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In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined.

Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$?

After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S^\bullet/R} \otimes_{S^\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation.

Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why. I think it is because $R_{(\Phi)}$ is defined to be $R$ as a ring, and the map $R_{(\Phi)} \to S_{(\Phi)}$ as rings should be the same as the map $R \to S$ as rings, so their modules of Kähler differentials should be the same.

In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined.

Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$?

After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S_\bullet/R} \otimes_{S_\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation.

Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why. I think it is because $R_{(\Phi)}$ is defined to be $R$ as a ring, and the map $R_{(\Phi)} \to S_{(\Phi)}$ as rings should be the same as the map $R \to S$ as rings, so their modules of Kähler differentials should be the same.

P.S. I don't understand why Scholze uses $S_\bullet$ in the definition of $\Bbb L_{S/R}$ when he uses $S^\bullet$. Indeed, Gabber–Romero uses $P^\bullet$ in Lemma 6.5.9.

In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined.

Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$?

After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S^\bullet/R} \otimes_{S^\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation.

Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why. I think it is because $R_(\Phi)$ is defined to be $R$ as a ring, and the map $R_{(\Phi)} \to S_{(\Phi)}$ as rings should be the same as the map $R \to S$ as rings, so their modules of Kähler differentials should be the same.

In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined.

Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$?

After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S^\bullet/R} \otimes_{S^\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation.

Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why. I think it is because $R_{(\Phi)}$ is defined to be $R$ as a ring, and the map $R_{(\Phi)} \to S_{(\Phi)}$ as rings should be the same as the map $R \to S$ as rings, so their modules of Kähler differentials should be the same.

In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined.

Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$?

After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S^\bullet/R} \otimes_{S^\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation.

Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why.

In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined.

Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$?

After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S^\bullet/R} \otimes_{S^\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation.

Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why. I think it is because $R_(\Phi)$ is defined to be $R$ as a ring, and the map $R_{(\Phi)} \to S_{(\Phi)}$ as rings should be the same as the map $R \to S$ as rings, so their modules of Kähler differentials should be the same.