## 39.3 Equivalence relations

Recall that a *relation* $R$ on a set $A$ is just a subset of $R \subset A \times A$. We usually write $a R b$ to indicate $(a, b) \in R$. We say the relation is *transitive* if $a R b, b R c \Rightarrow a R c$. We say the relation is *reflexive* if $a R a$ for all $a \in A$. We say the relation is *symmetric* if $a R b \Rightarrow b R a$. A relation is called an *equivalence relation* if it is transitive, reflexive and symmetric.

In the setting of schemes we are going to relax the notion of a relation a little bit and just require $R \to A \times A$ to be a map. Here is the definition.

Definition 39.3.1. Let $S$ be a scheme. Let $U$ be a scheme over $S$.

A *pre-relation* on $U$ over $S$ is any morphism of schemes $j : R \to U \times _ S U$. In this case we set $t = \text{pr}_0 \circ j$ and $s = \text{pr}_1 \circ j$, so that $j = (t, s)$.

A *relation* on $U$ over $S$ is a monomorphism of schemes $j : R \to U \times _ S U$.

A *pre-equivalence relation* is a pre-relation $j : R \to U \times _ S U$ such that the image of $j : R(T) \to U(T) \times U(T)$ is an equivalence relation for all $T/S$.

We say a morphism $R \to U \times _ S U$ of schemes is an *equivalence relation on $U$ over $S$* if and only if for every scheme $T$ over $S$ the $T$-valued points of $R$ define an equivalence relation on the set of $T$-valued points of $U$.

In other words, an equivalence relation is a pre-equivalence relation such that $j$ is a relation.

Lemma 39.3.2. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j : R \to U \times _ S U$ be a pre-relation. Let $g : U' \to U$ be a morphism of schemes. Finally, set

\[ R' = (U' \times _ S U')\times _{U \times _ S U} R \xrightarrow {j'} U' \times _ S U' \]

Then $j'$ is a pre-relation on $U'$ over $S$. If $j$ is a relation, then $j'$ is a relation. If $j$ is a pre-equivalence relation, then $j'$ is a pre-equivalence relation. If $j$ is an equivalence relation, then $j'$ is an equivalence relation.

**Proof.**
Omitted.
$\square$

Definition 39.3.3. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j : R \to U \times _ S U$ be a pre-relation. Let $g : U' \to U$ be a morphism of schemes. The pre-relation $j' : R' \to U' \times _ S U'$ is called the *restriction*, or *pullback* of the pre-relation $j$ to $U'$. In this situation we sometimes write $R' = R|_{U'}$.

Lemma 39.3.4. Let $j : R \to U \times _ S U$ be a pre-relation. Consider the relation on points of the scheme $U$ defined by the rule

\[ x \sim y \Leftrightarrow \exists \ r \in R : t(r) = x, s(r) = y. \]

If $j$ is a pre-equivalence relation then this is an equivalence relation.

**Proof.**
Suppose that $x \sim y$ and $y \sim z$. Pick $r \in R$ with $t(r) = x$, $s(r) = y$ and pick $r' \in R$ with $t(r') = y$, $s(r') = z$. Pick a field $K$ fitting into the following commutative diagram

\[ \xymatrix{ \kappa (r) \ar[r] & K \\ \kappa (y) \ar[u] \ar[r] & \kappa (r') \ar[u] } \]

Denote $x_ K, y_ K, z_ K : \mathop{\mathrm{Spec}}(K) \to U$ the morphisms

\[ \begin{matrix} \mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(\kappa (r)) \to \mathop{\mathrm{Spec}}(\kappa (x)) \to U
\\ \mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(\kappa (r)) \to \mathop{\mathrm{Spec}}(\kappa (y)) \to U
\\ \mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(\kappa (r')) \to \mathop{\mathrm{Spec}}(\kappa (z)) \to U
\end{matrix} \]

By construction $(x_ K, y_ K) \in j(R(K))$ and $(y_ K, z_ K) \in j(R(K))$. Since $j$ is a pre-equivalence relation we see that also $(x_ K, z_ K) \in j(R(K))$. This clearly implies that $x \sim z$.

The proof that $\sim $ is reflexive and symmetric is omitted.
$\square$

Lemma 39.3.5. Let $j : R \to U \times _ S U$ be a pre-relation. Assume

$s, t$ are unramified,

for any algebraically closed field $k$ over $S$ the map $R(k) \to U(k) \times U(k)$ is an equivalence relation,

there are morphisms $e : U \to R$, $i : R \to R$, $c : R \times _{s, U, t} R \to R$ such that

\[ \xymatrix{ U \ar[r]_ e \ar[d]_\Delta & R \ar[d]_ j & R \ar[d]^ j \ar[r]_ i & R \ar[d]^ j & R \times _{s, U, t} R \ar[d]^{j \times j} \ar[r]_ c & R \ar[d]^ j \\ U \times _ S U \ar[r] & U \times _ S U & U \times _ S U \ar[r]^{flip} & U \times _ S U & U \times _ S U \times _ S U \ar[r]^{\text{pr}_{02}} & U \times _ S U } \]

are commutative.

Then $j$ is an equivalence relation.

**Proof.**
By condition (1) and Morphisms, Lemma 29.35.16 we see that $j$ is a unramified. Then $\Delta _ j : R \to R \times _{U \times _ S U} R$ is an open immersion by Morphisms, Lemma 29.35.13. However, then condition (2) says $\Delta _ j$ is bijective on $k$-valued points, hence $\Delta _ j$ is an isomorphism, hence $j$ is a monomorphism. Then it easily follows from the commutative diagrams that $R(T) \subset U(T) \times U(T)$ is an equivalence relation for all schemes $T$ over $S$.
$\square$

## Comments (6)

Comment #40 by Pieter Belmans on

Comment #41 by Johan on

Comment #2240 by clarifications on

Comment #2275 by Johan on

Comment #6572 by Hans Schoutens on

Comment #6573 by Johan on