## 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.

1. 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.

2. 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. $\square$

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 $K = M[z]/(z^ p - b_1/b_2)$ for some $b_1, b_2 \in B$ with $b_2 \not= 0$ and $b_1/b_2$ not a $p$th power in $M$. Choose $a \in A$ nonzero such that $az \in A$. After replacing $z$ by $b_2 a^ p z$ we obtain $K = M[z]/(z^ p - b)$ with $z \in A$ and $b \in B$ 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:

1. fields,

2. Noetherian complete local rings,

3. $\mathbf{Z}$,

4. Noetherian local rings of dimension $1$,

5. Nagata rings of dimension $1$,

6. Dedekind domains with fraction field of characteristic zero,

7. 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 $L/\kappa (\mathfrak p)$ 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. $\square$

Comment #6542 by Yijin Wang on

In lemma 15.48.6,the last line,'not containing D(f)'should be 'not contaning f'.

Comment #6543 by 嘉然今晚吃什么 on

In lemma 15.48.6,the last third line,'D(f) not equal to 0'should be 'D(b) not equal to 0'

Comment #6544 by Yijin Wang on

I'm sorry there is a typo in my comment,In lemma 15.48.6,the last line,'not containing D(f)'should be 'not contaning D(b)'

Comment #6654 by 国蝻 on

I'm sorry but how can I deduce that A=B[z]/(z^p-b) or A\subset B[z]/(z^p-b) in the proof of lemma 07PI? (So we can use lemma07PG or lemma07PA ?)

Comment #6655 by on

@国蝻 Yes, good question! I guess I should have said that we may and do choose $z$ to be an element of $A$. We apply Lemma 15.48.4 to the ring extension $A' = B[z]/(z^p - b)$ of $B$ and we see that $A'$ is J-0. Then $A' \subset A$ induces an isomorphism of fraction fields. Hence $A$ is J-0 by Lemma 15.47.5. Do you agree?

Comment #6656 by 国蝻 on

@ Johan Yes I totally agree. Thank you very much.

Comment #6657 by 国蝻 on

I think there is still one point in the proof of lemma 07PI: Assume that K=M[a] with a\in A, to guarantee that a^p\in B I think it is better to take B as the integral closure of A in M.

Comment #6659 by on

@#6657. You can do that but you don't need to do that because you can always get $a^p \in B$ by clearing denominators. Namely, if $a^p = b_1/b_2$ for $b_i \in B$ with $b_2$ nonzero, then you have $(b_2 a)^p \in B$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).