Bourbaki

Proposition 14

Tag 066Y, permanent. Cite it as [Bourbaki, Tag 066Y].

Algebra, chapter III, § 10. Derivations

*Let AA be a commutative ring, MM an AA-module, BB the AA-algebra T(M)T(M) (resp. S(M)S(M), resp. Λ(M)\Lambda(M)) and EE a (B,B)(B, B)-bimodule. Let d0:A→Ed_0 : A \to E be a derivation of the ring A\mathbf{A} into the A\mathbf{A}-module E\mathbf{E} and d1:M→Ed_1 : M \to \mathbf{E} an additive group homomorphism such that, for all a∈Aa \in \mathbf{A} and all x∈Mx \in M,

(29)d1(ax)=ad1(x)+d0(a)⋅x, (29) \quad d_1(ax) = ad_1(x) + d_0(a) \cdot x,

and further, when B=S(M)B = S(M),

(30)x⋅d1(y)+d1(x)⋅y=y⋅d1(x)+d1(y)⋅x (30) \quad x \cdot d_1(y) + d_1(x) \cdot y = y \cdot d_1(x) + d_1(y) \cdot x

for all x,yx, y in MM, and, when B=⋀(M)B = \bigwedge(M),

(31)x⋅d1(x)+d1(x)⋅x=0 (31) \quad x \cdot d_1(x) + d_1(x) \cdot x = 0

for all x∈Mx \in M. Then there exists one and only one derivation dd of BB (considered as a Z\mathbf{Z}-algebra) into the (B,B)(B, B)-bimodule E\mathbf{E} such that d∣A=d0d|_{\mathbf{A}} = d_0 and d∣M=d1d|_{M} = d_1.

We take on the Z\mathbf{Z}-module B⊕EB \oplus \mathbf{E} the associative Z\mathbf{Z}-algebra structure defined by

(b,t)(b′,t′)=(bb′,bt′+tb′) (b, t)(b', t') = (bb', bt' + tb')

which has (1,0)(1, 0) as unit element (no. 8, Proposition 13). Under the canonical injection t↦(0,t)t \mapsto (0, t), E\mathbf{E} is identified with a two-sided ideal of B⊕EB \oplus \mathbf{E} such that E2={0}\mathbf{E}^2 = \{0\}. On the other hand, the mapping h0:B⊕Eh_0 : B \oplus \mathbf{E} defined by h0(a)=(a,d0(a))h_0(a) = (a, d_0(a)) is a unital ring homomorphism (no. 8, Proposition 12); under this mapping, B⊕EB \oplus \mathbf{E} then becomes an A\mathbf{A}-algebra. Moreover, if, for all x∈Mx \in M, we write h1(x)=(x,d1(x))h_1(x) = (x, d_1(x)), it follows from the definition of h0h_0 and (29) that h1(ax)=h0(a)h1(x)h_1(ax) = h_0(a)h_1(x); in other words h1h_1 is an A\mathbf{A}-linear mapping of MM into B⊕EB \oplus \mathbf{E}. Then there exists one and only one A\mathbf{A}-algebra homomorphism, h:B→B⊕Eh : B \to B \oplus \mathbf{E} such that h∣M=h1h|_M = h_1 (and necessarily h∣A=h0h|_{\mathbf{A}} = h_0): for, if B=T(M)B = T(M), this follows from §5§ 5, no. 1, Proposition 1; if B=S(M)B = S(M), condition (30) shows that h(x)h(y)=h(y)h(x)h(x)h(y) = h(y)h(x) for all x,yx, y in MM and the conclusion follows from §6§ 6, no. 1, Proposition 2; finally if B=⋀(M)B = \bigwedge(M), condition (31) shows that (h(x))2=0(h(x))^2 = 0 for all x∈Mx \in M, since x∧x=0x \wedge x = 0 and the conclusion follows from §7§ 7, no. 1, Proposition 1. The homomorphism hh is such that the composition p∘h:B→Bp \circ h : B \to B with the augmentation ρ:B⊕E→B\rho : B \oplus \mathbf{E} \to B is the identity lBl_B, for p∘hp \circ h and lBl_B coincide by definition for the elements of A\mathbf{A} and those of MM and the set of these elements is a generating system of BB. We can therefore write h(b)=(b,d(b))h(b) = (b, d(b)) for all b∈Bb \in B and the mapping b↦d(b)b \mapsto d(b) of BB into E\mathbf{E} is a derivation with the required properties, by virtue of Proposition 12 of no. 8.

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

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