Does anything happen if I forget and tensor back up along a highly connective map of $E\_{\infty}$-rings?


Here's what I mean precisely: Let $f \colon A \to B$ be a $n$-connective map between connective $E\_{\infty}$-rings, with $n \geq 1$. Here $n$-connective means that $\pi\_i (fib(f))=0$ for $i < n$. Roughly, $f$ is a very surjective map, and if we let $fib(f)=I$, then $B$ is roughly $A/I$.


Now let's take a $B$-module $M$. We can forget the $B$-module structure and view it as an $A$-module. Then we can tensor it back up again to get $B \otimes \_A M$. By adjunction there is a canonical map $m \colon B \otimes \_A M \to M$, which is the multiplication map. How far is this map from being an equivalence?


The case I am really interested is when $f \colon A \to B$ is a square-zero extension obtained from a $n$-connective derivation $\eta \colon L\_B \to M[1]$. Then $fib(f)$ can be identified with $M$, and so additonally has the structure of a $B$-module. Does anything special happen in this case?



Forgetting and tensoring up for very connective maps of $E_{\infty}$-rings
(Derived) tensoring preserves fiber sequences. In particular, there is a fiber sequence
$$
fib(f) \otimes\_A M \to M \to B \otimes\_A M.
$$
If you allow additional assumptions, you can say more about this. For instance, if $A$ and $B$ are connective, then the Kunneth/hypertor spectral sequence
$$
Tor^{\pi\_\*A} (\pi\_\* fib(f), \pi\_\* M) \Rightarrow \pi\_\*(fib(f) \otimes\_A M)
$$
will tell you that the connectivity of the fiber is the sum of the connectivity of $M$ and the connectivity of $f$ (in the terminology that you're using). Roughly, then, you find that the map $M \to B \otimes\_A M$ is an isomorphism on the same number of homotopy groups that $A \to B$ is.


In the special case where $fib(f)$ is $M$ itself, you're getting a fiber sequence
$$
M \otimes\_A M \to M \to B \otimes\_A M.
$$
However, the map on the left is the multiplication map, which is zero by the square-zero assumption, and so this degenerates to an equivalence
$$
B \otimes\_A M \simeq M \vee (M \otimes\_A M)[1].
$$


