Lemma 29.46.11. Let $A \to B$ be a ring map.

1. If $A \to B$ induces a universal homeomorphism on spectra, then $B = \mathop{\mathrm{colim}}\nolimits B_ i$ is a filtered colimit of finitely presented $A$-algebras $B_ i$ such that $A \to B_ i$ induces a universal homeomorphism on spectra.

2. If $A \to B$ induces isomorphisms on residue fields and a universal homeomorphism on spectra, then $B = \mathop{\mathrm{colim}}\nolimits B_ i$ is a filtered colimit of finitely presented $A$-algebras $B_ i$ such that $A \to B_ i$ induces isomorphisms on residue fields and a universal homeomorphism on spectra.

Proof. Proof of (1). We will use the criterion of Algebra, Lemma 10.127.4. Let $A \to C$ be of finite presentation and let $\varphi : C \to B$ be an $A$-algebra map. Let $B' = \varphi (C) \subset B$ be the image. Then $A \to B'$ induces a universal homeomorphism on spectra by Lemma 29.46.1. By Algebra, Lemma 10.127.2 we can write $B' = \mathop{\mathrm{colim}}\nolimits _{i \in I} B_ i$ with $A \to B_ i$ of finite presentation and surjective transition maps. By Algebra, Lemma 10.127.3 we can choose an index $0 \in I$ and a factorization $C \to B_0 \to B'$ of the map $C \to B'$. We claim that $\mathop{\mathrm{Spec}}(B_ i) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism for $i$ sufficiently large. The claim finishes the proof of (1).

Proof of the claim. By Lemma 29.45.6 the ring map $A_{red} \to B'_{red}$ induces a universal homeomorphism on spectra. Thus $A_{red} \subset B'_{red}$ by Algebra, Lemma 10.30.6. Setting $A' = \mathop{\mathrm{Im}}(A \to B')$ we have surjections $A \to A' \to A_{red}$ inducing bijections $\mathop{\mathrm{Spec}}(A_{red}) = \mathop{\mathrm{Spec}}(A') = \mathop{\mathrm{Spec}}(A)$. Thus $A' \subset B'$ induces a universal homeomorphism on spectra. By Proposition 29.46.8 and the fact that $B'$ is finite type over $A'$ we can find $n$ and $b'_1, \ldots , b'_ n \in B'$ such that $B' = A'[b'_1, \ldots , b'_ n]$ and such that for $j = 1, \ldots , n$ we have

1. $(b'_ j)^2, (b'_ j)^3 \in A'[b'_1, \ldots , b'_{j - 1}]$, or

2. there exists a prime number $p$ with $pb'_ j, (b'_ j)^ p \in A'[b'_1, \ldots , b'_{j - 1}]$.

Choose $b_1, \ldots , b_ n \in B_0$ lifting $b'_1, \ldots , b'_ n$. For $i \geq 0$ denote $b_{j, i}$ the image of $b_ j$ in $B_ i$. For large enough $i$ we will have for $j = 1, \ldots , n$

1. $b_{j, i}^2, b_{j, i}^3 \in A_ i[b_{1, i}, \ldots , b_{j - 1, i}]$, or

2. there exists a prime number $p$ with $pb_{j, i}, b_{j, i}^ p \in A_ i[b_{1, i}, \ldots , b_{j - 1, i}]$.

Here $A_ i \subset B_ i$ is the image of $A \to B_ i$. Observe that $A \to A_ i$ is a surjective ring map whose kernel is a locally nilpotent ideal. After increasing $i$ more if necessary, we may assume $B_ i$ is generated by $b_1, \ldots , b_ n$ over $A_ i$, in other words $B_ i = A_ i[b_1, \ldots , b_ n]$. By Algebra, Lemmas 10.46.7 and 10.46.4 we conclude that $A \to A_ i \to A_ i[b_1] \to \ldots \to A_ i[b_1, \ldots , b_ n] = B_ i$ induce universal homeomorphisms on spectra. This finishes the proof of the claim.

The proof of (2) is exactly the same. $\square$

Comment #4335 by Elden on

I think in the statement Lemma 28.44.11 you want to say that $A \rightarrow B_i$ is of finite presentation and furthermore induces a universal homeomorphism on $Spec$.

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