Lemma 86.15.5. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$. If $A/I$ is Jacobson for some (equivalently any) ideal of definition $I \subset A$ and $\varphi $ is naively rig-flat, then $\varphi $ is rig-flat.

**Proof.**
Assume $\varphi $ is naively rig-flat. We first state some obvious consequences of the assumptions. Namely, let $f \in A$. Then $A, B, A_{\{ f\} }, B_{\{ f\} }$ are Noetherian adic topological rings. The maps $A \to A_{\{ f\} } \to B_{\{ f\} }$ and $A \to B \to B_{\{ f\} }$ are adic and topologically of finite type. The ring maps $A \to A_{\{ f\} }$ and $B \to B_{\{ f\} }$ are flat as compositions of $A \to A_ f$ and $B \to B_ f$ and the completion maps which are flat by Algebra, Lemma 10.97.2. The quotients of each of the rings $A, B, A_{\{ f\} }, B_{\{ f\} }$ by $I$ is of finite type over $A/I$ and hence Jacobson too (Algebra, Proposition 10.35.19).

Let $\mathfrak q' \subset B_{\{ f\} }$ be rig-closed. It suffices to prove that $(B_{\{ f\} })_{\mathfrak q'}$ is flat over $A_{\{ f\} }$, see Lemma 86.15.1. By Lemma 86.14.5 the primes $\mathfrak q \subset B$ and $\mathfrak p' \subset A_{\{ f\} }$ and $\mathfrak p \subset A$ lying under $\mathfrak q'$ are rig-closed. We are going to apply Algebra, Lemma 10.100.2 to the diagram

with $M = B_\mathfrak q$. The only assumption that hasn't been checked yet is the fact that $\mathfrak p$ generates the maximal ideal of $(A_{\{ f\} })_{\mathfrak p'}$. This follows from Lemma 86.14.8; here we use that $\mathfrak p$ and $\mathfrak p'$ are rig-closed to see that $f$ maps to a unit of $A/\mathfrak p$ (this is the only step in the proof that fails without the Jacobson assumption). Namely, this tells us that $A/\mathfrak p \to A_{\{ f\} }/\mathfrak p'$ is a finite inclusion of local rings (Lemma 86.14.4) and $f$ maps to a unit in the second one. $\square$

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