In this section we fix a prime number $p$. If $A$ is a ring with $p = 0$ in $A$, then $F_ A : A \to A$ denotes the Frobenius endomorphism $a \mapsto a^ p$.
Lemma 92.10.1. Let $A \to B$ be a ring map with $p = 0$ in $A$. Let $P_\bullet $ be the standard resolution of $B$ over $A$. The map $P_\bullet \to P_\bullet $ induced by the diagram discussed in Section 92.6 is homotopic to the Frobenius endomorphism $P_\bullet \to P_\bullet $ given by Frobenius on each $P_ n$.
\[ \xymatrix{ B \ar[r]_{F_ B} & B \\ A \ar[u] \ar[r]^{F_ A} & A \ar[u] } \]
Proof. Let $\mathcal{A}$ be the category of $\mathbf{F}_ p$-algebra maps $A \to B$. Let $\mathcal{S}$ be the category of pairs $(A, E)$ where $A$ is an $\mathbf{F}_ p$-algebra and $E$ is a set. Consider the adjoint functors
and
Let $X$ be the simplicial object in in the category of functors from $\mathcal{A}$ to $\mathcal{A}$ constructed in Simplicial, Section 14.34. It is clear that $P_\bullet = X(A \to B)$ because if we fix $A$ then.
Set $Y = U \circ V$. Recall that $X$ is constructed from $Y$ and certain maps and has terms $X_ n = Y \circ \ldots \circ Y$ with $n + 1$ terms; the construction is given in Simplicial, Example 14.33.1 and please see proof of Simplicial, Lemma 14.34.2 for details.
Let $f : \text{id}_\mathcal {A} \to \text{id}_\mathcal {A}$ be the Frobenius endomorphism of the identity functor. In other words, we set $f_{A \to B} = (F_ A, F_ B) : (A \to B) \to (A \to B)$. Then our two maps on $X(A \to B)$ are given by the natural transformations $f \star 1_ X$ and $1_ X \star f$. Details omitted. Thus we conclude by Simplicial, Lemma 14.33.6. $\square$
Lemma 92.10.2. Let $p$ be a prime number. Let $A \to B$ be a ring homomorphism and assume that $p = 0$ in $A$. The map $L_{B/A} \to L_{B/A}$ of Section 92.6 induced by the Frobenius maps $F_ A$ and $F_ B$ is homotopic to zero.
Proof. Let $P_\bullet $ be the standard resolution of $B$ over $A$. By Lemma 92.10.1 the map $P_\bullet \to P_\bullet $ induced by $F_ A$ and $F_ B$ is homotopic to the map $F_{P_\bullet } : P_\bullet \to P_\bullet $ given by Frobenius on each term. Hence we obtain what we want as clearly $F_{P_\bullet }$ induces the zero map $\Omega _{P_ n/A} \to \Omega _{P_ n/A}$ (since the derivative of a $p$th power is zero). $\square$
Lemma 92.10.3. Let $p$ be a prime number. Let $A \to B$ be a ring homomorphism and assume that $p = 0$ in $A$. If $A$ and $B$ are perfect, then $L_{B/A}$ is zero in $D(B)$.
Proof. The map $(F_ A, F_ B) : (A \to B) \to (A \to B)$ is an isomorphism hence induces an isomorphism on $L_{B/A}$ and on the other hand induces zero on $L_{B/A}$ by Lemma 92.10.2. $\square$
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