cat}$ and $(X/R)^{cat}$ is normal.

In the second case, let $Z\to (X/R)^{cat}$ be a finite morphism which is a universal homeomorphism (\[univ.homeo.defn\]). As before, we get liftings $\pi_Z\circ p_1,\pi_Z\circ p_2:R\rightrightarrows X\to Z$ which agree on closed points. Since $R$ is reduced, we conclude that $\pi_Z\circ p_1=\pi_Z\circ p_2$, thus $(X/R)^{cat}$ is seminormal.

The following result goes back at least to E. Noether.

\[inv.of.fin.gps\] Let $A$ be a Noetherian ring, $R$ a Noetherian $A$-algebra and $G$ a finite group of $A$-automorphisms of $R$. Let $R^G\subset R$ denote the subalgebra of $G$-invariant elements. Assume that one of the following holds:

1.  $\frac1{|G|}\in A$,

2.  $R$ is essentially of finite type over $A$, or

3.  $R$ is finite over $A[R^p]$ for every prime $p$ that divides $|G|$.

Then $R^G$ is Noetherian and $R$ is finite over $R^G$.

Proof. Assume first that $R$ is a localization of a finitely generated $A$ algebra $A[r_1,\dots, r_m]\subset R$. We may assume that $G$ permutes the $r_j$. Let $\sigma_{ij}$ denote the $j$th elementary symmetric polynomial of the $\{g(r_i):g\in G\}$. Then $$A\bigl[\sigma_{ij}\bigr]\subset A[r_1,\dots, r_m]^G\subset R^G$$ and, with $n:=|G|$, each $r_i$ satisfies the equation $$r_i^n-\sigma_{i1}r_i^{n-1}+\sigma_{i2}r_i^{n-2}-+\cdots =0.$$ Thus $A[r_1,\dots, r_m]$ is integral over $A\bigl[\sigma_{ij}\bigr]$, hence also over the larger ring $A[r_1,\dots, r_m]^G$.

By assumption $R=U^{-1}A[r_1,\dots, r_m]$ where $U$ is a subgroup of units in $A[r_1,\dots, r_m]$. We may assume that $U$ is $G$-invariant. If $r/u\in R$ where $r\in A[r_1,\dots, r_m]$ and $u$ a unit in $A[r_1,\dots, r_m]$, then $$\frac{r}{u}=\frac{r\prod_{g\neq 1} g(u)}{u\prod_{g\neq 1} g(u)},$$ where the product is over the non-identity elements of $G$. Thus $r/u=r'/u'$ where $r'\in A[r_1,\dots, r_m]$ and $u'$ is a $G$-invariant unit in $A[r_1,\dots, r_m]$. Therefore, $$R=\bigl(U^G\bigr)^{-1}A[r_1,\dots, r_m]
{\quad\mbox{is finite over}\quad}
\bigl(U^G\bigr)^{-1}A\bigl[\sigma_{ij}\bigr].$$ Since $R^G$ is an $\bigl(U^G\bigr)^{-1}A\bigl[\sigma_{ij}\bigr]$-submodule of $R$, it is also finite over $\bigl(U^G\bigr)^{-1}A\bigl[\sigma_{ij}\bigr]$, hence the localization of a finitely generated algebra.

Assume next that $|G|$ is invertible in $A$. We claim that $JR\cap R^G=J$ for any ideal $J\subset R^G$. Indeed, if $a_i\in R^G$, $r_i\in R$ and $\sum r_ia_i\in R^G$ then $$|G|\cdot \sum_i r_ia_i=\sum_{g\in G}\sum_ig(r_i)g(a_i)=
\sum_ia_i\sum_{g\in G}g(r_i)\in \sum_ia_iR^G.$$ If $|G|$ is invertible, this gives that $R^G\cap\sum a_iR =\sum a_iR^G$. Thus the map $J\mapsto JR$ from the ideals of $R^G$ to the ideals of $R$ is an injection which preserves inclusions. Therefore $R^G$ is Noetherian if $R$ is.

If $R$ is an integral domain, then $R$ is finite over $R^G$ by (\[r/rg.finite.general\]). The general case, which we do not use, is left to the reader.

The arguments in case (3) are quite involved, see [@fogarty].

\[r/rg.finite.general\] Let $R$ be an integral domain and $G$ a finite group of automorphisms of $R$. Then $R$ is contained in a finite $R^G$-module. Thus, if $R^G$ is Noetherian, then $R$ is finite over $R^G$.

Proof. Let $K\supset R$ and $K^G\supset R^G$ denote the quotient fields. $K/K^G$ is a Galois extension with group $G$. Pick $r_1,\dots, r_n\in R$ that form a $K^G$-basis of $K$. Then any $r\in R$ can be written as $$r=\sum_i a_i r_i{\quad\mbox{where $a_i\in K^G$.}\quad}$$ Applying any $g\in G$ to it, we get a system of equations $$\sum_i g(r_i)a_i= g(r){\quad\mbox{for $g\in G$.}\quad}$$ We can view these as linear equations with unknowns $a_i$. The system determinant is $D:=\det_{i,g}\bigl(g(r_i)\bigr)$, which is nonzero since its square is the discriminant of $K/K^G$. $D$ is $G$-invariant up to sign, thus $D^2$ is $G$-invariant hence in $R^G$. By Kramer’s rule, $a_i\in D^{-2}R^G$, hence $R\subset D^{-2}\sum_i r_i R^G$.

In the opposite case, when the equivalence relation is nontrivial only on a proper subscheme, we have the following general result.

\[quot.by.pushout\] Let $X$ be a reduced scheme, $Z\subset X$ a closed, reduced subscheme and $R\rightrightarrows X$ a finite, set theoretic equivalence relation. Assume that $R$ is the identity on $R\setminus Z$ and that the geometric quotient $Z/R|_Z$ exists. Then $X/R$ exists and is given by the universal push-out diagram $$\begin{array}{ccc}
Z & \into & X\\
 \downarrow && \downarrow \\
Z/R|_Z & \into & X/R.
\end{array}$$

Proof. Let $Y$ denote the universal push-out (\[glue.thm.asp\]). Then $X\to Y$ is finite and so $X/R$ exists and we have a natural map $X/R\to Y$ by (\[quot.X/S.finite.lem\]). On the other hand, there is a natural map $Z/R|_Z\to X/R$ by (\[quot.of.sub\]), hence the universal property of the push-out gives the inverse $Y\to X/R$.

Inductive plan for constructing quotients {#induct.plan.sect}
=========================================

\[up-down.eq.defn\] Let $R\rightrightarrows X$ be a finite, set theoretic equivalence relation and $g:Y\to X$ a finite morphism. Then $$g^*R:= R\times_{(X\times X)}(Y\times Y)\rightrightarrows Y$$ defines a finite, set theoretic equivalence relation on $Y$. It is called the [*pull-back*]{} of $R\rightrightarrows X$. (Strictly speaking, it should be denoted by $(g\times g)^*R$.)

Note that the $g^*R$-equivalence classes on the geometric points of $Y$ map injectively to the $R$-equivalence classes on the geometric points of $X$.

If $X/R$ exists then, by (\[quot.X/S.finite.lem\]), $Y/g^*R$ also exists and the natural morphism $Y/g^*R\to X/R$ is injective on geometric points. If, in addition, $g$ is surjective then $Y/g^*R\to X/R$ is a finite and universal homeomorphism (\[univ.homeo.defn\]). Thus, if $X$ is seminormal and the characteristic is 0, then $Y/g^*R\cong X/R$.

Let $h:X\to Z$ be a finite morphism. If the geometric fibers of $h$ are subsets of $R$-equivalence classes, then the composite $R\rightrightarrows X\to Z$ defines a finite, set theoretic pre-equivalence relation $$h_*R:=(h\times h)(R)\subset Z\times Z,$$ called the [*push forward*]{} of $R\rightrightarrows X$. If $Z/R$ exists, then, by (\[quot.X/S.finite.lem\]), $X/R$ also exists and the natural morphism $X/R\to Z/R$ is a finite and universal homeomorphism.

\[norm.quot=quot\] Let $X$ be weakly normal, excellent and $R\rightrightarrows X$ a finite, set theoretic equivalence relation. Let $\pi:X^n\to X$ be the normalization and $R^n\rightrightarrows X^n$ the pull back of $R$ to $X^n$. If $X^n/R^n$ exists then $X/R$ also exists and $X/R=X^n/R^n$.

Proof. Let $X^*\subset \bigl(X^n/R^n\bigr)\times_S X$ be the image of $X^n$ under the diagonal morphism. Since $X^n\to X$ is a finite surjection, $X^*$ is a closed subscheme of $\bigl(X^n/R^n\bigr)\times_S X$ and $X^*\to X$ is a finite surjection. Moreover, for any geometric point $\bar x\to X$, its preimages $\bar x_i\to X^n$ are $R^n$-equivalent, hence they map to the same point in $\bigl(X^n/R^n\bigr)\times_S X$. Thus $X^*\to X$ is a finite and one-to-one on geometric points, so it is a finite and universal homeomorphism (\[univ.homeo.defn\]). $X^n\to X$ is a local isomorphism at the generic point of every irreducible component of $X$, hence $X^*\to X$ is also a local isomorphism at the generic point of every irreducible component of $X$. Since $X$ is weakly normal, $X^*\cong X$ and we have a morphism $X\to X^n/R^n$ and thus $X/R=X^n/R^n$.

\[normal.subrel.lem\] Let $X$ be normal and of pure dimension $d$. Let $\sigma:R\rightrightarrows X$ be a finite, set theoretic equivalence relation and $R^d\subset R$ its $d$-dimensional part. Then $\sigma^d:R^d\rightrightarrows X$ is also an equivalence relation.

Proof. The only question is transitivity. Since $X$ is normal, the maps $\sigma^d:R^d\rightrightarrows X$ are both universally open by Chevalley’s criterion, see [@EGA IV.14.4.4]. Thus the fiber product $R^d\times_XR^d\to X$ is also universally open and hence its irreducible components have pure dimension $d$.

Let $C$ be a curve with an involution $\tau$. Pick $p,q\in C$ with $q$ different from $p$ and $\tau(p)$. Let $C'$ be the nodal curve obtained from $C$ by identifying $p$ and $q$. The equivalence relation generated by $\tau$ on $C'$ consists of the diagonal, the graph of $\tau$ plus the pairs $\bigl(\tau(p),\tau(q)\bigr)$ and $\bigl(\tau(q),\tau(p)\bigr)$. The 1-dimensional parts of the equivalence relation do not form an equivalence relation.

\[induct.plan\] Let $X$ be an excellent scheme that satisfies one of the conditions (\[quot.pure.dim.lem\].1–3) and $R\rightrightarrows X$ a finite, set theoretic equivalence relation. We aim to construct the geometric quotient $X/R$ in two steps. First we construct a space that, roughly speaking, should be the normalization of $X/R$ and then we try to go from the normalization to the geometric quotient itself.

[*Step 1.*]{} Let $X^n\to X$ be the normalization of $X$ and $R^n\rightrightarrows X^n$ the pull back of $R$ to $X^n$. Set $d=\dim X$ and let $X^{nd}\subset X^n$ (resp. $R^{nd}\subset R^n$) denote the union of the $d$-dimensional irreducible components. By (\[normal.subrel.lem\]), $R^{nd}\rightrightarrows X^{nd}$ is a pure dimensional, finite, set theoretic equivalence relation and the geometric quotient $X^{nd}/R^{nd}$ exists by (\[quot.pure.dim.lem\]).

There is a closed, reduced subscheme $Z\subset X^n$ of dimension $<d$ such that $Z$ is closed under $R^n$ and the two equivalence relations $$R^n|_{X^n\setminus Z}{\quad\mbox{and}\quad}
R^{nd}|_{X^n\setminus Z} {\quad\mbox{coincide.}\quad}$$ Let $Z_1\subset  X^{nd}/R^{nd}$ denote the image of $Z$. $R^n|_Z\rightrightarrows Z$ gives a finite set theoretic equivalence relation on $Z$. Since the geometric fibers of $Z\to Z_1$ are subsets of $R^n$-equivalence classes, by (\[up-down.eq.defn\]), the composite maps $R^n|_Z\rightrightarrows Z\to Z_1$ define a finite set theoretic pre-equivalence relation on $Z_1$.

[*Step 2.*]{} In order to go from $  X^{nd}/R^{nd}$ to $X/R$, we make the following

[*Inductive assumption*]{} (\[induct.plan\].2.1). The geometric quotient $Z_1/\bigl(R^{n}|_Z\bigr)$ exists.

Then, by (\[quot.by.pushout\]) $X^n/R^n$ exists and is given as the universal push-out of the following diagram: $$\begin{array}{ccc}
Z_1 & \into & X^n/R^{nd}\\
\downarrow &&\downarrow \\
Z_1/\bigl(R^{n}|_Z\bigr) &\into & X^n/R^n.
\end{array}$$

As in (\[norm.quot=quot\]), let $X^*\subset \bigl(X^n/R^n\bigr)\times_S X$ be the image of $X^n$ under the diagonal morphism. We have established that $X^*\to X$ is a finite and universal homeomorphism (\[univ.homeo.defn\]) sitting in the following diagram: $$\begin{array}{ccccccc}
Z_1 & \into & X^n/R^{nd} & \leftarrow & X^n &&\\
\downarrow && \downarrow &\swarrow & \downarrow & \searrow &\\
Z_1/\bigl(R^{n}|_Z\bigr) & \to & X^n/R^n & \leftarrow & X^* & \to & X
\end{array}
\eqno{(\ref{induct.plan}.2.2)}$$

There are now two ways to proceed.

[*Positive characteristic*]{} (\[induct.plan\].2.3). Most finite and universal homeomorphisms can be inverted, up to a power of the Frobenius (\[factor.through.frob.prop\]), and so we obtain a morphism $$X\to \bigl(X^*\bigr)^{(q)}\to \bigl(X^n/R^n\bigr)^{(q)}$$ for some $q=p^m$. $X/R$ is then obtained using (\[quot.X/S.finite.lem\]). This is discussed in Section \[pos.char.sec\].

In this case the inductive assumption (\[induct.plan\].2.1) poses no extra problems.

[*Zero characteristic*]{} (\[induct.plan\].2.4). As the examples of Section \[first.exmp.sec\] show, finite and universal homeomorphisms cause a substantial problem. The easiest way to overcome these difficulties is to assume to start with that $X$ is seminormal. In this case, by (\[norm.quot=quot\]), we obtain that $X/R=X^n/R^n$.

Unfortunately, the inductive assumption (\[induct.plan\].2.1) becomes quite restrictive. By construction $Z_1$ is reduced, but it need not be seminormal in general. Thus we get the induction going only if we can guarantee that $Z_1$ is seminormal. Note that, because of the inductive set-up, seminormality needs to hold not only for $X$ and $Z_1$, but on further schemes that one obtains in applying the inductive proof to $R^{n}|_Z\rightrightarrows Z_1$, and so on.

It turns out, however, that the above inductive plan works when gluing semi-log-canonical schemes. This will be treated elsewhere.

\[monom.defn\] A morphism of schemes $f:X\to Y$ is a [*monomorphism*]{} if for every scheme $Z$ the induced map of sets ${\operatorname{Mor}}(Z,X)\to {\operatorname{Mor}}(Z,Y)$ is an injection.

By [@EGA IV.17.2.6] this is equivalent to assuming that $f$ be universally injective and unramified.

A proper monomorphism $f:Y\to X$ is a closed embedding. Indeed, a proper monomorphism is injective on geometric points, hence finite. Thus it is a closed embedding iff ${{\mathcal O}}_X\to f_*{{\mathcal O}}_Y$ is onto. By the Nakayama lemma this is equivalent to $f_x:f^{-1}(x)\to x$ being an isomorphism for every $x\in f(Y)$. By passing to geometric points, we are down to the case when $X={\operatorname{Spec}}k$, $k$ is algebraically closed and $Y={\operatorname{Spec}}A$ where $A$ is an Artin $k$-algebra.

If $A\neq k$ then there are at least 2 different $k$ maps $A\to k[\epsilon]$, thus ${\operatorname{Spec}}A\to {\operatorname{Spec}}k$ is not a monomorphism.

\[univ.homeo.defn\] We say that a morphism of schemes $g:U\to V$ is a [*universal homeomorphism*]{} if it is a homeomorphism and for every $W\to V$ the induced morphism $U\times_VW\to W$ is again a homeomorphism. The definition extends to morphisms of algebraic spaces the usual way [@Knutson71 II.3].

A simple example of a homeomorphism which is not a universal homeomorphism is ${\operatorname{Spec}}K\to {\operatorname{Spec}}L$ where $L/K$ is a finite field extension and $L\neq K$. A more interesting example is given by the normalization of the nodal curve $\bigl(y^2=x^2(x+1)\bigr)$ with one of the preimages of the node removed: $${{\mathbb A}}^1\setminus \{-1\}\to \bigl(y^2=x^2(x+1)\bigr)
{\quad\mbox{given by}\quad} t\mapsto \bigl(t^2-1, t(t^2-1)\bigr).$$ When $g$ is finite, the notion is pretty much set theoretic since a continuous proper map of topological spaces which is injective and surjective is a homeomorphism. Thus we see that for a finite and surjective morphism of algebraic spaces $g:U\to V$ the following are equivalent (cf.  [@ega71 I.3.7–8])

1.  $g$ is a universal homeomorphism.

2.  $g$ is surjective and universally injective.

3.  For every $v\in V$ the fiber $g^{-1}(v)$ has a single point $v'$ and $k(v')$ is a purely inseparable field extension of $k(v)$.

4.  $g$ is surjective and injective on geometric points.

One of the most important properties of these morphisms is that taking the fiber product induces an equivalence between the categories $$(\mbox{\'etale morphisms: $\ast\to V$})
\stackrel{\ast\mapsto \ast\times_VU}{\longrightarrow}
(\mbox{\'etale morphisms: $\ast\to U$}).$$ See [@sga1 IX.4.10] for a proof. We do not use this in the sequel.

In low dimensions one can start the method (\[induct.plan\]) and it gives the following. These results are sufficient to deal with the moduli problem for surfaces.

\[quot.by.R.lowdim\] Let $S$ be a Noetherian scheme over a field of characteristic 0 and $X$ an algebraic space of finite type over $S$. Let $R\rightrightarrows X$ be a finite, set theoretic equivalence relation. Assume that one of the following holds:

1.  $X$ is 1-dimensional and reduced,

2.  $X$ is 2-dimensional and seminormal,

3.  $X$ is 3-dimensional, normal and there is a closed, seminormal $Z\subset X$ such that $R$ is the identity on $X\setminus Z$.

Then the geometric quotient $X/R$ exists.

Proof. Consider first the case when $\dim X=1$. Let $\pi:X^n\to X$ be the normalization. We construct $X^n/R^{nd}$ as in (\[induct.plan\]). Note that since $Z$ is zero dimensional, it is finite over $S$. Let $V\subset S$ be its image. Next we make a different choice for $Z_1$. Instead, we take a subscheme $Z_2\subset X^n/R^{nd}$ whose support is $Z_1$ such that the pull back of its ideal sheaf $I(Z_2)$ to $X^n$ is a subsheaf of the inverse image sheaf $\pi^{-1}{{\mathcal O}}_X\subset {{\mathcal O}}_{X^n}$.

Then we consider the push-out diagram $$V\leftarrow Z_2\into X^n/R^{nd}$$ with universal push-out $Y$. Then $X\to Y$ is a finite morphism and $X/R$ exists by (\[quot.X/S.finite.lem\]).

The case when $\dim X=2$ and $X$ is seminormal is a direct consequence of (\[induct.plan\].2.4) since the inductive assumption (\[induct.plan\].2.1) is guaranteed by (\[quot.by.R.lowdim\].1).

If $\dim X=3$, then $X$ is already normal and $Z$ is seminormal by assumption. Thus $Z/\bigl(R|_Z\bigr)$ exists by (\[quot.by.R.lowdim\].2). Therefore $X/R$ is given by the push-out of $Z/\bigl(R|_Z\bigr)\leftarrow Z\into X$.

Quotients in positive characteristic {#pos.char.sec}
====================================

The main result of this section is the proof of (\[quot.by.R.charp\]).

[@sga5 XIV] \[frob.say\] Let $S$ be an ${{\mathbb F}}_p$-scheme. Fix $q=p^r$ for some natural number $r$. Then $a\mapsto a^q$ defines an ${{\mathbb F}}_p$-morphism $F^q:S\to S$. This can be extended to polynomials by the formula $$f=\sum a_Ix^I\mapsto f^{(q)}:= \sum a^q_Ix^I.$$

Let $U={\operatorname{Spec}}R$ be an affine scheme over $S$. Write $R={{\mathcal O}}_S[x_1,\dots,x_m]/(f_1,\dots,f_n)$ and set $$R^{(q)}:= {{\mathcal O}}_S[x^{(q)}_1,\dots,x^{(q)}_m]/(f^{(q)}_1,\dots,f^{(q)}_n)
{\quad\mbox{and}\quad}  U^{(q)}:=
{\operatorname{Spec}}R^{(q)},$$ where the $x^{(q)}_i$ are new variables. There are natural morphisms $$F^q:U\to U^{(q)}{\quad\mbox{and}\quad} (F^q)^*:R^{(q)}\to R{\quad\mbox{given by}\quad}
(F^q)^*(x^{(q)}_i)=x_i^q.$$

It is easy to see that these are independent of the choices made. Thus $F^q$ gives a functor from algebraic spaces over $S$ to algebraic spaces over $S$. One can define $X^{(q)}$ intrinsically as $$X^{(q)}=X\times_{S,F^q} S.$$

If $X$ is an algebraic space which is essentially of finite type over ${{\mathbb F}}_p$ then $F^q:X\to X^{(q)}$ is a finite and universal homeomorphism.

For us the most important feature of the Frobenius morphism is the following universal property:

\[factor.through.frob.prop\] Let $S$ be a scheme essentially of finite type over ${{\mathbb F}}_p$ and $X,Y$ algebraic spaces which are essentially of finite type over $S$. Let $g:X\to Y$ be a finite and universal homeomorphism. Then for $q=p^r\gg 1$ the map $F^q$ can be factored as $$F^q:X\stackrel{g}{\to}Y \stackrel{\bar g}{\to} X^{(q)}.$$

Moreover, for large enough $q$ (depending on $g:X\to Y$), there is a functorial choice of the factorization in the sense that if $$\begin{array}{ccc}
X_1& \stackrel{g_1}{\to} & Y_1\\
\downarrow && \downarrow\\
X_2& \stackrel{g_2}{\to} & Y_2\\
\end{array}$$ is a commutative diagram where the $g_i$ are finite and universal homeomorphisms, then, for $q\gg 1$ (depending on the $g_i:X_i\to Y_i$) the factorization gives a commutative diagram $$\begin{array}{ccccc}
X_1& \stackrel{g_1}{\to} & Y_1& \stackrel{\bar g_1}{\to} & X_1^{(q)}\\
\downarrow && \downarrow && \downarrow\\
X_2& \stackrel{g_2}{\to} & Y_2& \stackrel{\bar g_2}{\to} & X_2^{(q)}.
\end{array}$$

Proof. It is sufficient to construct the functorial choice of the factorization in case $X$ and $Y$ are affine schemes over an affine scheme ${\operatorname{Spec}}C$. Thus we have a ring homomorphism $g^*:A\to B$ where $A$ and $B$ are finitely generated $C$-algebras. We can decompose $g^*$ into $A\twoheadrightarrow B_1$ and $B_1\hookrightarrow
B$. We deal with them separately.

First consider $B_1\subset B$. In this case there is no choice involved and we need to show that there is a $q$ such that $B^q\subset B_1$, where $B^q$ denotes the $C$-algebra generated by the $q$-th powers of all elements. The proof is by Noetherian induction.

First consider the case when $B$ is Artinian. The residue field of $B$ is finite and purely inseparable over the residue field of $B_1$, hence $B^q$ is contained in a field of representatives of $B_1$ for large enough $q$.

In the general, we can use the Artinian case over the generic points to obtain that $B_1\subset B_1B^q$ is an isomorphism at all generic points for $q\gg 1$. Let $I\subset B_1$ denote the conductor of this extension. That is, $IB_1B^q=I$. By induction we know that there is a $q'$ such that $(B_1B^q/I)^{q'}\subset B_1/I$. Thus we get that $$B^{(qq')}\to B^{qq'}\subset (B_1B^q)^{q'}\subset B_1+IB_1B^q=B_1.$$

Next consider $A\twoheadrightarrow  B_1$. Here we have to make a good choice. The kernel is a nilpotent ideal $I\subset A$, say $I^m=0$. Choose $q'$ such that $q'\geq
m$. For $b_1\in B_1$ let $b_1'\in A$ be any preimage. Then $(b'_1)^{q'}$ depends only on $b_1$. The map $$b_1\mapsto (b'_1)^{q'}{\quad\mbox{defines a factorization}\quad} 
B_1^{(q')}\to A\to B_1.$$ Combining the map $B^{(q)}\to B_1$ with $B_1^{(q')}\to A$ we obtain $B^{(qq')}\to A$.

The question is local on $S$, hence we may assume that $S$ is affine. $X$ and $R$ are defined over a finitely generated subring of ${{\mathcal O}}_S$, hence we may assume that $S$ is of finite type over ${{\mathbb F}}_p$.

The proof is by induction on $\dim X$. We follow the inductive plan in (\[induct.plan\]) and use its notation.

If $\dim X=0$ then $X$ is finite over $S$ and the assertion follows from (\[quot.X/S.finite.lem\]).

In going from dimension $d-1$ to $d$, the assumption (\[induct.plan\].2.1) holds by induction. Thus (\[induct.plan\].2.3) shows that $X^n/R^n$ exists.

Let $X^*\subset (X^n/R^n)\times_S X$ be the image of $X^n$ under the diagonal morphism. As we noted in (\[induct.plan\]), $X^*\to X$ is a finite and universal homeomorphism. Thus, by (\[factor.through.frob.prop\]), there is a factorization $$X^*\to X\to {X^*}^{(q)}\to \bigl(X^n/R^n\bigr)^{(q)}.$$ Here $X\to \bigl(X^n/R^n\bigr)^{(q)}$ is finite and $R$ is an equivalence relation on $X$ over the base scheme $\bigl(X^n/R^n\bigr)^{(q)}$. Hence, by (\[quot.X/S.finite.lem\]), the geometric quotient $X/R$ exists.

Some of the scheme theoretic aspects of the purely inseparable case are treated in [@eke] and [@sga3 Exp.V].

Gluing or Pinching {#glue.sec}
==================

The aim of this section is to give an elementary proof of the following.

[@artin70 Thm.3.1] \[glue.thm.asp\] Let $X$ be a Noetherian algebraic space over a Noetherian base scheme $S$. Let $Z\subset X$ be a closed subspace. Let $g:Z\to V$ be a finite surjection. Then there is a universal push-out diagram of algebraic spaces $$\begin{array}{ccl}
Z & \into & X\\
g \downarrow\hphantom{g} && \ \downarrow \pi\\
V & \into & Y:=X/(Z\to V)
\end{array}$$ Furthermore,

1.  $Y$ is a Noetherian algebraic space over $S$

2.  $V\to Y$ is a closed embedding and $Z=\pi^{-1}(V)$,

3.  the natural map $\ker\bigl[{{\mathcal O}}_Y\to {{\mathcal O}}_V\bigr] \to \pi_*\ker\bigl[{{\mathcal O}}_X\to {{\mathcal O}}_Z\bigr]$ is an isomorphism, and

4.  if $X$ is of finite type over $S$ then so is $Y$.

If $X$ is of finite type over $A$ and $A$ itself is of finite type over a field or an excellent Dedekind ring, then this is an easy consequence of the contraction results [@artin70 Thm.3.1]. The more general case above follows using the later approximation results [@pop]. The main point of [@artin70] is to understand the case when $Z\to V$ is proper but not finite. This is much harder than the finite case we are dealing with. An elementary approach following [@ferrand] and [@raoult] is discussed below.

\[glue.lem.affine\] The affine case of (\[glue.thm.asp\]) is simple algebra. Indeed, let $R={{\mathcal O}}_X$, $I=I(Z)$, $q:{{\mathcal O}}_X\to {{\mathcal O}}_Z$ the restriction and $S={{\mathcal O}}_V$. By (\[eak-nag\]), $q^{-1}(S)$ is Noetherian. Set $Y:={\operatorname{Spec}}q^{-1}(S)$.

If $\bar r_i\in {{\mathcal O}}_X/I(Z)$ generate ${{\mathcal O}}_X/I(Z)$ as an ${{\mathcal O}}_V$-module then $r_i\in {{\mathcal O}}_X$ and $I(Z)$ generate ${{\mathcal O}}_X$ as a $q^{-1}({{\mathcal O}}_V)$-module. Since $I(Z)\subset q^{-1}(S)$, we obtain that $r_i\in {{\mathcal O}}_X$ and $1\in {{\mathcal O}}_X$ generate ${{\mathcal O}}_X$ as a $q^{-1}(S)$-module. Applying (\[eak-nag\]) to $R_1={{\mathcal O}}_X$ and $R_2=q^{-1}({{\mathcal O}}_V)$ gives the rest.

For the proof of the following result, see [@mats-cr Thm.3.7] and the proof of (\[inv.of.fin.gps\]).

\[eak-nag\] Let $R_1\supset R_2$ be $A$-algebras with $A$ Noetherian. Assume that $R_1$ is finite over $R_2$.

1.  If $R_1$ is Noetherian then so is $R_2$.

2.  If $R_1$ is a finitely generated $A$-algebra then so is $R_2$.

Gluing for algebraic spaces, following [@raoult], is easier than the quasi projective case.

\[pf.of.glue.thm.asp\] For every $p\in V$ we construct a commutative diagram $$\begin{array}{ccccc}
V_p & \stackrel{g_p}{\leftarrow} & Z_p & \to & X_p\\
\tau_V\downarrow\hphantom{\tau_V} && 
\hphantom{\tau_Z}\downarrow\tau_Z &&\hphantom{\tau_X}\downarrow\tau_X \\
V & \stackrel{g}{\leftarrow} & Z & \to & X
\end{array}$$ where

1.  $V_p, Z_p, X_p$ are affine,

2.  $g_p$ is finite and $Z_p  \to  X_p$ is a closed embedding,

3.  $V_p$ (resp. $Z_p, X_p$) is an étale neighborhood of $p$ (resp.  $g^{-1}(p)$) and

4.  both squares are fiber products.

Affine gluing (\[glue.lem.affine\]) then gives $Y_p:=X_p/(Z_p\to V_p)$ and (\[glue.etloc.lem\]) shows that the $Y_p$ are étale charts on $Y=X/(Z\to V)$.

Start with affine, étale neighborhoods $V_1\to V$ of $p$ and $X_1\to X$ of $g^{-1}(p)$. Set $Z_1:=Z\times_XX_1\subset X_1$. By (\[et.nbhds.say\]) we may assume that there is a (necessarily étale) morphism $Z\times_VV_1\to Z_1$. In general there is no étale neighborhood $X'\to X_1$ extending $Z\times_VV_1\to Z_1$, but there is an affine, étale neighborhood $X_2\to X_1$ extending $Z\times_VV_1\to Z_1$ over a Zariski neighborhood of $g^{-1}(p)$ (\[et.nbhds.say\]).

Thus we have affine, étale neighborhoods $V_2\to V$ of $p$, $X_2\to X$ of $g^{-1}(p)$ and an open embedding $ Z\times_XX_2\into Z_2:=Z\times_VV_2$. Our only remaining problem is that $Z_2\neq Z\times_XX_2$, hence $Z_2$ is not a subscheme of $X_2$. We achieve this by further shrinking $V_2$ and $X_2$.

The complement $B_2:=Z_2\setminus Z\times_XX_2$ is closed, thus $g(B_2)\subset V_2$ is a closed subset not containing $p$. Pick $\phi\in \Gamma({{\mathcal O}}_{V_2})$ that vanishes on $g(B_2)$ such that $\phi(p)\neq 0$. Then $\phi\circ g$ is a function on $Z_2$ that vanishes on $B_2$ but is nowhere zero on $g^{-1}(p)$. We can thus extend $\phi\circ g$ to a function $\Phi$ on $X_2$. Thus $V_P:=V_2\setminus (\phi=0), Z_P:=Z_2\setminus (\phi\circ g=0)$ and $X_P:=X_2\setminus (\Phi=0)$ have the required properties.

\[et.nbhds.say\] During the proof we have used two basic properties of étale neighborhoods.

First, if $\pi:X\to Y$ is finite then for every étale neighborhood $(u\in U)\to (x\in X)$ there is an étale neighborhood $(v\in V)\to (\pi(x)\in Y)$ and a connected component $(v'\in V')\subset X\times_YV$ such that there is a lifting $(v'\in V')\to (u\in U)$.

Second, if $\pi:X\to Y$ is a closed embedding, $U\to X$ is étale and $P\subset U$ is a finite set of points then we can find an étale $V\to Y$ such that $P\subset V$ and there is an open embedding $(P\subset X\times_YV)\to (P\subset U)$.

For proofs see [@milne 3.14 and 4.2–3].

The next result shows that gluing commutes with flat morphisms.

\[glue.etloc.lem\] For $i=1,2$, let $X_i$ be Noetherian affine $A$-schemes, $Z_i\subset X_i$ closed subschemes and $g_i:Z_i\to V_i$ finite surjections with universal push-outs $Y_i$. Assume that in the diagram below both squares are fiber products. $$\begin{array}{ccccc}
V_1 & \stackrel{g_1}{\leftarrow} & Z_1 & \to & X_1\\
\downarrow && \downarrow && \downarrow \\
V_2 & \stackrel{g_2}{\leftarrow} & Z_2 & \to & X_2
\end{array}$$

1.  If the vertical maps are flat then $Y_1 \to Y_2$ is also flat.

2.  If the vertical maps are smooth then $Y_1 \to Y_2$ is also smooth.

Proof. We may assume that all occurring schemes are affine. Thus we have $I_i\subset R_i$ and $S_i\subset R_i/I_i$. Furthermore, $R_1$ is flat over $R_2$, $I_1=I_2R_1$ and $S_1$ is flat over $S_2$. We may also assume that $R_2$ is local. The key point is the isomorphism $$\bigl( R_1/I_1\bigr)\cong 
\bigl( R_2/I_2\bigr)\otimes_{R_2}R_1\cong \bigl( R_2/I_2\bigr)\otimes_{S_2}S_1.
\eqno{(\ref{glue.etloc.lem}.3)}$$ Note that this isomorphism is not naturally given, see (\[twist-glue.rem\]).

We check the local criterion of flatness (cf. [@mats-cr Thm.22.3]). The first condition we need is that $q_1^{-1}(S_1)/I_1\cong S_1$ be flat over $q_2^{-1}(S_2)/I_2\cong S_2$. This holds by assumption. Second, we need that the maps $$\bigl( I_2^n/I_2^{n+1}\bigr)\otimes_{S_2}S_1\to 
I_2^nR_1/I_2^{n+1}R_1$$ be isomorphisms. Since $R_1$ is flat over $R_2$, the right hand side is isomorphic to $$\bigl(I_2^n/I_2^{n+1}\bigr)\otimes_{R_2/I_2} \bigl(R_1/I_1\bigr).$$ Using (\[glue.etloc.lem\].3), we get that $$\bigl(I_2^n/I_2^{n+1}\bigr)\otimes_{R_2/I_2} \bigl(R_1/I_1\bigr)\cong
\bigl(I_2^n/I_2^{n+1}\bigr)\otimes_{R_2/I_2} \bigl(R_2/I_2\bigr)
\otimes_{S_2}S_1\cong
\bigl(I_2^n/I_2^{n+1}\bigr)\otimes_{S_2}S_1.$$ This settles flatness. In order to prove the smooth case, we just need to check that the fibers of $Y_1 \to Y_2$ are smooth. Outside $V_1\to V_2$ we have the same fibers as before and $V_1\to V_2$ is smooth by assumption.

\[twist-glue.rem\] Note that there is some subtlety in (\[glue.etloc.lem\]). Consider the simple case when $X_2$ is a smooth curve over a field $k$, $Z_2=\{p,q\}$ two $k$-points and $V_2={\operatorname{Spec}}k$. Then $Y_2$ is a nodal curve where $p$ and $q$ are identified.

Let now $X_1=X_2\times\{0,1\}$ as 2 disjoint copies. Then $Z_1$ consists of 4 points $p_0,q_0,p_1,q_1$ and $V_1$ is 2 copies of ${\operatorname{Spec}}k$. There are two distinct way to arrange $g_1$. Namely,

1.  either $g'_1(p_0)=g'_1(q_0)$ and $g'_1(p_1)=g'_1(q_1)$ and then $Y'_1$ consists of 2 disjoint nodal curves,

2.  or $g''_1(p_0)=g''_1(q_1)$ and $g''_1(p_1)=g''_1(q_0)$ and then $Y''_1$ consists of a connected curve with 2 nodes and 2 irreducible components.

Both of these are étale double covers of $Y_2$.

As in (\[pf.of.glue.thm.asp\]), the next lemma will be used to reduce quasi projective gluing to the affine case.

\[affine.red.lem\] Let $X$ be an $A$-scheme, $Z\subset X$ a closed subscheme and $g:Z\to V$ a finite surjection.

Let $P\subset V$ be a finite subset and assume that there are open affine subsets $P\subset V_1\subset V$ and $g^{-1}(P)\subset X_1\subset X$.

Then there are open affine subsets $P\subset V_P\subset V_1$ and $g^{-1}(P)\subset X_P\subset X_1$ such that $g$ restricts to a finite morphism $g: Z\cap X_P\to V_P$.

Proof. There is an affine subset $g^{-1}(P)\subset X_2\subset X_1$ such that $g^{-1}(V\setminus V_1)$ is disjoint from $X_2$. Thus $g$ maps $Z\cap X_2$ to $V_1$. The problem is that $(Z\cap X_2)\to V_1$ is only quasi finite in general. The set $Z\setminus X_2$ is closed in $X$ and so $g( Z\setminus X_2)$ is closed in $V$. Since $V_1$ is affine, there is a function $f_P$ on $V_1$ which vanishes on $g( Z\setminus X_2)\cap V_1$ but does not vanish on $P$. Then $f_P\circ g$ is a function on $g^{-1}(V_1)$ which vanishes on $(Z\setminus X_2)\cap g^{-1}(V_1)$ but does not vanish at any point of $g^{-1}(P)$. Since $Z\cap X_1$ is affine, $f_P\circ g$ can be extended to a regular function $F_p$ on $X_2$.

Set $V_P:=V_1\setminus (f_P=0)$ and $X_P:=X_2\setminus (F_P=0)$. The restriction $(Z\cap X_P)\to V_P$ is finite since, by construction, $X_P\cap Z$ is the preimage of $V_P$.

We say that an algebraic space $X$ has the [*Chevalley-Kleiman property*]{} if $X$ is separated and every finite subscheme is contained in an open affine subscheme. In particular, $X$ is necessarily a scheme.

These methods give the following interesting corollary.

\[CK.prop.desc\] Let $\pi:X\to Y$ be a finite and surjective morphism of separated, excellent algebraic spaces. Then $X$ has the Chevalley-Kleiman property iff $Y$ has.

Proof. Assume that $Y$ has the Chevalley-Kleiman property and let $P\subset X$ be a finite subset. Since $\pi(P)\subset Y$ is finite, there is an open affine subset $Y_P\subset Y$ containing $\pi(P)$. Then $g^{-1}(Y_P)\subset X$ is an open affine subset containing $P$.

Conversely, assume that $X$ has the Chevalley-Kleiman property. By the already established direction, we may assume that $X$ is normal. Next let $ Y^n$ be the normalization of $Y$. Then $X\to  Y^n$ is finite and dominant. Fix irreducible components $X_1\subset X$ and $Y_1\subset Y^n$ such that the induced map $X_1\to Y_1$ is finite and dominant. Let $\pi'_1:X'_1\to X_1\to Y_1$ be the Galois closure of $X_1/Y_1$ with Galois group $G$. We already know that $X'_1$ has the Chevalley-Kleiman property, hence there is an open affine subset $X'_P\subset X_1$ containing $\bigl(\pi'_1\bigr)^{-1}(P)$. Then $U'_P:=\cap_{g\in G}g(X_P)\subset  X'_1$ is affine, Galois invariant and $\bigl(\pi'_1\bigr)^{-1}\bigl(\pi'_1(U'_P)\bigr)=U'_P$.

Thus $U'_P\to \pi'_1(U'_P)$ is finite and, by Chevalley’s theorem [@hartsh Exrc.III.4.2], $\pi'_1(U'_P)\subset  Y_1$ is an open affine subset containing $P$. Thus $Y^n$ has the Chevalley-Kleiman property.

Next consider the normalization map $g:Y^n\to {\operatorname{red}}Y$. There are lower dimensional closed subschemes $P\subset V\subset {\operatorname{red}}Y$ and $Z:=g^{-1}(V)\subset Y^n$ such that $g:Y^n\setminus Z\cong {\operatorname{red}}Y\setminus V$ is an isomorphism. By induction on the dimension, $V$ has the Chevalley-Kleiman property.

By (\[affine.red.lem\]) there are open affine subsets $P\subset V_P\subset V$ and $g^{-1}(P)\subset Y^n_P\subset Y^n$ such that $g$ restricts to a finite morphism $g: Z\cap Y^n_P\to V_P$. Thus, by (\[glue.lem.affine\]), $g\bigl(Y^n_P\bigr)\subset {\operatorname{red}}Y$ is open, affine and it contains $P$. Thus ${\operatorname{red}}Y$ has the Chevalley-Kleiman property.

Finally, ${\operatorname{red}}Y\to Y$ is a homeomorphism, thus if $U\subset {\operatorname{red}}Y$ is an affine open subset and $U'\subset Y$ the “same” open subset of $Y$ then $U'$ is also affine by Chevalley’s theorem and so $Y$ has the Chevalley-Kleiman property.

Let $E$ be an elliptic curve and set $S:=E\times {{\mathbb P}}^1$. Pick a general $p\in E$ and $g:E\times \{0,1\}\to E$ be the identity on $E_0:=E\times \{0\}$ and translation by $-p$ on $E_1:=E\times \{1\}$. Where are the affine charts on the quotient $Y$?

If $P_i\subset E_i$ are 0-cycles then there is an ample divisor $H$ on $S$ such that $(H\cdot E_i)=P_i$ iff ${{\mathcal O}}_{E_0}(P_0)={{\mathcal O}}_{E_1}(P_1)$ under the identity map $E_0\cong E_1$.

Pick any $a,b\in E_0$ and let $a+p,b+p\in E_1$ be obtained by translation by $p$. Assume next that $2a+b=a+p+2(b+p)$, or, equivalently, that $3p=a-b$. Let $H(a,b)$ be an ample divisor on $S$ such that $H(a,b)\cap E_0=\{a,b\}$ and $H(a,b)\cap E_1=\{a+p,b+p\}$. Then $U(a,b):=S\setminus H(a,b)$ is affine and $g$ maps $E_i\cap U(a,b)$ isomorphically onto $E\setminus\{a,b\}$ for $i=0,1$. As we vary $a,b$ (subject to $3p=a-b$) we get an affine covering of $Y$.

Note however that the curves $H(a,b)$ do not give Cartier divisors on $Y$. In fact, for non-torsion $p\in E$, every line bundle on $Y$ pull back from the nodal curve obtained from the ${{\mathbb P}}^1$ factor by gluing the points $0$ and $1$ together.

Appendix by Claudiu Raicu {#appendix-by-claudiu-raicu .unnumbered}
=========================

\[raicu\] Let $A$ be a noetherian commutative ring and $X={{\mathbb A}}_S^n$ the $n$-dimensional affine space over $S={\operatorname{Spec}}A$. Then ${{\mathcal O}}_X\simeq A[{\textit{\textbf{x}}}]$, where ${\textit{\textbf{x}}}=(x_1,\dots,x_n)$. To give a finite equivalence relation $R\subset X\times_S X$ is equivalent to giving an ideal $I({\textit{\textbf{x}}},{\textit{\textbf{y}}})\subset A[{\textit{\textbf{x}}},{\textit{\textbf{y}}}]$ which satisfies the following properties:

1.  (reflexivity) $I({\textit{\textbf{x}}},{\textit{\textbf{y}}})\subset (x_1-y_1,\dots,x_n-y_n)$.

2.  (symmetry) $I({\textit{\textbf{x}}},{\textit{\textbf{y}}})=I({\textit{\textbf{y}}},{\textit{\textbf{x}}}).$

3.  (transitivity) $I({\textit{\textbf{x}}},{\textit{\textbf{z}}})\subset I({\textit{\textbf{x}}},{\textit{\textbf{y}}})+I({\textit{\textbf{y}}},{\textit{\textbf{z}}})$ in $A[{\textit{\textbf{x}}},{\textit{\textbf{y}}},{\textit{\textbf{z}}}]$.

4.  (finiteness) $A[{\textit{\textbf{x}}},{\textit{\textbf{y}}}]/I({\textit{\textbf{x}}},{\textit{\textbf{y}}})$ is finite over $A[{\textit{\textbf{x}}}]$.

Suppose now that we have an ideal $I({\textit{\textbf{x}}},{\textit{\textbf{y}}})$ satisfying (1-4) and let $R$ be the equivalence relation it defines. If the geometric quotient exists, then by (\[cat.geom.quot\]) it is of the form ${\operatorname{Spec}}A[f_1,\dots,f_m]$ for some polynomials $f_1,\dots,f_m\in A[{\textit{\textbf{x}}}]$. It follows that $$I({\textit{\textbf{x}}},{\textit{\textbf{y}}})\supset (f_i({\textit{\textbf{x}}})-f_i({\textit{\textbf{y}}}):i=1,2,\dots,m)$$ and $R$ is [*effective*]{} iff equality holds.

We are mainly interested in the case when $A$ is ${{\mathbb Z}}$ or some field $k$ and $I$ is homogeneous. Write $I$ as $I({\textit{\textbf{x}}},{\textit{\textbf{y}}})=\bigl(J({\textit{\textbf{x}}},{\textit{\textbf{y}}}),f({\textit{\textbf{x}}},{\textit{\textbf{y}}})\bigr)$, where $J$ is an ideal of the form $$J({\textit{\textbf{x}}},{\textit{\textbf{y}}})=\bigl(f_i({\textit{\textbf{x}}})-f_i({\textit{\textbf{y}}}):i=1,2,\dots,m\bigr),$$ with homogeneous $f_i\in A[{\textit{\textbf{x}}}]$ such that $A[{\textit{\textbf{x}}}]$ is a finite module over $A[f_1,\dots,f_m]$ and $f\in A[{\textit{\textbf{x}}},{\textit{\textbf{y}}}]$ a homogeneous polynomial that satisfies the cocycle condition $$f({\textit{\textbf{x}}},{\textit{\textbf{y}}})+f({\textit{\textbf{y}}},{\textit{\textbf{z}}})-f({\textit{\textbf{x}}},{\textit{\textbf{z}}})\in J({\textit{\textbf{x}}},{\textit{\textbf{y}}})+J({\textit{\textbf{y}}},{\textit{\textbf{z}}})\subset A[{\textit{\textbf{x}}},{\textit{\textbf{y}}},{\textit{\textbf{z}}}].
\eqno{(\ref{raicu}.1)}$$ The reason we call (\[raicu\].1) a cocycle condition is the following. If we let $B=A[f_1,\dots,f_m]$, $C=A[{\textit{\textbf{x}}}]$ and consider the complex (starting in degree zero) $$C\to C\otimes_B C\to\cdots\to C^{\otimes_B m}\to\cdots
\eqno{(\ref{raicu}.2)}$$ with differentials given by the formula $$d_{m-1}(c_1\otimes c_2\otimes\cdots\otimes c_m)=
\sum_{i=1}^{m+1}(-1)^i c_1\otimes\cdots\otimes c_{i-1}\otimes 1\otimes c_i\otimes\cdots\otimes c_m,$$ then $C\otimes_B C\simeq A[{\textit{\textbf{x}}},{\textit{\textbf{y}}}]/J({\textit{\textbf{x}}},{\textit{\textbf{y}}})$, $C\otimes_B C\otimes_B C\simeq A[{\textit{\textbf{x}}},{\textit{\textbf{y}}},{\textit{\textbf{z}}}]/(J({\textit{\textbf{x}}},{\textit{\textbf{y}}})+J({\textit{\textbf{y}}},{\textit{\textbf{z}}}))$, and if the polynomial $f({\textit{\textbf{x}}},{\textit{\textbf{y}}})$ satisfies (\[raicu\].1), then its class in $C\otimes_B C$ is a 1-cocycle in the complex (\[raicu\].2).

Any ideal $I({\textit{\textbf{x}}},{\textit{\textbf{y}}})$ defined as above is the ideal of an equivalence relation. To show that the equivalence relation it defines is noneffective it suffices to check that $f({\textit{\textbf{x}}},{\textit{\textbf{y}}})$ is not congruent to a difference modulo $J({\textit{\textbf{x}}},{\textit{\textbf{y}}})$. This can be done using a computer algebra system by computing the finite $A$-module $U$ of homogeneous forms of the same degree as $f$ which are congruent to differences modulo $J$, and checking that $f$ is not contained in $U$. We used Macaulay 2 to check that the following example gives a noneffective equivalence relation (we took $A={{\mathbb Z}}$ and $n=2$):

$$f_1({\textit{\textbf{x}}})={\textit{\textbf{x}}}_1^2,\ f_2({\textit{\textbf{x}}})={\textit{\textbf{x}}}_1{\textit{\textbf{x}}}_2-{\textit{\textbf{x}}}_2^2,\ f_3({\textit{\textbf{x}}})={\textit{\textbf{x}}}_2^3,$$ $$f({\textit{\textbf{x}}},{\textit{\textbf{y}}})=({\textit{\textbf{x}}}_1{\textit{\textbf{y}}}_2-{\textit{\textbf{x}}}_2{\textit{\textbf{y}}}_1){\textit{\textbf{y}}}_2^3.$$ $$I({\textit{\textbf{x}}},{\textit{\textbf{y}}})=({\textit{\textbf{x}}}_1^2-{\textit{\textbf{y}}}_1^2,{\textit{\textbf{x}}}_1{\textit{\textbf{x}}}_2-{\textit{\textbf{x}}}_2^2-{\textit{\textbf{y}}}_1{\textit{\textbf{y}}}_2+{\textit{\textbf{y}}}_2^2,
{\textit{\textbf{x}}}_2^3-{\textit{\textbf{y}}}_2^3,({\textit{\textbf{x}}}_1{\textit{\textbf{y}}}_2-{\textit{\textbf{x}}}_2{\textit{\textbf{y}}}_1){\textit{\textbf{y}}}_2^3).$$

We also claim that this example remains noneffective after any base change. Indeed, the $A$-module $V$ generated by the forms of degree $5(=\deg(f))$ in $I$ and the differences $g({\textit{\textbf{x}}})-g({\textit{\textbf{y}}})$ with $g$ homogeneous of degree $5$, is a direct summand in $U$. Elements of $V$ correspond to 0-coboundaries in (\[raicu\].2). The module $W$ consisting of elements of $k[{\textit{\textbf{x}}},{\textit{\textbf{y}}}]_5$ whose classes in $k[{\textit{\textbf{x}}},{\textit{\textbf{y}}}]/J$ are 1-cocycles is also a direct summand in $U$. The quotient $W/V$ is a free ${{\mathbb Z}}$-module $H$ generated by the class of $f({\textit{\textbf{x}}},{\textit{\textbf{y}}})$. This shows that $W=V\oplus H$, hence for any field $k$ we have $W_k=V_k\oplus H_k$, where for an abelian group $G$ we let $G_k=G\otimes_{{{\mathbb Z}}}k$. If we denote by $d_i^k$ the differentials in the complex obtained from (\[raicu\].2) by base changing from ${{\mathbb Z}}$ to $k$, then we get that ${\operatorname{im}}d_0^k=V_k$ and $\ker d_1^k\supset W_k$. It follows that the nonzero elements of $H_k$ will represent nonzero cohomology classes in (\[raicu\].2) for any field $k$, hence our example is indeed universal.

One can see [@CR Lem.4.3] that all homogeneous noneffective equivalence relations are contained in a homogeneous noneffective equivalence relation constructed as above.

In the positive direction, we have the following result in the toric case, where a *toric equivalence relation* (over a field $k$) is a scheme theoretic equivalence relation $R$ on a (not necessarily normal) toric variety $X/k$ that is invariant under the diagonal action of the torus.

\[thmamit\] [@CR Thm.4.2] Let $k$ be a field, $X/k$ an affine toric variety, and $R$ a toric equivalence relation on $X$. Then there exists an affine toric variety $Y/k$ together with a toric map $X\to Y$ such that $R\simeq X\times_Y X$.

Notice that we do not require the equivalence relation to be finite.

We thank D. Eisenbud, M. Hashimoto, C. Huneke, K. Kurano, M. Lieblich, P. Roberts, Ch. Rotthaus, D. Rydh and R. Skjelnes for useful comments, corrections and references. Partial financial support was provided by the NSF under grant number DMS-0758275.

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Princeton University, Princeton NJ 08544

    kollar@math.princeton.edu

University of California, Berkeley, CA 94720-3840

Institute of Mathematics “Simion Stoilow" of the Romanian Academy

    claudiu@math.berkeley.edu
