Given a commutative ring, the rank of a free module is unique. This is the well known statement that commutative rings have invariant basis numbers. Does an analogue of this property hold for free modules over $\mathbb{E}\_\infty$-ring spectra?


More precisely, I want to know the following: let $R$ be an $\mathbb{E}\_\infty$ ring spectrum. Is it true that for all natural numbers $n\neq m$ there exists no homotopy equivalence $R^n\simeq R^m$ of $R$-module spectra (i.e. no equivalence in the $\infty$-category of $R$-modules as defined in Lurie's Higher Algebra)?



Is the rank of free module spectra unique?
$\pi\_0: Sp\to Ab$ is a direct sum preserving functor, and it sends $E\_1$-ring spectra to rings, and modules over them to modules over them.


In particular you get a functor $\pi\_0: Mod\_R\to Mod\_{\pi\_0(R)}$. If $R^n\simeq R^m$ as $R$-modules, then $\pi\_0(R)^n\cong \pi\_0(R)^m$ as $\pi\_0(R)$-modules. So if $\pi\_0(R)$ is commutative and nonzero (which is a much weaker hypothesis than $R$ having an $E\_\infty$-structure), this implies $n=m$.


More generally, it suffices that $\pi\_0(R)$ have the invariant basis property.


Note that conversely, if $\pi\_0(R)$ does *not* have the invariant basis property, we can find inverse nonsquare matrices $M,N$ with coefficients in $\pi\_0(R)$, and you can view them as elements of $\pi\_0map\_R(R^n,R^m)$ ($\pi\_0map\_R(R^m,R^n)$ respectively), and their matrix product corresponds to the composition up to homotopy, so that $R^n\simeq R^m$ as $R$-modules.


So it's an "if and only if" situation with $\pi\_0(R)$.


A related claim is the fact that group-completion $K$-theory (here meaning only $K\_0$) only sees $\pi\_0$, namely if $R$ is a ring spectrum, then the group-completion $K$-theory of projective $R$-modules (summands of $R^n$ for some finite $n$, no shifts) is the same as that of $\tau\_{\geq 0}R$ (this is true for the whole group completion K-theory space), which is the same as that of $\pi\_0(R)$ (this is true only for $K\_0$)


