Lemma 86.14.9. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$ which is adic and topologically of finite type. Let $\mathfrak q \subset B$ be rig-closed. There exists an $f \in A$ which maps to a unit in $B/\mathfrak q$ such that we obtain a diagram

\[ \vcenter { \xymatrix{ B \ar[r] & B_{\{ f\} } \\ A \ar[r] \ar[u]_\varphi & A_{\{ f\} } \ar[u]_{\varphi _{\{ f\} }} } } \quad \text{with primes}\quad \vcenter { \xymatrix{ \mathfrak q \ar@{-}[r] \ar@{-}[d] & \mathfrak q' \ar@{-}[d] \ar@{=}[r] & \mathfrak q B_{\{ f\} } \\ \mathfrak p \ar@{-}[r] & \mathfrak p' } } \]

such that $\mathfrak p'$ is rig-closed, i.e., the map $A_{\{ f\} } \to B_{\{ f\} }$ and the prime ideals $\mathfrak q'$ and $\mathfrak p'$ satisfy the equivalent conditions of Lemma 86.14.4.

**Proof.**
Please see Lemma 86.14.8 for the description of $\mathfrak q'$. The only assertion the lemma makes is that for a suitable choice of $f$ the prime ideal $\mathfrak p'$ has the property $\dim ((A_ f)^\wedge /\mathfrak p') = 1$. By Lemma 86.14.4 this in turn just means that the residue field $\kappa $ of $B/\mathfrak q = (B_ f)^\wedge /\mathfrak q'$ is finite over $(A_ f)^\wedge /\mathfrak a' = (A/\mathfrak a)_ f$. By Lemma 86.14.3 we know that $A/\mathfrak a \to \kappa $ is a finite type algebra homomorphism. By the Hilbert Nullstellensatz in the form of Algebra, Lemma 10.34.2 we can find an $f \in A$ which maps to a unit in $\kappa $ such that $\kappa $ is finite over $A_ f$. This finishes the proof.
$\square$

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