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. $\square$

Proof. Flatness of the completion is Algebra, Lemma 10.97.2. To show that the map is formally unramified, it suffices to show that $\Omega _{A^\wedge /A} = 0$, see Algebra, Lemma 10.148.2.

We sketch a proof. Choose $x_1, \ldots , x_ r \in A$ which map to a $p$-basis $\overline{x}_1, \ldots , \overline{x}_ r$ of $\kappa$, i.e., such that $\kappa$ is minimally generated by $\overline{x}_ i$ over $\kappa ^ p$. Choose a minimal set of generators $y_1, \ldots , y_ s$ of $\mathfrak m$. For each $n$ the elements $x_1, \ldots , x_ r, y_1, \ldots , y_ s$ generate $A/\mathfrak m^ n$ over $(A/\mathfrak m^ n)^ p$ by Frobenius. Some details omitted. We conclude that $F : A^\wedge \to A^\wedge$ is finite. Hence $\Omega _{A^\wedge /A}$ is a finite $A^\wedge$-module. On the other hand, for any $a \in A^\wedge$ and $n$ we can find $a_0 \in A$ such that $a - a_0 \in \mathfrak m^ nA^\wedge$. We conclude that $\text{d}(a) \in \bigcap \mathfrak m^ n \Omega _{A^\wedge /A}$ which implies that $\text{d}(a)$ is zero by Algebra, Lemma 10.51.4. Thus $\Omega _{A^\wedge /A} = 0$.

Suppose $A$ is regular. Then, using the Cohen structure theorem $x_1, \ldots , x_ r, y_1, \ldots , y_ s$ is a $p$-basis for the ring $A^\wedge$, i.e., we have

with $I = (i_1, \ldots , i_ r)$, $J = (j_1, \ldots , j_ s)$ and $0 \leq i_ a, j_ b \leq p - 1$. Details omitted. In particular, we see that $\Omega _{A^\wedge }$ is a free $A^\wedge$-module with basis $\text{d}(x_1), \ldots , \text{d}(x_ r), \text{d}(y_1), \ldots , \text{d}(y_ s)$.

Now if $A \to A^\wedge$ is formally étale or even just formally smooth, then we see that $\mathop{N\! L}\nolimits _{A^\wedge /A}$ has vanishing cohomology in degrees $-1, 0$ by Algebra, Proposition 10.138.8. It follows from the Jacobi-Zariski sequence (Algebra, Lemma 10.134.4) for the ring maps $\mathbf{F}_ p \to A \to A^\wedge$ that we get an isomorphism $\Omega _ A \otimes _ A A^\wedge \cong \Omega _{A^\wedge }$. Hence we find that $\Omega _ A$ is free on $\text{d}(x_1), \ldots , \text{d}(x_ r), \text{d}(y_1), \ldots , \text{d}(y_ s)$. Looking at fraction fields and using that $A$ is normal we conclude that $a \in A$ is a $p$th power if and only if its image in $A^\wedge$ is a $p$th power (details omitted; use Algebra, Lemma 10.158.2). A second consequence is that the operators $\partial /\partial x_ a$ and $\partial /\partial y_ b$ are defined on $A$.

We will show that the above lead to the conclusion that $A$ is finite over $A^ p$ with $p$-basis $x_1, \ldots , x_ r, y_1, \ldots , y_ s$. This will contradict the non-excellency of $A$ by a result of Kunz, see [Corollary 2.6, Kun76]. Namely, say $a \in A$ and write

with $a_{I, J} \in A^\wedge$. To finish the proof it suffices to show that $a_{I, J} \in A$. Applying the operator

to both sides we conclude that $a_{I, J}^ p \in A$ where $I = (p - 1, \ldots , p - 1)$ and $J = (p - 1, \ldots , p - 1)$. By our remark above, this also implies $a_{I, J} \in A$. After replacing $a$ by $a' = a - a_{I, J}^ p x^ I y^ J$ we can use a $1$-order lower differential operators to get another coefficient $a_{I, J}$ to be in $A$. Etc. $\square$

Proof. The backward implication follows from Lemma 110.44.2. For the forward implication, note that we already know from Lemma 110.44.2 that $A \to A^\wedge$ is formally unramified or equivalently that $\Omega _{A^\wedge /A}$ is zero. Thus, it suffices to show that the completion map is formally smooth when $A$ is excellent. By Néron-Popescu desingularization $A \to A^\wedge$ can be written as a filtered colimit of smooth $A$-algebras (Smoothing Ring Maps, Theorem 16.12.1). Hence $\mathop{N\! L}\nolimits _{A^\wedge /A}$ has vanishing cohomology in degree $-1$. Thus $A \to A^\wedge$ is formally smooth by Algebra, Proposition 10.138.8. $\square$