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What are some general results describing the Krull dimension of the completion of a non-Noetherian ring with a "nice" topology?

An example of the sort of "nice" topological ring I'm looking for is a perfection of an affinoid algebra over a perfectoid (probably just complete non-archimedean with a non-discrete valuation is good enough) field $K$ in characteristic $p > 0$.

So let $R$ be a quotient of a Tate algebra $K\langle T_1, \ldots, T_n \rangle$ of power series which converge on the adic closed (i.e. the coefficients tend to $0$). The topology is induced from the non-archimedean valuation topology on $K$ and is given by the Gauss norm $\|\sum_\alpha a_\alpha T^\alpha\| = \sup_\alpha |a_\alpha|$. So in particular $R$ is a noetherian topological ring which is "Tate" (meaning its topology is defined by a topologically nilpotent unit $\varpi$, and there's an open subring on which the topology is $\varpi$-adic).

We can form its perfection $R^{\mathrm{perf}} = \varinjlim_m R^{1/p^m}$, which is a quotient of the perfection of the Tate algebra $\cup_m K \langle T_1^{1/p^m}, \ldots, T_n^{1/p^m} \rangle$. The perfection comes equipped with the direct limit topology, also compatible with the Gauss norm and "induced from the topology on $K$" in an appropriate sense. The completion of $R^{\mathrm{perf}}$ with respect to this topology is what is sometimes called the "completed perfection" of $R$.

I'd like to show that the Krull dimension of this ring agrees with that of $R^{\mathrm{perf}}$, and therefore with $R$.

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Unfortunately, the Krull dimension often behaves in strange ways for such non-noetherian rings. My impression is that the results of Lang--Ludwig should adapt to the present situation, and show that this ring has infinite Krull dimension.

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