Lemma 88.14.8. Let $A$ be an adic Noetherian topological ring. Let $\mathfrak p \subset A$ be a prime ideal. Let $f \in A$ be an element mapping to a unit in $A/\mathfrak p$. Then
is a prime ideal with quotient
Lemma 88.14.8. Let $A$ be an adic Noetherian topological ring. Let $\mathfrak p \subset A$ be a prime ideal. Let $f \in A$ be an element mapping to a unit in $A/\mathfrak p$. Then
is a prime ideal with quotient
Proof. Since $A_ f$ is Noetherian the ring map $A \to A_ f \to (A_ f)^\wedge $ is flat. For any finite $A$-module $M$ we see that $M \otimes _ A (A_ f)^\wedge $ is the completion of $M_ f$. If $f$ is a unit on $M$, then $M_ f = M$ is already complete. See discussion in Algebra, Section 10.97. From these observations the results follow easily. $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)