The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GGG. Beware of the difference between the letter 'O' and the digit '0'.