Lemma 86.14.8. 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 86.14.8. 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

\[ \mathfrak p A_{\{ f\} } = \mathfrak p(A_ f)^\wedge = \mathfrak p \otimes _ A (A_ f)^\wedge = (\mathfrak p_ f)^\wedge \]

is a prime ideal with quotient

\[ A/\mathfrak p = (A/\mathfrak p) \otimes _ A (A_ f)^\wedge = (A_ f)^\wedge / \mathfrak p (A_ f)^\wedge = A_{\{ f\} }/\mathfrak p A_{\{ f\} } \]

**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.
$\square$

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