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 $(