The Stacks project

Lemma 26.18.2. Let $S$ be a scheme. Let $f : X \to Y$ be an immersion (resp. closed immersion, resp. open immersion) of schemes over $S$. Then any base change of $f$ is an immersion (resp. closed immersion, resp. open immersion).

Proof. We can think of the base change of $f$ via the morphism $S' \to S$ as the top left vertical arrow in the following commutative diagram:

\[ \xymatrix{ X_{S'} \ar[r] \ar[d] & X \ar[d] \ar@/^4ex/[dd] \\ Y_{S'} \ar[r] \ar[d] & Y \ar[d] \\ S' \ar[r] & S } \]

The diagram implies $X_{S'} \cong Y_{S'} \times _ Y X$, and the lemma follows from Lemma 26.17.6. $\square$

Comments (2)

Comment #8466 by on

I think it could be nice to reference to make explicit the formal nature of the argument (I believe such little hints are quite useful for those recently initiated to algebraic geometry—and often to category theory too—reading this webpage, like the version of myself one year ago).

An alternative reformulation of the proof would be by placing this lemma after remark and then just invoking the remark plus 26.17.6.

There are also:

  • 12 comment(s) on Section 26.18: Base change in algebraic geometry

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Your email address will not be published.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 01JY. Beware of the difference between the letter 'O' and the digit '0'.