## Tag `01JT`

Chapter 25: Schemes > Section 25.17: Fibre products of schemes

Lemma 25.17.5. 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 triple as above. Recall that we may think of $z$ as a morphism $\mathop{\rm Spec}(\kappa(z)) \to X \times_S Y$, see Lemma 25.13.3. This morphism corresponds to morphisms $a : \mathop{\rm Spec}(\kappa(z)) \to X$ and $b : \mathop{\rm Spec}(\kappa(z)) \to Y$ 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 $$ \xymatrix{ X \times_S Y \ar@/_/[dddr] \ar@/^/[rrrd] & & & \\ & \mathop{\rm Spec}(\kappa(x) \otimes_{\kappa(s)} \kappa(y)/\mathfrak p) \ar[r] \ar[d] \ar@{-->}[lu] & \mathop{\rm Spec}(\kappa(y)) \ar[d] \ar[r] & Y \ar[dd] \\ & \mathop{\rm Spec}(\kappa(x)) \ar[r] \ar[d] & \mathop{\rm Spec}(\kappa(s)) \ar[rd] & \\ & X \ar[rr] & & S } $$ see the discussion in Section 25.13. Thus we get the dotted arrow. The corresponding point $z$ of $X \times_S Y$ is the image of the generic point of $\mathop{\rm Spec}(\kappa(x) \otimes_{\kappa(s)} \kappa(y)/\mathfrak p)$. We omit the verification that the two constructions are inverse to each other. $\square$

The code snippet corresponding to this tag is a part of the file `schemes.tex` and is located in lines 3098–3111 (see updates for more information).

```
\begin{lemma}
\label{lemma-points-fibre-product}
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 $\mathfrak p$.
\end{lemma}
\begin{proof}
Let $z$ be a point of $X \times_S Y$ and let us construct a
triple as above. Recall that we may think of $z$ as a morphism
$\Spec(\kappa(z)) \to X \times_S Y$, see
Lemma \ref{lemma-characterize-points}. This morphism corresponds
to morphisms $a : \Spec(\kappa(z)) \to X$
and $b : \Spec(\kappa(z)) \to Y$ 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.
\medskip\noindent
Conversely, given a quadruple $(x, y, s, \mathfrak p)$ we get a
commutative solid diagram
$$
\xymatrix{
X \times_S Y
\ar@/_/[dddr] \ar@/^/[rrrd]
& & & \\
&
\Spec(\kappa(x) \otimes_{\kappa(s)} \kappa(y)/\mathfrak p)
\ar[r] \ar[d] \ar@{-->}[lu]
&
\Spec(\kappa(y)) \ar[d] \ar[r] &
Y \ar[dd] \\
&
\Spec(\kappa(x)) \ar[r] \ar[d] &
\Spec(\kappa(s)) \ar[rd] &
\\
&
X \ar[rr] &
&
S
}
$$
see the discussion in Section \ref{section-points}. Thus we get the
dotted arrow. The corresponding point $z$ of $X \times_S Y$ is the
image of the generic point of
$\Spec(\kappa(x) \otimes_{\kappa(s)} \kappa(y)/\mathfrak p)$.
We omit the verification that the two constructions are inverse
to each other.
\end{proof}
```

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