# Dependence of completion on the base ring

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 $${_A}M^\wedge_I=M$$.

• What happens if we ask the question just for ordinary rings and modules instead of spectra? Commented Jul 30, 2022 at 4:29
• @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 :) Commented Jul 30, 2022 at 15:06
• In this case, I'd suspect $_AM/I^n$ depends on $A$ but that the completion does not. Commented Jul 31, 2022 at 13:45
• What do you mean exactly when you say "depends on $A$"? Could you elaborate a little? Commented Aug 8, 2022 at 21:39
• @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 $n=2$. Commented Aug 10, 2022 at 3:42