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Let $M$ be a module over an $E_\infty$ ring, $A$. Let $I$ be an $A$-non unital commutative algebra together with an associative map $I \wedge_A M \to M$.

Define ${_A}(M/I^n)$ as the cofiber of $I^{\wedge_A n} \wedge_A M \to M$, and the completion ${_A}M^\wedge_I$ by $\text{lim}_n {_A}M/I^n$.

I want to understand the dependence of the underlying space of $\text{lim}_n {_A}M/I^n$ on $A$. Note that the underlying space of ${_A}M/I^n$ depends on $A$. But sometimes the underlying space of ${_A}M^\wedge_I$ doesn't depend on $A$ since sometimes ${_A}M^\wedge_I=M$.

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    $\begingroup$ What happens if we ask the question just for ordinary rings and modules instead of spectra? $\endgroup$
    – user43326
    Commented Jul 30, 2022 at 4:29
  • $\begingroup$ @user43326 If I interpret discrete in a very naive sense, then its true: if I interpret cofiber as modding out by the image then $_A M/I^n=M/\text{ image } I^{\otimes_A n} \otimes_A M=M/I^nM$ doesn't depend on $A$. But if I interpret discrete as meaning eilenberg maclane spectra, I'm not sure :) $\endgroup$
    – Hari Rau-Murthy
    Commented Jul 30, 2022 at 15:06
  • $\begingroup$ In this case, I'd suspect $_AM/I^n$ depends on $A$ but that the completion does not. $\endgroup$
    – Hari Rau-Murthy
    Commented Jul 31, 2022 at 13:45
  • $\begingroup$ What do you mean exactly when you say "depends on $A$"? Could you elaborate a little? $\endgroup$
    – Simone Virili
    Commented Aug 8, 2022 at 21:39
  • $\begingroup$ @SimoneVirili If one has a module $M$, and a nonunital commutative algebra that are simultaneously over $B$ and over $A$(e.g. if $B$ is an $A$ algebra),the homotopy type of the spaces $_AM/I^n$ and $_BM/I^n$ are different. A good example is with $M=I$ and $n=2$. $\endgroup$
    – Hari Rau-Murthy
    Commented Aug 10, 2022 at 3:42

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