# Cotangent complex of perfect algebra over a perfect field

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:

1. Does it perhaps vanish?
2. 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.

• If I recall correctly finiteness results are true under the hypotheses that $A$ is a finitely presented $E_\infty$-algebra over $\kappa$. I don't think that's true very often unfortunately – Denis Nardin Dec 30 '17 at 9:29
• There is certainly Theorem HA.7.4.3.18 (HA = Higher Algebra), saying that $L_{B/A}$ is (almost) perfect if $A\to B$ is quite generally an (almost) finitely presented map of connected $\mathbb E_\infty$-rings. However the converse is only stated under the assumption that $\pi_0A\to \pi_0B$ if finitely presented. Based on that, there might still be a chance for $L_{B/A}$ to be (almost) perfect even if $A\to B$ fails the finiteness condition, so long as the underlying map $\pi_0A\to \pi_0B$ also fails it. – A Rock and a Hard Place Dec 30 '17 at 11:23

Let me explain why the $E_\infty$-cotangent complex $L_{B/A}$ vanishes for any map $A \to B$ of perfect rings over $\mathbf{F}_p$. (I do not know the answer to the more general question at the end.)

The proof uses formal properties of the cotangent complex (Kunneth formula, transitivity triangle) and relies on the following two observations (where all tensor products are derived):

1) If $R \to S$ is map of $E_\infty$-rings with $S \otimes_R S \simeq S$ via the multiplication map, then $L_{S/R} \simeq 0$. Indeed, we always have $L_{S \otimes_R S/R} \simeq p_1^* L_{S/R} \oplus p_2^* L_{S/R}$ by the Kunneth formula. If the multiplication map is an isomorphism, then we get $L_{S/R} \oplus L_{S/R} \simeq L_{S/R}$ via the sum map, which means $L_{S/R} \simeq 0$. (This is the classical proof that the cotangent complex of an open immersion is $0$.)

2) If $R \to S$ is any map of perfect rings, then $\pi_i(S \otimes_R S) =: \mathrm{Tor}^i_R(S,S)$ vanishes for $i > 0$. See, for example, Lemma 3.16 in https://arxiv.org/abs/1507.06490.

Now say $A \to B$ is a map of perfect rings. Consider the multiplication map $R := B \otimes_A B \to S := B$. Then $S \otimes_R S \simeq S$ via the multiplication map: this is clear on $\pi_0$ and thus follows from (2) as everything is perfect. Then (1) implies that $L_{B/B \otimes_A B} \simeq 0$. But the Kunneth formula and the transitivity triangle for $A \to B \otimes_A B \to B$ show that $L_{B/B \otimes_A B} \simeq L_{B/A}[1]$, and thus $L_{B/A} \simeq 0$.

• Thanks, that's just what I was after! It was fact 2) that I was missing before, so thank you for teaching it to me! – A Rock and a Hard Place Dec 30 '17 at 19:07
• @ARockandaHardPlace By the way, I want to point out that this argument also generalizes to the case where $L_{B/A}^{\mathrm{alg}}\simeq0$. The point is that, $S\otimes_RS$ in question is just the Hochshild (or Shukla) homology $\operatorname{HH}(B/A)$, but this is $B$ if the cotangent complex vanishes, because of the HKR filtration. – Yai0Phah Jun 20 '19 at 22:08

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.

• Dear Jacob, thank you very much for this very interesting answer! Is this something we can look forward to in the SAG.VII (or is it perhaps already written out somewhere?)? – A Rock and a Hard Place Jan 3 '18 at 19:32
• Also I'm a little confused about how $B^+$ can fail to be commutative if it is equivalent to $B\otimes \mathbf Z$, and both $B$ and $\mathbf Z$ are commutative. Probably I'm missing something obvious, sorry. – A Rock and a Hard Place Jan 3 '18 at 19:38
• The ring of quaternions $\mathbf{H}$ contains commutative subfields $\mathbf{R}[i]$ and $\mathbf{R}[j]$, and the inclusion maps $\mathbf{R}[i] \hookrightarrow \mathbf{H} \hookleftarrow \mathbf{R}[j]$ induce an isomorphism $\mathbf{R}[i] \otimes_{\mathbf{R}} \mathbf{R}[j] \simeq \mathbf{H}$. But $\mathbf{H}$ is not commutative. – Jacob Lurie Jan 4 '18 at 9:23
• But both $\mathbf R[i]$ and $\mathbf R[j]$ are isomorphic to $\mathbf C$ as $\mathbf R$-algebras, so shouldn't it be that $\mathbf R[i]\otimes_{\mathbf R}\mathbf R[j]\simeq \mathbf C\otimes_{\mathbf R}{\mathbf C}\simeq \mathbf C^2$? – A Rock and a Hard Place Jan 5 '18 at 10:32
• The vector space $\mathbf{R}[i] \otimes_{ \mathbf{R} } \mathbf{R}[j]$ does not have a unique $\mathbf{R}$-algebra structure, even if you require the inclusion maps $\mathbf{R}[i] \hookrightarrow \mathbf{R}[i] \otimes_{ \mathbf{R} } \mathbf{R}[j] \hookleftarrow \mathbf{R}[j]$ to be maps of $\mathbf{R}$-algebras. There's a unique algebra structure in which $\mathbf{R}[i]$ and $\mathbf{R}[j]$ commute with each other, but also algebra structures where they do not (like $\mathbf{H}$). The ring spectrum $B^{+}$ is like the latter. – Jacob Lurie Jan 5 '18 at 15:19