Bourbaki

Corollary

Tag 0520, permanent. Cite it as [Bourbaki, Tag 0520].

Commutative Algebra, chapter VII, § 3. Factorial domains

With the hypotheses and notation of Proposition 5, suppose that s≥1s \geq 1, so that f∈Bm+BXf \in Bm + BX. Then the A-homomorphism hh of B′=A[[T]]B' = A[[T]] to B=A[[X]]B = A[[X]] such that h(T)=fh(T) = f (Chapter III, § 2, no. 9, Proposition 11 (a)) defines on B a free B′B'-module structure admitting {1,X,…,Xs−1}\{1, X, \ldots, X^{s-1}\} as basis. In particular hh is injective.

Let the B′B'-module B be given the (T)-adic filtration, which consists of the fnBf^n B for n≥0n \geq 0 (Chapter 111, § 2, no. 1). Then B/fBB/fB is a free module over the ring A=B′/TB′A = B'/TB' and the images of the XiX^i (0≤i≤s−10 \leq i \leq s-1) in this A-module form a basis of it (Proposition 5); as moreover ff is not a divisor of zero in B (Proposition 5), Bfn/Bfn+1Bf^n/Bf^{n+1} is also a free (B′/TB′)(B'/TB')-module of rank s, so that condition (GR) of Chapter 111, § 2, no. 8 is satisfied (replacing A by B' and M by B). On the other hand, since B' is Hausdorff and complete with respect to the (T)-adic filtration and gr(B) is a finitely generated gr(B')-module by the above, it is seen first (Chapter III, § 2, no. 9, Corollary 1 to Proposition 12) that B is a finitely generated B' module. The first assertion of the corollary then follows from Chapter III, § 2, no. 9, Proposition 13. The second follows immediately from it.

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.