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
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
Denote $x_ K, y_ K, z_ K : \mathop{\mathrm{Spec}}(K) \to U$ the morphisms
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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: