For \(A\) an analytic ring and \(q \in A^{\gas}\) a unit, define
\begin{equation}
A^{\triangleright} \left \langle \frac{f_1,\dots,f_n}{g} \right\rangle^{\dagger}_q :=
\frac{A^{\triangleright} \left \langle T_1,\dots,T_n \right\rangle^\dagger_q}{(f_1 - g T_1, \dots, f_n - g T_n)}\text{.}\tag{17.2}
\end{equation}
Set
\begin{equation*}
S := \{ \tfrac{f_1^m}{g^m} q, \dots, \tfrac{f_n^m}{g^m} q \colon m=1,2,\dots\}
\end{equation*}
and let \(A\left \langle \frac{f_1,\dots,f_n}{g} \right\rangle^{\dagger}_q\) denote the analytic ring
\begin{equation*}
\left(A^{\triangleright} \left \langle \frac{f_1,\dots,f_n}{g} \right\rangle^{\dagger}_q, \Mod_{A^{\triangleright} \left \langle \frac{f_1,\dots,f_n}{g} \right\rangle^{\dagger}_q \liquid S}
\times_{\Mod_{A^{\triangleright}}} \Mod_A \right)\text{.}
\end{equation*}
When
\(q\) is topologically nilpotent, the construction does not depend on
\(q\) (by
Lemma 17.1.5 and
Corollary 10.4.7) and so we omit
\(q\) from the notation.
When
\(f_1,\dots,f_n\) generate the unit ideal, we call this construction the
dagger localization of
\(A\) with respect to
\(f_1,\dots,f_n,g\text{.}\) By calculating as in
Remark 15.3.7, we see that
\(A \to A\left \langle \frac{f_1,\dots,f_n}{g} \right\rangle^{\dagger}_q\) is an idempotent morphism provided that the quotient in
(17.2) is resolved by the corresponding Koszul complex.