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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 https://stacks.math.columbia.edu/tag/001U#comment-3413 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 https://stacks.math.columbia.edu/tag/01JW#comment-8464 and then just invoking the remark plus 26.17.6.

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

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