Remark 13.1.1.
In the following discussion, at certain points we will need to assume that we are working with a solid condensed ring \(A\) which admits a topologically nilpotent unit \(q\text{.}\) When this is not the case, we can formally push ourselves back into that situation by replacing \(A\) with \(A \otimes_{\underline{\ZZ} \solid} \underline{\ZZ((q))}\text{,}\) noting that the map \(A \to A \otimes_{\underline{\ZZ} \solid} \underline{\ZZ((q))}\) is split in \(\Mod_A \times_{\CAb} \CAb_\solid\) by the constant coefficient map.
We can make a similar construction starting with a Huber ring \(A\text{:}\) choose a ring of definition \(A_0\) and an ideal of definition \(I_0\text{,}\) complete \(A_0[q]\) with respect to the \((I_0,q)\)-adic topology, then tensor over \(A_0[q]\) with \(A[q^{\pm}]\text{.}\)