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. $\square$

Comment #3524 by Dario Weißmann on

Don't we need that $g\in R$ is a nonzerodivisor in $R'$? For the last step in the proof we regard $R'$ as a subring of $R'_g$ right?

Comment #3758 by Laurent Moret-Bailly on

Typo in last sentence of statement: "an finite type extension".

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