Bourbaki

Proposition 8

Tag 0525, permanent. Cite it as [Bourbaki, Tag 0525].

Commutative Algebra, chapter VII, § 3. Factorial domains

Let C be a ring which is either a field or a discrete valuation ring. Then the domain of formal power series C[[X1,…,Xn]]C[[X_1, \ldots, X_n]] is factorial.

Let pp be the maximal ideal of C and xx a generator of pp (if C is a field, then π=0\pi = 0). Let C be given the pp-adic topology, which is Hausdorff. As C is a Noetherian local ring, B=C[[X1,…,Xn]]B = C[[X_1, \ldots, X_n]] is a Noetherian local ring and its completion is C^[[X1,…,Xn]]\hat{C}[[X_1, \ldots, X_n]] (Chapter III, §2, no. 6, Proposition 6). By the Corollary to Proposition 4 (no. 6), it suffices to prove that C^[[X1,…,Xn]]\hat{C}[[X_1, \ldots, X_n]] is factorial. Now, if C is a field, then C^=C\hat{C} = C; if C is a discrete valuation ring, the same is true of C^\hat{C} (Chapter VI, §5, no. 3, Proposition 5). We shall therefore assume in the remainder of the proof that C is complete.

Arguing by induction starting with the trivial case n=0n = 0, we shall assume that it has been proved that A=C[[X1,…,Xn−1]]A = C[[X_1, \ldots, X_{n-1}]] is factorial. We shall identify B with A[[Xn]]A[[X_n]] and denote by m the maximal ideal of A (generated by π,X1,…,Xn−1\pi, X_1, \ldots, X_{n-1}). We shall prove that every non-zero element g of B is, in an essentially unique way, a product of extremal elements.

Let K be the field C/CπC/C\pi; as B/BπB/B\pi is identified with K[[X1,…,Xn]]K[[X_1, \ldots, X_n]], the ideal BxBx is prime and xx is extremal. If x≠0x \neq 0, BBπB_{B\pi} is therefore the ring of a normed discrete valuation w (Chapter VI, §3, no. 6, Proposition 9); every non-zero element g of B may therefore be written as g=πw(g)fg = \pi^{w(g)} f, where f∈Bf \in B and f is not a multiple of π\pi. It will therefore suffice to show that ff is an essentially unique product of extremal elements. Now the canonical image of ff in K[[X1,…,Xn]]K[[X_1, \ldots, X_n]] is not zero; Lemma 3 (no. 7) therefore shows that there exists an automorphism of B which maps ff to an element f′f' such that the coefficients of f′(0,…,0,Xn)f'(0, \ldots, 0, X_n) are not all in CxCx; this means that the coefficients of the series f′f', considered as a formal power series in XnX_n, are not all in m. It will suffice to prove our assertion for f′f'.

In what follows, all the elements of B will be considered as formal power series in XnX_n with coefficients in A. By Proposition 6 of no. 8 (applicable since C and therefore A are separable and complete and the reduced series off' is ≠0\neq 0), f′f' is associated, in B, with a unique distinguished polynomial F. By Proposition 7 of no. 8, every series which divides f′f' (or, what amounts to the same, which divides F) is associated with a distinguished polynomial which divides F and every decomposition off' is, to within invertible factors, of the form f′=uF1⋯Fqf' = u F_1 \cdots F_q, where uu is invertible and the FiF_i are extremal distinguished polynomials (in B) such that F=F1⋯FqF = F_1 \cdots F_q. By the Corollary to Proposition 7 of no. 8, the FiF_i are also extremal in A[Xn]\mathbf{A}[X_n]. Now, as A\mathbf{A} is factorial by the induction hypothesis, so is A[Xn]\mathbf{A}[X_n] (Theorem 2, no. 5); hence, since they are monic, the FiF_i are uniquely determined by FF (up to a permutation). This shows the uniqueness of the decomposition f′=uF1…Frf' = u F_1 \ldots F_r; its existence follows from the fact that BB is Noetherian, which completes the proof.

Remarks (1) There exist factorial rings A\mathbf{A} such that the ring A[[X]]\mathbf{A}[[X]] is not factorial (Exercise 8). However, if A\mathbf{A} is a principal ideal domain, A[[X1,…,Xn]]\mathbf{A}[[X_1, \ldots, X_n]] is factorial (Exercise 9). (2) \* We shall see later, by homological methods, that every regular local ring is factorial (cf. § 4, no. 7, Corollary 3 to Proposition 16). This will give another proof, conceptually simpler, of Proposition 8. \*

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

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