Bourbaki

Proposition 8

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

Algebra, chapter IV, § 4. Formal power series

— Let X=(Xi)iX = (X_i)_i, be a finite family of indeterminates. (i) Every derivation of the ring of formal power series A[[X]]\mathbf{A}[[X]] is continuous. (ii) Every derivation of the polynomial ring A[X]\mathbf{A}[X] into the ring of formal power series A[[X]]\mathbf{A}[[X]] extends in a unique fashion to a derivation of the ring A[[X]]\mathbf{A}[[X]]. (iii) The family (Di)i∈I(D_i)_{i \in I} is a basis of the A[[X]]\mathbf{A}[[X]]-module of A-derivations of A[[X]]\mathbf{A}[[X]] into itself. Let bnb_n be the set of all formal power series of order ≥n\geq n. It is clear that bnb_n is an ideal in the ring A[[X]]\mathbf{A}[[X]], generated by the monomials of degree nn. Hence bnb_n consists of finite sums of products of nn formal power series without constant terms; if DD is a derivation of A[[X]]\mathbf{A}[[X]], we have

D(f1⋯fn)=∑i=1nf1⋯fi−1D(fi)fi+1⋯fn, D(f_1 \cdots f_n) = \sum_{i=1}^n f_1 \cdots f_{i-1} D(f_i) f_{i+1} \cdots f_n ,

whence it follows at once that Dbn⊂bn−1Db_n \subset b_{n-1} for n≥1n \geq 1. Since the sequence (bn)n≥0(b_n)_{n \geq 0} is a fundamental system of neighbourhoods of 0 in A[[X]]\mathbf{A}[[X]] (IV, p. 26 remark), D is continuous and (i) is proved.

Let AA be a derivation of A[X]A[X] into A[[X]]A[[X]]. Arguing as before, we can show that A(h)A(h) belongs to b,−1b, -_1, for every homogeneous polynomial hh of degree n≥1n \geq 1. Now let u∈A[[X]]u \in A[[X]] and let unu_n be the homogeneous component of degree nn of uu. Since A(un)∈b,−1A(u_n) \in b, -_1 for n≥1n \geq 1, the family (A(u)),…(A(u)), \ldots, is summable in A[[X]]A[[X]] and we can define a derivation DD of A[[X]]A[[X]] into itself by

D(u)=∑n≥0Δ(un) D(u) = \sum_{n \geq 0} \Delta(u_n)

We have D(bn)⊂bn−1D(b_n) \subset b_{n-1}, hence DD is a continuous endomorphism of the additive group of A[[X]]A[[X]]. The mapping Φ:(u,v)↦D(uv)−uD(v)−D(u)v\Phi : (u, v) \mapsto D(uv) - uD(v) - D(u)v of A[[X]]×A[[X]]A[[X]] \times A[[X]] into A[[X]]A[[X]] is continuous and zero on A[X]×A[X]A[X] \times A[X]. Since A[X]A[X] is dense in A[[X]]A[[X]], we have Φ=0\Phi = 0; in other words, DD is a derivation of A[[X]]A[[X]] into itself, extending AA.

Finally, A[X]A[X] is dense in A[[X]]A[[X]] and every derivation of A[[X]]A[[X]] is continuous by (i); hence there exists a unique extension of AA to a derivation of A[[X]]A[[X]]. This proves (ii).

It remains to prove (iii). Formula (18) (IV, p. 33) shows that the family (Di)i∈I(D_i)_{i \in I} is linearly independent over A[[X]]A[[X]], and formula (16) (IV, p. 32), applied in the case E=A[[X]]E = A[[X]], shows that every AA-derivation is a linear combination of the DiD_i with coefficients in A[[X]]A[[X]].

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

Cited by

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Cites

Transcribed from the printed edition, 2003, Springer. Not translated.