Bourbaki

Proposition 16

Tag 0671, permanent. Cite it as [Bourbaki, Tag 0671].

Algebra, chapter III, § 10. Derivations

Let AA be a commutative KK-algebra, MiM_i (1≤i≤n1 \leq i \leq n) and PP AA-modules and HH the AA-module of AA-multilinear mappings of M1×M2×⋯×MnM_1 \times M_2 \times \cdots \times M_n into P. Suppose that there is given a K-derivation d0:A→Ad_0 : A \to A of the algebra A, for each i, a K-linear mapping di:Mi→Mid_i : M_i \to M_i and a K-linear mapping D:P→PD : P \to P, so that, for 1≤i≤n1 \leq i \leq n, (d0,di,di)(d_0, d_i, d_i) is a K-derivation of (A,Mi,Mi)(A, M_i, M_i) into itself and (d0,D,D)(d_0, D, D) is a K-derivation of (A,P,P)(A, P, P) into itself. Then there exists a K-linear mapping D′:H→HD' : H \to H such that (d0,D′,D′)(d_0, D', D') is a K-derivation of (A,H,H)(A, H, H) into itself and

(37) D(f(x1,…,xn))D(f(x_1, \ldots, x_n))

=(D′f)(x1,…,xn)+∑i=1nf(x1,…,xi−1,dixi,xi+1,…,xn) = (D'f)(x_1, \ldots, x_n) + \sum_{i=1}^n f(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)

for all xi∈Mix_i \in M_i for 1≤i≤n1 \leq i \leq n and f∈Hf \in H.

We show that for f∈Hf \in H, the mapping D′fD'f of M1×M2×⋯×MnM_1 \times M_2 \times \cdots \times M_n into P defined by (37) is A-multilinear. For a∈Aa \in A,

(D′f)(ax1,x2,…,xn)=D(af(x1,…,xn))−f(d1(ax1),x2,…,xn) (D'f)(ax_1, x_2, \ldots, x_n) = D(af(x_1, \ldots, x_n)) - f(d_1(ax_1), x_2, \ldots, x_n)
−a∑i=2nf(x1,…,xi−1,dixi,xi+1,…,xn) -a \sum_{i=2}^n f(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)

and by hypothesis

D(af(x1,…,xn))=(d0a)f(x1,…,xn)+aD(f(x1,…,xn)) D(af(x_1, \ldots, x_n)) = (d_0a)f(x_1, \ldots, x_n) + aD(f(x_1, \ldots, x_n))

and d1(ax1)=(d0a)x1+a.d1x1d_1(ax_1) = (d_0a)x_1 + a.d_1x_1, which gives

(D′f)(ax1,x2,…,xn)=a.(D′f)(x1,…,xn) (D'f)(ax_1, x_2, \ldots, x_n) = a.(D'f)(x_1, \ldots, x_n)

and linearity in each of the xix_i is proved similarly. On the other hand,

(D′(af))(x1,…,xn)=D(af(x1,…,xn)) (D'(af))(x_1, \ldots, x_n) = D(af(x_1, \ldots, x_n))
−∑i=1naf(x1,…,xi−1,dixi,xi+1,…,xn) - \sum_{i=1}^n af(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)
=(d0a)f(x1,…,xn))+aD(f(x1,…,xn)) = (d_0a)f(x_1, \ldots, x_n)) + aD(f(x_1, \ldots, x_n))
−∑i=1naf(x1,…,xi−1,dixi,xi+1,…,xn) - \sum_{i=1}^n af(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)
=(d0a)f(x1,…,xn)+a(D′f)(x1,…,xn) = (d_0a)f(x_1, \ldots, x_n) + a(D'f)(x_1, \ldots, x_n)

in other words

D′(af)=(d0a)f+a(D′f) D'(af) = (d_0a)f + a(D'f)

which establishes the proposition.

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

Cited by

Cites

Nothing.

Transcribed from the printed edition, 1998, Springer. Not translated.