Lemma 29.46.2. Let $A$ be a ring. Let $B = \mathop{\mathrm{colim}}\nolimits B_\lambda$ be a filtered colimit of $A$-algebras. If each $f_\lambda : \mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, resp. a universal homeomorphism inducing isomorphisms on residue fields, resp. universally closed, resp. universally closed and universally injective, then the same thing is true for $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$.

Proof. If $f_\lambda$ is universally closed, then $B_\lambda$ 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 each $f_\lambda$ is universally closed.

For a prime $\mathfrak q \subset B$ lying over $\mathfrak p \subset A$ denote $\mathfrak q_\lambda \subset B_\lambda$ the inverse image. Then $\kappa (\mathfrak q) = \mathop{\mathrm{colim}}\nolimits \kappa (\mathfrak q_\lambda )$. Thus if $A \to B_\lambda$ induces purely inseparable extensions of residue fields, then the same is true for $A \to B$. This proves the case where $f_\lambda$ is universally closed and universally injective, see Lemma 29.10.2.

The case where $f$ is a universal homeomorphism follows from the remarks above and Lemma 29.45.5 combined with the fact that prime ideals in $B$ are the same thing as compatible sequences of prime ideals in all of the $B_\lambda$.

If $A \to B_\lambda$ 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. $\square$

There are also:

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

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).