The Stacks project

Proof. Let $z$ be a point of $X \times _ S Y$ and let us construct a quadruple as above. Recall that we may think of $z$ as a morphism $\mathop{\mathrm{Spec}}(\kappa (z)) \to X \times _ S Y$, see Lemma 26.13.3. This morphism corresponds to morphisms $a : \mathop{\mathrm{Spec}}(\kappa (z)) \to X$ and $b : \mathop{\mathrm{Spec}}(\kappa (z)) \to Y$ such that $f \circ a = g \circ b$. By the same lemma again we get points $x \in X$, $y \in Y$ lying over the same point $s \in S$ as well as field maps $\kappa (x) \to \kappa (z)$, $\kappa (y) \to \kappa (z)$ such that the compositions $\kappa (s) \to \kappa (x) \to \kappa (z)$ and $\kappa (s) \to \kappa (y) \to \kappa (z)$ are the same. In other words we get a ring map $\kappa (x) \otimes _{\kappa (s)} \kappa (y) \to \kappa (z)$. We let $\mathfrak p$ be the kernel of this map.

Conversely, given a quadruple $(x, y, s, \mathfrak p)$ we get a commutative solid diagram

see the discussion in Section 26.13. Thus we get the dotted arrow. The corresponding point $z$ of $X \times _ S Y$ is the image of the generic point of $\mathop{\mathrm{Spec}}(\kappa (x) \otimes _{\kappa (s)} \kappa (y)/\mathfrak p)$. We omit the verification that the two constructions are inverse to each other. $\square$