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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 $S$.

  1. 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 $j = (t, s)$.

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

  3. 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 $T/S$.

  4. 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

  1. $s, t$ are unramified,

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

  3. 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 on

There is a space missing at the start of the second sentence in this chapter.

Comment #2240 by clarifications on

In Definition 38.3.1:

  1. Is R a scheme?
  2. Is an \emph{equivalence relation} a relation on U such that ...? (Or is it a pre-relation such that ...?)

Comment #2275 by on

Dear Clarifications, your first question has an affirmative answer. In the chapters on schemes (Schemes, Properties, Morphisms, etc) we often say "let be a morphism" without saying explicitly that and are schemes. Most of the time there should be no possible confusion. Since this caused you to want to clarify I have added the relevant words in this commit.

For your second question: I have added that is a morphism of schemes, etc. Then it is a small lemma that indeed an equivalence relation is a relation (in fact is is also a pre-relation and a pre-equivalence relation of course).

Comment #6572 by Hans Schoutens on

Reply to comment #2275: the issue seems to be why does (4) imply that is a monomorphism (which is implied by the phrase "In other words". If it is a small Lemma, and doesn't seem to be obvious enough to not inlcude the lemma.

Comment #6573 by on

In any category: monomorphism map of representable presheaves injective

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