The Stacks project

Lemma 29.46.1. Let $A \to B$ be a ring map such that the induced morphism of schemes $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, resp. a universal homeomorphism inducing isomorphisms on residue fields, resp. universally closed, resp. universally closed and universally injective. Then for any $A$-subalgebra $B' \subset B$ the same thing is true for $f' : \mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$.

Proof. If $f$ is universally closed, then $B$ is integral over $A$ by Lemma 29.44.7. Hence $B'$ is integral over $A$ and $f'$ is universally closed (by the same lemma). This proves the case where $f$ is universally closed.

Continuing, we see that $B$ is integral over $B'$ (Algebra, Lemma 10.36.15) which implies $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ is surjective (Algebra, Lemma 10.36.17). Thus if $A \to B$ induces purely inseparable extensions of residue fields, then the same is true for $A \to B'$. This proves the case where $f$ is universally closed and universally injective, see Lemma 29.10.2.

The case where $f$ is a universal homeomorphism follows from the remarks above, Lemma 29.45.5, and the obvious observation that if $f$ is surjective, then so is $f'$.

If $A \to B$ induces isomorphisms on residue fields, then so does $A \to B'$ (see argument in second paragraph). In this way we see that the lemma holds in the remaining case. $\square$

Comments (2)

Comment #2225 by Matthew Emerton on

At the beginning of line 2 of para. 2 of the proof, I think B' and B should be swapped.

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  • 2 comment(s) on Section 29.46: Universal homeomorphisms of affine schemes

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