## 15.48 Regularity and derivations

Let $R \to S$ be a ring map. Let $D : R \to R$ be a derivation. We say that $D$ *extends to* $S$ if there exists a derivation $D' : S \to S$ such that

\[ \xymatrix{ S \ar[r]_{D'} & S \\ R \ar[u] \ar[r]^ D & R \ar[u] } \]

is commutative.

Lemma 15.48.1. Let $R$ be a ring. Let $D : R \to R$ be a derivation.

For any ideal $I \subset R$ the derivation $D$ extends canonically to a derivation $D^\wedge : R^\wedge \to R^\wedge $ on the $I$-adic completion.

For any multiplicative subset $S \subset R$ the derivation $D$ extends uniquely to the localization $S^{-1}R$ of $R$.

If $R \subset R'$ is a finite type extension of rings such that $R_ g \cong R'_ g$ for some $g \in R$ which is a nonzerodivisor in $R'$, then $g^ ND$ extends to $R'$ for some $N \geq 0$.

**Proof.**
Proof of (1). For $n \geq 2$ we have $D(I^ n) \subset I^{n - 1}$ by the Leibniz rule. Hence $D$ induces maps $D_ n : R/I^ n \to R/I^{n - 1}$. Taking the limit we obtain $D^\wedge $. We omit the verification that $D^\wedge $ is a derivation.

Proof of (2). To extend $D$ to $S^{-1}R$ just set $D(r/s) = D(r)/s - rD(s)/s^2$ and check the axioms.

Proof of the final statement. Let $x_1, \ldots , x_ n \in R'$ be generators of $R'$ over $R$. Choose an $N$ such that $g^ Nx_ i \in R$. Consider $g^{N + 1}D$. By (2) this extends to $R_ g$. Moreover, by the Leibniz rule and our construction of the extension above we have

\[ g^{N + 1}D(x_ i) = g^{N + 1}D(g^{-N} g^ Nx_ i) = -Ng^ Nx_ iD(g) + gD(g^ Nx_ i) \]

and both terms are in $R$. This implies that

\[ g^{N + 1}D(x_1^{e_1} \ldots x_ n^{e_ n}) = \sum e_ i x_1^{e_1} \ldots x_ i^{e_ i - 1} \ldots x_ n^{e_ n} g^{N + 1}D(x_ i) \]

is an element of $R'$. Hence every element of $R'$ (which can be written as a sum of monomials in the $x_ i$ with coefficients in $R$) is mapped to an element of $R'$ by $g^{N + 1}D$ and we win.
$\square$

slogan
Lemma 15.48.2. Let $R$ be a regular ring. Let $f \in R$. Assume there exists a derivation $D : R \to R$ such that $D(f)$ is a unit of $R/(f)$. Then $R/(f)$ is regular.

**Proof.**
It suffices to prove this when $R$ is a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa $. In this case it suffices to prove that $f \not\in \mathfrak m^2$, see Algebra, Lemma 10.106.3. However, if $f \in \mathfrak m^2$ then $D(f) \in \mathfrak m$ by the Leibniz rule, a contradiction.
$\square$

Lemma 15.48.3. Let $(R, \mathfrak m, \kappa )$ be a regular local ring. Let $m \geq 1$. Let $f_1, \ldots , f_ m \in \mathfrak m$. Assume there exist derivations $D_1, \ldots , D_ m : R \to R$ such that $\det _{1 \leq i, j \leq m}(D_ i(f_ j))$ is a unit of $R$. Then $R/(f_1, \ldots , f_ m)$ is regular and $f_1, \ldots , f_ m$ is a regular sequence.

**Proof.**
It suffices to prove that $f_1, \ldots , f_ m$ are $\kappa $-linearly independent in $\mathfrak m/\mathfrak m^2$, see Algebra, Lemma 10.106.3. However, if there is a nontrivial linear relation the we get $\sum a_ i f_ i \in \mathfrak m^2$ for some $a_ i \in R$ but not all $a_ i \in \mathfrak m$. Observe that $D_ i(\mathfrak m^2) \subset \mathfrak m$ and $D_ i(a_ j f_ j) \equiv a_ j D_ i(f_ j) \bmod \mathfrak m$ by the Leibniz rule for derivations. Hence this would imply

\[ \sum a_ j D_ i(f_ j) \in \mathfrak m \]

which would contradict the assumption on the determinant.
$\square$

Lemma 15.48.4. Let $R$ be a regular ring. Let $f \in R$. Assume there exists a derivation $D : R \to R$ such that $D(f)$ is a unit of $R$. Then $R[z]/(z^ n - f)$ is regular for any integer $n \geq 1$. More generally, $R[z]/(p(z) - f)$ is regular for any $p \in \mathbf{Z}[z]$.

**Proof.**
By Algebra, Lemma 10.163.10 we see that $R[z]$ is a regular ring. Apply Lemma 15.48.2 to the extension of $D$ to $R[z]$ which maps $z$ to zero. This works because $D$ annihilates any polynomial with integer coefficients and sends $f$ to a unit.
$\square$

Lemma 15.48.5. Let $p$ be a prime number. Let $B$ be a domain with $p = 0$ in $B$. Let $f \in B$ be an element which is not a $p$th power in the fraction field of $B$. If $B$ is of finite type over a Noetherian complete local ring, then there exists a derivation $D : B \to B$ such that $D(f)$ is not zero.

**Proof.**
Let $R$ be a Noetherian complete local ring such that there exists a finite type ring map $R \to B$. Of course we may replace $R$ by its image in $B$, hence we may assume $R$ is a domain of characteristic $p > 0$ (as well as Noetherian complete local). By Algebra, Lemma 10.160.11 we can write $R$ as a finite extension of $k[[x_1, \ldots , x_ n]]$ for some field $k$ and integer $n$. Hence we may replace $R$ by $k[[x_1, \ldots , x_ n]]$. Next, we use Algebra, Lemma 10.115.7 to factor $R \to B$ as

\[ R \subset R[y_1, \ldots , y_ d] \subset B' \subset B \]

with $B'$ finite over $R[y_1, \ldots , y_ d]$ and $B'_ g \cong B_ g$ for some nonzero $g \in R$. Note that $f' = g^{pN} f \in B'$ for some large integer $N$. It is clear that $f'$ is not a $p$th power in the fraction field of $B'$. If we can find a derivation $D' : B' \to B'$ with $D'(f') \not= 0$, then Lemma 15.48.1 guarantees that $D = g^ MD'$ extends to $B$ for some $M > 0$. Then $D(f) = g^ ND'(f) = g^ MD'(g^{-pN}f') = g^{M - pN}D'(f')$ is nonzero. Thus it suffices to prove the lemma in case $B$ is a finite extension of $A = k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ m]$.

Assume $B$ is a finite extension of $A = k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ m]$. Denote $L$ the fraction field of $B$. Note that $\text{d}f$ is not zero in $\Omega _{L/\mathbf{F}_ p}$, see Algebra, Lemma 10.158.2. We apply Lemma 15.46.5 to find a subfield $k' \subset k$ of finite index such that with $A' = k'[[x_1^ p, \ldots , x_ n^ p]][y_1^ p, \ldots , y_ m^ p]$ the element $\text{d}f$ does not map to zero in $\Omega _{L/K'}$ where $K'$ is the fraction field of $A'$. Thus we can choose a $K'$-derivation $D' : L \to L$ with $D'(f) \not= 0$. Since $A' \subset A$ and $A \subset B$ are finite by construction we see that $A' \subset B$ is finite. Choose $b_1, \ldots , b_ t \in B$ which generate $B$ as an $A'$-module. Then $D'(b_ i) = f_ i/g_ i$ for some $f_ i, g_ i \in B$ with $g_ i \not= 0$. Setting $D = g_1 \ldots g_ t D'$ we win.
$\square$

Lemma 15.48.6. Let $A$ be a Noetherian complete local domain. Then $A$ is J-0.

**Proof.**
By Algebra, Lemma 10.160.11 we can find a regular subring $A_0 \subset A$ with $A$ finite over $A_0$. The induced extension $K/K_0$ of fraction fields is finite. If $K/K_0$ is separable, then we are done by Lemma 15.47.5. If not, then $A_0$ and $A$ have characteristic $p > 0$. For any subextension $K/M/K_0$ there exists a finite subextension $A_0 \subset B \subset A$ whose fraction field is $M$. Hence, arguing by induction on $[K : K_0]$ we may assume there exists $A_0 \subset B \subset A$ such that $B$ is J-0 and $K/M$ has no nontrivial subextensions. In this case, if $K/M$ is separable, then we see that $A$ is J-0 by Lemma 15.47.5. If not, then $K = M[z]/(z^ p - b)$ for some $b \in B$ which is not a $p$th power in $M$. By Lemma 15.48.5 we can find a derivation $D : B \to B$ with $D(b) \not= 0$. Applying Lemma 15.48.4 we see that $A_\mathfrak p$ is regular for any prime $\mathfrak p$ of $A$ lying over a regular prime of $B$ and not containing $D(b)$. As $B$ is J-0 we conclude $A$ is too.
$\square$

Proposition 15.48.7. The following types of rings are J-2:

fields,

Noetherian complete local rings,

$\mathbf{Z}$,

Noetherian local rings of dimension $1$,

Nagata rings of dimension $1$,

Dedekind domains with fraction field of characteristic zero,

finite type ring extensions of any of the above.

**Proof.**
For cases (1), (3), (5), and (6) this is proved by checking condition (4) of Lemma 15.47.6. We will only do this in case $R$ is a Nagata ring of dimension $1$. Let $\mathfrak p \subset R$ be a prime ideal and let $\kappa (\mathfrak p) \subset L$ be a finite purely inseparable extension. If $\mathfrak p \subset R$ is a maximal ideal, then $R \to L$ is finite and $L$ is a regular ring and we've checked the condition. If $\mathfrak p \subset R$ is a minimal prime, then the Nagata condition insures that the integral closure $R' \subset L$ of $R$ in $L$ is finite over $R$. Then $R'$ is a normal domain of dimension $1$ (Algebra, Lemma 10.112.3) hence regular (Algebra, Lemma 10.157.4) and we've checked the condition in this case as well.

For case (2), we will use condition (3) of Lemma 15.47.6. Let $R$ be a Noetherian complete local ring. Note that if $R \to R'$ is finite, then $R'$ is a product of Noetherian complete local rings, see Algebra, Lemma 10.160.2. Hence it suffices to prove that a Noetherian complete local ring which is a domain is J-0, which is Lemma 15.48.6.

For case (4), we also use condition (3) of Lemma 15.47.6. Namely, if $R$ is a local Noetherian ring of dimension $1$ and $R \to R'$ is finite, then $\mathop{\mathrm{Spec}}(R')$ is finite. Since the regular locus is stable under generalization, we see that $R'$ is J-1.
$\square$

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