The Stacks project

Proof. Note that under the Frobenius map $F_ A : A \to A$, the target copy of $A$ is a free-module over the domain with basis $\{ 1, T, \dots , T^{p - 1}\} $. Thus, $F_ A$ is faithfully flat, and consequently, so is $A \to A_{perf}$ since it is a colimit of faithfully flat maps. Since $A_{perf}$ is a perfect ring, the relative Frobenius $F_{A_{perf}/A}$ is a surjection. In other words, $A_{perf} = A[A_{perf}^ p]$, which readily implies $\Omega _{A_{perf}/A} = 0$. Then $A \rightarrow A_{perf}$ is formally unramified by More on Morphisms, Lemma 37.6.7

It suffices to show that $A \rightarrow A_{perf}$ is not formally smooth. Note that since $A$ is a smooth $\mathbb {F}_ p$-algebra, the cotangent complex $L_{A/\mathbb {F}_ P} \simeq \Omega _{A/\mathbb {F}_ p}[0]$ is concentrated in degree $0$, see Cotangent, Lemma 92.9.1. Moreover, $L_{A_{perf}/\mathbb {F}_ p} = 0$ in $D(A_{perf})$ by Cotangent, Lemma 92.10.3. Consider the distinguished triangle of cotangent complexes

in $D(A_{perf})$, see Cotangent, Section 92.7. We find $L_{A_{perf}/A} = \Omega _{A/\mathbb {F}_ p} \otimes _ A A_{perf}[1]$, that is, $L_{A_{perf}/A}$ is equal to a free rank $1$ $A_{perf}$ module placed in degree $-1$. Thus $A \rightarrow A_{perf}$ is not formally smooth by More on Morphisms, Lemma 37.13.5 and Cotangent, Lemma 92.11.3. $\square$