The Stacks project

Lemma 64.4.3. Let $S$ be a scheme. Let

\[ \xymatrix{ Z \times _ Y X \ar[r] \ar[d] & X \ar[d] \\ Z \ar[r] & Y } \]

be a cartesian diagram of algebraic spaces over $S$. Then the map of sets of points

\[ |Z \times _ Y X| \longrightarrow |Z| \times _{|Y|} |X| \]

is surjective.

Proof. Namely, suppose given fields $K$, $L$ and morphisms $\mathop{\mathrm{Spec}}(K) \to X$, $\mathop{\mathrm{Spec}}(L) \to Z$, then the assumption that they agree as elements of $|Y|$ means that there is a common extension $K \subset M$ and $L \subset M$ such that $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(K) \to X \to Y$ and $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(L) \to Z \to Y$ agree. And this is exactly the condition which says you get a morphism $\mathop{\mathrm{Spec}}(M) \to Z \times _ Y X$. $\square$


Comments (2)

Comment #2039 by Keenan Kidwell on

The scheme introduced in the first sentence is never mentioned again. Presumably one wants the algebraic spaces to be algebraic spaces over ?

There are also:

  • 2 comment(s) on Section 64.4: Points of algebraic spaces

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Your email address will not be published.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 03H4. Beware of the difference between the letter 'O' and the digit '0'.