TITLE: Cotangent complex of perfect algebra over a perfect field
QUESTION [11 upvotes]: Let $A$ be a perfect $\kappa$-algebra over a perfect field $\kappa$ of positive characteristic $p$. Then the algebraic (= classical) cotangent complex  $L_{A/\kappa}^{\operatorname{alg}}$ is known to vanish, due to the Frobenious automorphism having  simultaneously to induce on the cotangent complex an automorphism and multiplication by $p$.
But we can also view $A$ and $\kappa$ as discrete $\mathbb E_\infty$-rings. The cotangent complex $L_{A/\kappa}$, which we obtain that way, is generally different from $L^{\operatorname{alg}}_{A/\kappa}$, since their homotopy groups give topological Andre-Quillen homology and (ordinary) Andre-Quillen homology respectively.
Q: Can we still say something about $L_{A/\kappa}$?
For instance:

Does it perhaps vanish?
Are there at least any finiteness results (e.g. when $A$ is a field, is $\dim_A \pi_n L_{A/\kappa} < \infty$)?

Perhaps a bit more broad afterquestion: what is in general the relationship between $L_{B/A}$ and $L^{\operatorname{alg}}_{B/A}$ for a discrete commutative $A$-algebra $B$? Other than that they coincide over the rationals and that $\pi_0$ of both is the module of Kähler differentials $\Omega_{\pi_0B/\pi_0A}$, of course.
Thanks in advance!

REPLY [7 votes]: If $f: A \rightarrow B$ is a morphism of simplicial commutative rings (for example, a morphism of ordinary commutative rings), then the "topological" cotangent complex comes with additional structure: the action
of $B$ on $L_{B/A}$ can be promoted to a (left) action of a certain
associative ring spectrum $B^{+}$. There are canonical maps of ring spectra
$B \rightarrow B^{+} \leftarrow \mathbf{Z}$ which induce (via the multiplication on $B^{+}$) an equivalence between $B^{+}$ and the smash product of $B$ with $\mathbf{Z}$ (beware that $B^{+}$ is not commutative and the order of the multiplication matters).
There's also a canonical map of ring spectra $B^{+} \rightarrow B$,
and the "algebraic" cotangent complex can be recovered as the tensor product $B \otimes_{ B^{+} } L_{B/A}$.
You can use this description (and the fact that $B^{+}$ is not too different from $B$) to answer a lot of the sorts of questions that you're asking. For example, $L_{B/A}$ is zero, or almost perfect, or connected through some range, if and only if the algebraic cotangent complex $L^{\mathrm{alg}}_{B/A}$ has the same property.