## 15.35 Geometric regularity

Let $k$ be a field. Let $(A, \mathfrak m, K)$ be a Noetherian local $k$-algebra. The Jacobi-Zariski sequence (Algebra, Lemma 10.134.4) is a canonical exact sequence

$H_1(L_{K/k}) \to \mathfrak m/\mathfrak m^2 \to \Omega _{A/k} \otimes _ A K \to \Omega _{K/k} \to 0$

because $H_1(L_{K/A}) = \mathfrak m/\mathfrak m^2$ by Algebra, Lemma 10.134.6. We will show that exactness on the left of this sequence characterizes whether or not a regular local ring $A$ is geometrically regular over $k$. We will link this to the notion of formal smoothness in Section 15.40.

Proposition 15.35.1. Let $k$ be a field of characteristic $p > 0$. Let $(A, \mathfrak m, K)$ be a Noetherian local $k$-algebra. The following are equivalent

1. $A$ is geometrically regular over $k$,

2. for all $k \subset k' \subset k^{1/p}$ finite over $k$ the ring $A \otimes _ k k'$ is regular,

3. $A$ is regular and the canonical map $H_1(L_{K/k}) \to \mathfrak m/\mathfrak m^2$ is injective, and

4. $A$ is regular and the map $\Omega _{k/\mathbf{F}_ p} \otimes _ k K \to \Omega _{A/\mathbf{F}_ p} \otimes _ A K$ is injective.

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 . 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. $\square$

Lemma 15.35.2. Let $k$ be a field of characteristic $p > 0$. Let $(A, \mathfrak m, K)$ be a Noetherian local $k$-algebra. Assume $A$ is geometrically regular over $k$. Let $K/F/k$ be a finitely generated subextension. Let $\varphi : k[y_1, \ldots , y_ m] \to A$ be a $k$-algebra map such that $y_ i$ maps to an element of $F$ in $K$ and such that $\text{d}y_1, \ldots , \text{d}y_ m$ map to a basis of $\Omega _{F/k}$. Set $\mathfrak p = \varphi ^{-1}(\mathfrak m)$. Then

$k[y_1, \ldots , y_ m]_\mathfrak p \to A$

is flat and $A/\mathfrak pA$ is regular.

Proof. Set $A_0 = k[y_1, \ldots , y_ m]_\mathfrak p$ with maximal ideal $\mathfrak m_0$ and residue field $K_0$. Note that $\Omega _{A_0/k}$ is free of rank $m$ and $\Omega _{A_0/k} \otimes K_0 \to \Omega _{K_0/k}$ is an isomorphism. It is clear that $A_0$ is geometrically regular over $k$. Hence $H_1(L_{K_0/k}) \to \mathfrak m_0/\mathfrak m_0^2$ is an isomorphism, see Proposition 15.35.1. Now consider

$\xymatrix{ H_1(L_{K_0/k}) \otimes K \ar[d] \ar[r] & \mathfrak m_0/\mathfrak m_0^2 \otimes K \ar[d] \\ H_1(L_{K/k}) \ar[r] & \mathfrak m/\mathfrak m^2 }$

Since the left vertical arrow is injective by Lemma 15.34.2 and the lower horizontal by Proposition 15.35.1 we conclude that the right vertical one is too. Hence a regular system of parameters in $A_0$ maps to part of a regular system of parameters in $A$. We win by Algebra, Lemmas 10.128.2 and 10.106.3. $\square$

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