**Proof.**
Proof of (3) $\Rightarrow $ (1). Assume (3). Let $k'/k$ be a finite purely inseparable extension. Set $A' = A \otimes _ k k'$. This is a local ring with maximal ideal $\mathfrak m'$. Set $K' = A'/\mathfrak m'$. We get a commutative diagram

\[ \xymatrix{ 0 \ar[r] & H_1(L_{K/k}) \otimes K' \ar[r] \ar[d]_\beta & \mathfrak m/\mathfrak m^2 \otimes K' \ar[r] \ar[d] & \Omega _{A/k} \otimes _ A K' \ar[r] \ar[d]_{\cong } & \Omega _{K/k} \otimes K' \ar[r] \ar[d]_\alpha & 0 \\ & H_1(L_{K'/k'}) \ar[r] & \mathfrak m'/(\mathfrak m')^2 \ar[r] & \Omega _{A'/k'} \otimes _{A'} K' \ar[r] & \Omega _{K'/k'} \ar[r] & 0 } \]

with exact rows. The third vertical arrow is an isomorphism by base change for modules of differentials (Algebra, Lemma 10.131.12). Thus $\alpha $ is surjective. By Lemma 15.34.3 we have

\[ \dim \mathop{\mathrm{Ker}}(\alpha ) - \dim \mathop{\mathrm{Ker}}(\beta ) + \dim \mathop{\mathrm{Coker}}(\beta ) = 0 \]

(and these dimensions are all finite). A diagram chase shows that $\dim \mathfrak m'/(\mathfrak m')^2 \leq \dim \mathfrak m/\mathfrak m^2$. However, since $A \to A'$ is finite flat we see that $\dim (A) = \dim (A')$, see Algebra, Lemma 10.112.6. Hence $A'$ is regular by definition.

Equivalence of (3) and (4). Consider the Jacobi-Zariski sequences for rows of the commutative diagram

\[ \xymatrix{ \mathbf{F}_ p \ar[r] & A \ar[r] & K \\ \mathbf{F}_ p \ar[r] \ar[u] & k \ar[r] \ar[u] & K \ar[u] } \]

to get a commutative diagram

\[ \xymatrix{ 0 \ar[r] & \mathfrak m/\mathfrak m^2 \ar[r] & \Omega _{A/\mathbf{F}_ p} \otimes _ A K \ar[r] & \Omega _{K/\mathbf{F}_ p} \ar[r] & 0 & \\ 0 \ar[r] & H_1(L_{K/k}) \ar[r] \ar[u] & \Omega _{k/\mathbf{F}_ p} \otimes _ k K \ar[r] \ar[u] & \Omega _{K/\mathbf{F}_ p} \ar[r] \ar[u] & \Omega _{K/k} \ar[r] \ar[u] & 0 } \]

with exact rows. We have used that $H_1(L_{K/A}) = \mathfrak m/\mathfrak m^2$ and that $H_1(L_{K/\mathbf{F}_ p}) = 0$ as $K/\mathbf{F}_ p$ is separable, see Algebra, Proposition 10.158.9. Thus it is clear that the kernels of $H_1(L_{K/k}) \to \mathfrak m/\mathfrak m^2$ and $\Omega _{k/\mathbf{F}_ p} \otimes _ k K \to \Omega _{A/\mathbf{F}_ p} \otimes _ A K$ have the same dimension.

Proof of (2) $\Rightarrow $ (4) following Faltings, see [Faltings-einfacher]. Let $a_1, \ldots , a_ n \in k$ be elements such that $\text{d}a_1, \ldots , \text{d}a_ n$ are linearly independent in $\Omega _{k/\mathbf{F}_ p}$. Consider the field extension $k' = k(a_1^{1/p}, \ldots , a_ n^{1/p})$. By Algebra, Lemma 10.158.3 we see that $k' = k[x_1, \ldots , x_ n]/(x_1^ p - a_1, \ldots , x_ n^ p - a_ n)$. In particular we see that the naive cotangent complex of $k'/k$ is homotopic to the complex $\bigoplus _{j = 1, \ldots , n} k' \rightarrow \bigoplus _{i = 1, \ldots , n} k'$ with the zero differential as $\text{d}(x_ j^ p - a_ j) = 0$ in $\Omega _{k[x_1, \ldots , x_ n]/k}$. Set $A' = A \otimes _ k k'$ and $K' = A'/\mathfrak m'$ as above. By Algebra, Lemma 10.134.8 we see that $\mathop{N\! L}\nolimits _{A'/A}$ is homotopy equivalent to the complex $\bigoplus _{j = 1, \ldots , n} A' \rightarrow \bigoplus _{i = 1, \ldots , n} A'$ with the zero differential, i.e., $H_1(L_{A'/A})$ and $\Omega _{A'/A}$ are free of rank $n$. The Jacobi-Zariski sequence for $\mathbf{F}_ p \to A \to A'$ is

\[ H_1(L_{A'/A}) \to \Omega _{A/\mathbf{F}_ p} \otimes _ A A' \to \Omega _{A'/\mathbf{F}_ p} \to \Omega _{A'/A} \to 0 \]

Using the presentation $A[x_1, \ldots , x_ n] \to A'$ with kernel $(x_ j^ p - a_ j)$ we see, unwinding the maps in Algebra, Lemma 10.134.4, that the $j$th basis vector of $H_1(L_{A'/A})$ maps to $\text{d}a_ j \otimes 1$ in $\Omega _{A/\mathbf{F}_ p} \otimes A'$. As $\Omega _{A'/A}$ is free (hence flat) we get on tensoring with $K'$ an exact sequence

\[ K'^{\oplus n} \to \Omega _{A/\mathbf{F}_ p} \otimes _ A K' \xrightarrow {\beta } \Omega _{A'/\mathbf{F}_ p} \otimes _{A'} K' \to K'^{\oplus n} \to 0 \]

We conclude that the elements $\text{d}a_ j \otimes 1$ generate $\mathop{\mathrm{Ker}}(\beta )$ and we have to show that are linearly independent, i.e., we have to show $\dim (\mathop{\mathrm{Ker}}(\beta )) = n$. Consider the following big diagram

\[ \xymatrix{ 0 \ar[r] & \mathfrak m'/(\mathfrak m')^2 \ar[r] & \Omega _{A'/\mathbf{F}_ p} \otimes K' \ar[r] & \Omega _{K'/\mathbf{F}_ p} \ar[r] & 0 \\ 0 \ar[r] & \mathfrak m/\mathfrak m^2 \otimes K' \ar[r] \ar[u]^\alpha & \Omega _{A/\mathbf{F}_ p} \otimes K' \ar[r] \ar[u]^\beta & \Omega _{K/\mathbf{F}_ p} \otimes K' \ar[r] \ar[u]^\gamma & 0 } \]

By Lemma 15.34.1 and the Jacobi-Zariski sequence for $\mathbf{F}_ p \to K \to K'$ we see that the kernel and cokernel of $\gamma $ have the same finite dimension. By assumption $A'$ is regular (and of the same dimension as $A$, see above) hence the kernel and cokernel of $\alpha $ have the same dimension. It follows that the kernel and cokernel of $\beta $ have the same dimension which is what we wanted to show.

The implication (1) $\Rightarrow $ (2) is trivial. This finishes the proof of the proposition.
$\square$

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