Let $A$ be a commutative ring and $I \subseteq A$ a finitely generated ideal (I am **not** assuming that $A$ is Noetherian). 


Assume that $A$ is derived $I$-complete, meaning, let's say, that $\mathrm{Hom}\_A(A\_f, A)=\mathrm{Ext}^1\_A(A\_f, A)=0$ for all $f \in I$, or equivalently that the map $A^{\times \mathbb{N}} \rightarrow A^{\times \mathbb{N}},\;\; (x\_n)\_n \mapsto (x\_n-fx\_{n+1})\_n$ is a bijection for all $f \in I$ (full definition and bunch of properties can be found at [Stacks Project](https://stacks.math.columbia.edu/tag/091N)). My question is



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> Assuming that $A$ is derived $I$-complete, is $I \subseteq \mathrm{rad}(A)?$
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Obviously, one can wlog assume that $I$ is principal. Something like this seems to be often implicitly used in Bhatt and Scholze's paper "[Prisms and prismatic cohomology](https://arxiv.org/abs/1905.08229)" (namely, throughout section 3 of the paper). 


I cannot find any reference or an argument how to prove it so far. 


I can prove it e.g. assuming that $A$ is Noetherian (then each simple module $N=A/\mathfrak{m}$ is finitely presented, hence derived $I$-complete, and then by "derived Nakayma" ($N=IN$ implies $N=0$ when $N$ is derived $I$-complete) we can only have $IN=0,$ i.e. $I \subseteq \mathfrak{m}$), but not in the general case (which seems to be important for prisms).


And if the assertion is not true in general, is there some other reason why the conclusion should be true in the case of prisms?



Jacobson radical of a derived $I$-complete ring
Never mind, I figured it out. Let me leave it here for potential benefit of other people trying to learn about derived completions:


Given $f \in I,$ there is a unique preimage $(y\_n)\_n$ to the element $(1, 1, 1, \dots)$ under the map $A^{\times \mathbb{N}} \rightarrow A^{\times \mathbb{N}}, (x\_n)\_n \mapsto (x\_n-fx\_{n+1})$. But the operation of shifting all the vectors by $1$ position (i.e. $(x\_n)\_n \mapsto (x\_{n+1})\_n$) leaves the map unchanged and does not change the vector $(1, 1, 1, \dots)$ either, so it follows that the unique solution remains also unchanged by the shift, i.e. $y\_{n+1}=y\_n=:y$ for all $n$. This means that $y(1-f)=1$. So $1+I$ consists entirely of units, hence $I \subseteq \mathrm{rad}(A)$.


