Lemma 26.17.5. Let $f : X \to S$ and $g : Y \to S$ be morphisms of schemes with the same target. Points $z$ of $X \times _ S Y$ are in bijective correspondence to quadruples

\[ (x, y, s, \mathfrak p) \]

where $x \in X$, $y \in Y$, $s \in S$ are points with $f(x) = s$, $g(y) = s$ and $\mathfrak p$ is a prime ideal of the ring $\kappa (x) \otimes _{\kappa (s)} \kappa (y)$. The residue field of $z$ corresponds to the residue field of the prime $\mathfrak p$.

**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

\[ \xymatrix{ X \times _ S Y \ar@/_/[dddr] \ar@/^/[rrrd] & & & \\ & \mathop{\mathrm{Spec}}(\kappa (x) \otimes _{\kappa (s)} \kappa (y)/\mathfrak p) \ar[r] \ar[d] \ar@{-->}[lu] & \mathop{\mathrm{Spec}}(\kappa (y)) \ar[d] \ar[r] & Y \ar[dd] \\ & \mathop{\mathrm{Spec}}(\kappa (x)) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(\kappa (s)) \ar[rd] & \\ & X \ar[rr] & & S } \]

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.
$\square$

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