Proposition 4

Tag 06BQ, permanent. Cite it as [Bourbaki, Tag 06BQ].

Algebra, chapter IV, § 3. Rational fractions

— *Let $E$ be an associative, commutative and unital $K$ -algebra, $x = (x_i)_{i, I}$ a family of elements of $E$ and $f \in K((X_i)_{i, I})$ . Suppose that $x$ is substitutable in $f$ and $y = f(x)$ . (i) For every derivation $A$ of $K((X_i)_{i, I})$ which maps $K[(X_i)_{i, I}]$ into itself, $x$ is substitutable in $Af$ . (ii) For every derivation $D$ of $E$ into an $E$ -module we have

Dy = \sum_{i \in I} (D_i f)(x) \cdot Dx_i .

Let $f = \frac{u}{v}$ with $u, v \in K[(X_i)_{i \in I}]$ and $v(x)$ invertible in E. Let A be a derivation of $K((X_i)_{i \in I})$ mapping $K[(X_i)_{i \in I}]$ into itself. We have

\Delta f = \frac{(\Delta u)\ v - u(\Delta v)}{v^2}

and

v^2(x)

is invertible, hence x is substitutable in

Af

. Secondly put

r = u(x)

s = v(x)

; we have

y = s^{-1}r

, hence for every derivation D of E into an E-module we have

\begin{align*} Dy &= s^{-2}(s(Dr) - r(Ds)) \\ &= s^{-2}\left( s \sum_{i \in I} (D_iu)(x) \cdot Dx_i - r \sum_{i \in I} (D_iv)(x) \cdot Dx_i \right) \end{align*}

by Prop. 4 of IV, p. 6. Thus

Dy = \sum_{i \in I} w_i \cdot Dx_i

with

w_i = v(x)^{-2}(v(x)(D_iu)(x) - u(x)(D_iv)(x)) = (D_if)(x).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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