Cochains on Eilenberg-MacLane Spaces

Let $p$ be a prime number, let $k$ be a commutative ring in which $p=0$, and let $X = K( {\mathbb Z}/p {\mathbb Z}, n)$ be an Eilenberg-MacLane space. Let $F$ be the free $E_{\infty}$-algebra over $k$ generated by a class $\eta$ in (homological) degree $-n$. A result of Mandell asserts that there is a cofiber sequence of $E_{\infty}$-algebras over $k$ $$F \stackrel{A}{\rightarrow} F \rightarrow C^{\ast}(X;k)$$ where $A$ is the Artin-Schreier'' map which carries $\eta$ to $\eta - P^0(\eta)$. In other words, as an $E_{\infty}$-algebra, the cochain complex $C^{\ast}(X;k)$ can be described by one generator (a class in degree $-n$) and one relation (the class should be fixed by $P^0$).

Is it possible to prove this result without explicitly computing the homotopy groups of the cofiber of $A$? Let's denote this cofiber by $R$. I can reduce to the problem of showing that $\pi_i R \simeq 0$ for $i > 0$ and that $\pi_0 R \simeq k$ (both of which are obvious consequences of Mandell's theorem). Is there some way to show this directly, without computing the other homotopy groups?

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Bertrand Toen doesn't seem to do much calculation in Champs affines.

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He also doesn't prove this theorem. Unless I misunderstand, he works in the setting of cosimplicial algebras (where the analogous statement is easy) and uses it to prove variants of Mandell's results. –  Jacob Lurie Dec 3 '10 at 17:11
oh............. –  Ben Wieland Dec 3 '10 at 18:18
Toen's technique seems to involve an inductive approach, using the result for $X=K(Z/pZ,n)$ to prove it for $BX=K(Z/pZ,n+1)$. Could that be used here to reduce to the case of $n=1$ or $n=0$? –  Charles Rezk Dec 3 '10 at 19:51
That's the strategy I had in mind. When n=0 you can prove it using deformation theory, so let's try induction on n. Let R(n) be the cofiber and let R'(n) be the cochains on K(Z/pZ,n). Then doing a bar construction on R(n) produces R(n-1), and similarly for R'(n). So the I.H. tells you that the map R(n) -> R'(n) is an equivalence after applying the bar construction. If you knew that R(n) had no positive homotopy and that pi_0 R(n) = k (statements which are obvious for R'(n)), then the bar construction doesn't lose any information and you are done. But a priori R(n) is a big mess. –  Jacob Lurie Dec 3 '10 at 20:14