I want to prove the following: Let $R$ be a perfectoid ring and $\varpi$ a pseudo uniformizer in $R$ which admits all $p$-th power roots, then a module over $R^\circ$ is flat if and only if it has no $\varpi$ nontrivial torsion.
(I know this is true when $R$ is a perfectoid field.)



Flatness over a perfectoid ring
Even in the case $R$ and $S$ are perfectoid algebras over a perfectoid field, it's not the case that $S$ is $R$-flat in many situations, eg. perf $R$, take a higher rank point in $\text{Spa}(R,R^0)$, look at the completed res field $\kappa$, at the $\kappa$-normalization $\kappa^+$ of $R^0$, and finally at the map $R^0\to\kappa^+$.


If $R$ is a perfectoid algebra over a perfectoid field $K$, $K^0$-flatness conveniently comes for free for $R^0$. You want to use it to check that ``derived relative perfectness'':


$$A\_{\varphi}\otimes\_{\varphi, A}^{\mathbf{L}} B\xrightarrow{\simeq} B\_{\varphi}[0]$$


in $\text{D}(B)$, $A$, $B$ $\mathbf{F}\_p$-algebras, $\varphi$ the abs Frobenius, $A\_{\varphi}$ is $A$ as an $A$-algebra under $\varphi$, same for $B$, is met when $A = K^0/\varpi$ and $B = R^0/\varpi$. This is obvious in deg $0$, and if $K$ is a perfectoid field also in higher degs by $K^0$-flatness of $R^0$. From here it's an easy lemma that $\mathbf{L}\_{(R^0/\varpi)/(K^0/\varpi)}\simeq 0$ in $\text{D}(R^0/\varpi)$.


For $K$ a perfectoid ring, you can't directly invoke derived rel perfectness and such vanishing lemma, and it's honestly not so clear why you should hunt flatness at all costs to make it available again.


Rather, you should reduce to the char $p$ case where perfect $\mathbf{F}\_p$-algebras are derived relatively perfect. From this you show that if $R$ is perfectoid and $S$ is a perfectoid $R$-algebra, then the analytic cotangent complex of $R^0\to S^0$ as introduced in Gabber and Ramero's book on Almost Ring Theory, always vanishes.


