The Stacks project

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. $\square$

Comments (4)

Comment #5532 by Chi Zhang on

Maybe $\mathfrak{p}$ is more precisely a maximal ideal? Since it is the kernel of the ring morphism $\kappa(x)\otimes_{\kappa(x)}\kappa(y)\to \kappa(z)$.

Comment #5722 by on

Dear Chi Zhang, no! It is easy to find nonmaximal prime ideals in where and are transcendentals.

Comment #7874 by Zhiyu YUAN on

A little typo: in the first line of the proof, for "triple" read "quadruple".

There are also:

  • 8 comment(s) on Section 26.17: Fibre products of schemes

Post a comment

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01JT. Beware of the difference between the letter 'O' and the digit '0'.