We shall denote by K the field of fractions of A and by $\mathscr{P}^*$ (or $\mathscr{P}^*(A)$) the ordered group of non-zero fractional principal ideals of A. The proof will be carried out by proving the following implications:

(a) ⇔ (b) ⇔ (c) ⇔ (d) ⇔ (e)

We show that (a) implies (b); if A is factorial, $\mathscr{P}^*$ is isomorphic to the group of divisors of A and hence to a direct sum of groups $\mathbf{Z}$ ($§ 1$, no. 3, Theorem 2).

Note now that the relation "the intersection of two integral principal ideals of A is a principal ideal" means that every ordered pair of elements of A admits a $lcm$, that is that $\mathscr{P}^*$ is a lattice-ordered group (Algebra, Chapter VI, $§ 1$, no. 9, Proposition 8). The fact that (b) implies (c) (and even is equivalent to it) therefore follows from Algebra, Chapter VI, $§ 1$, no. 13, Theorem 2. The fact that (c) implies (d) follows from Algebra, Chapter VI, $§ 1$, no. 13, Proposition 14 (DIV).

The fact that (d) implies (b) follows from Algebra, Chapter VI, $§ 1$, no. 13, Theorem 2 applied to the group $\mathscr{S}^*$.

We show that (b) implies (e). If (b) holds, there is an isomorphism of $\mathscr{P}^*$ onto $\mathbf{Z}^{(l)}$; let $(v_i(x))_{i \in I}$ denote the element of $\mathbf{Z}^{(l)}$ corresponding to the ideal $Ax$ ($x \in K^*$). It is seen immediately that each $v_i$ is a discrete valuation on K, that A is the intersection of the rings of the $v_i$ and that, for $x \in K^*$, $v_i(x) = 0$ except for a finite number of indices $i$; hence A is a Krull domain. On the other hand, let q be a prime ideal of A of height 1; it contains a non-zero element a which is necessarily not invertible and hence also (by definition of a prime ideal) one of the extremal elements of A; as $Ap$ is prime and non zero, $q = Ap$, which proves that q is principal.

Finally we show that (e) implies (a). Let a be a divisorial ideal of A. There exist prime ideals $p_i$ of A of height 1 such that $\operatorname{div} a = \sum_i n_i \operatorname{div} p_i$ where $n_i \in \mathbf{Z}$. If (e) holds, $p_i$ is of the form $Ap_i$, whence $\operatorname{div} a = \operatorname{div} \left( \prod_i Ap_i^{n_i} \right)$ and hence $a = \prod_i Ap_i^{n_i}$ since a is divisorial.

Suppose that every divisorial ideal of $A$ is invertible; as $A_{,,}$ is a Krull domain ($§ 1$, no. 4, Proposition 6), every divisorial ideal $a$ of $A_{,,}$ is the intersection of two principal fractional ideals ($§ 1$, no. 5, Corollary 2 to Proposition 9); hence $a = bA_m$, where $b$ is a divisorial ideal of $A$ (Chapter II, $§ 2$, no. 4); as $b$ is invertible by hypothesis, we deduce from Chapter II, $§ 5$, no. 6, Theorem 4 that $a$ is principal and hence $A_{,,}$ is a factorial domain (no. 1, Definition 1). Conversely, if all the $A_{,,}$ are factorial and $c$ is a finitely generated divisorial ideal of $A$, $cA_m$ is a divisorial ideal of $A_{,,}$, as follows from $§ 1$, no. 5, Corollary 2 to Proposition 9 and Chapter II, $§ 2$, no. 4; by hypothesis $cA_m$ is principal and hence it follows from Chapter 11, $§ 5$, no. 6, Theorem 4 that $c$ is invertible.

If $P$ satisfies this condition, clearly all its elements are extremal and every extremal element of $A$ is associated with a unique element of $P$. Recall that $P$ is then called a representative system of extremal elements of $A$ (Algebra, Chapter VII, $§ 1$, no. 3, Definition 2).

Suppose always that $A$ is factorial. It has been seen (no. 2, Theorem 1) that the group $\mathscr{P}^*$ is a lattice. We may therefore apply the results of Algebra, Chapter VI, $§ 1$, nos. 9 and 13. In particular, every element of $K^*$ may be written in an essentially unique way in the form of an irreducible fraction. Any two elements $a, b$ of $K^*$ have a g.c.d. and a l.c.m.; if $a = u \prod_{p \in P} p^{n(p)}$ and

are decompositions of $a$ and $b$ as products of extremal elements, then:

where $w, w'$ are in U. We recover, in particular, the results of Algebra, Chapter VIII, § 1, no. 3.

For all $p \in P$, the mapping $a \mapsto n(p)$ is a discrete valuation $v_p$ on K whose ring is obviously $A_{A_p}$. It follows from Theorem 1(e) that the $v_p$ are just the essential valuations of A and that the ideals $Ap$ ($p \in P$) are just the prime ideals of A of height 1.

(i) If A is factorial, $S^{-1}A$ is factorial. (ii) If S is generated by a family of elements $p_i$ such that the principal ideals $Ap_i$ are prime and $S^{-1}A$ is factorial, then A is factorial.

This follows immediately from Definition 1 of no. 1 and § 1, no. 10, Proposition 17.

Let d be a content of $ff'$. For every valuation $\nu$ on K which is essential for A, let $\bar{\nu}$ denote its canonical extension to $K[X]$. Then

Hence $cc'd^{-1}$ is an invertible element of A.

Let $f$ be a non-zero element of $A[X]$. In the ring $K[X]$ $f$ can be decomposed uniquely in the form:

Lemma 1 proves that $a$ is a content off. Hence $a \in A$. As $A$ is factorial, $a$ can be decomposed uniquely in the form:

Whence the existence and uniqueness of the decomposition:

Note that this theorem proves that every element of $A$ admits the same decomposition into extremal elements in $A$ and $A[X]$. The g.c.d. of a family of elements of $A$ is therefore the same in $A$ and in $A[X]$.

We may also use Proposition 18 of § 1, no. 10 to show that $A[X]$ is a factorial domain if and only if $A$ is a factorial domain.

Argue by induction on $n$.

This corollary may be extended to the case of an infinite family of indeterminates (cf. Exercise 2).

This follows from no. 1, Definition 1 and § 1, no. 10, Proposition 16.

Suppose that integers $u(i) \geqslant 1$ ($1 \leqslant i \leqslant k - 1$) are determined such that $f(X_n^{u(1)}, \ldots, X_n^{u(k-1)}, X_k, \ldots, X_n) \neq 0$. We shall determine an integer $u(k) \geqslant 1$ such that

The lemma will then be proved by induction.

Observe that the series $f(X_n^{u(1)}, \ldots, X_n^{u(k-1)}, X_k, \ldots, X)$ can be considered as a series in $X_k$ and $X_n$ with coefficients in $E[[X_{k+1}, \ldots, X_{n-1}]]$. Thus we see that it suffices to establish the lemma for $n = 2$.

Therefore let

where $f \neq 0$. Let $G \subset \mathbf{N} \times \mathbf{N}$ be the non-empty set of ordered pairs $(i, j)$ such that $e_{ij} \neq 0$. Let $\mathbf{N} \times \mathbf{N}$ be given the lexicographical ordering. Let $(c, d)$ be the least element of $G$. Choose an integer $p > d$. In the expansion of we look for the terms of degree $cp + d$. If $ip + j = cp + d, i \geq c + 1$ is impossible, for this would give nor is $i < c$ possible, for $(c, d)$ is the least element of $G$; therefore $i = c$ and then $j = d$. The term of degree $cp + d$ in $f(T^p, T)$ is therefore $e_{cd} T^{cp + d}$. Since $e_{cd} \neq 0, f(T^p, T) \neq 0$. Whence the lemma.

In the ring $E[[X_1, \ldots, X_n]]$, let $a$ be the ideal of formal power series without constant term. If $w_1, \ldots, w_n$ are elements of $a$, recall that the mapping $f(X_1, \ldots, X_n) \mapsto f(w_1, \ldots, w_n)$ is the unique endomorphism $s$ of the ring $E[[X_1, \ldots, X_n]]$ such that $s(X_i) = w_i$ for $1 \leq i \leq n$ (Chapter III, § 4, no. 5, Proposition 6).

We take $w_1 = X_1 + X_n^{u(1)}, \ldots, w_{n-1} = X_{n-1} + X_n^{u(n-1)}, w_n = X_n$, where the $u(i)$ are integers $\geq 1$. Let $s'$ be the endomorphism of $E[[X_1, \ldots, X_n]]$ which maps $X_1$ to $X_1 - X_n^{u(1)}, \ldots, X_{n-1}$ to $X_{n-1} - X_n^{u(n-1)}$ and $X_n$ to $X_n$. Then $s'(s(X_i)) = X_i$ for $1 \leq i \leq n$ and hence $s' \circ s$ is the identity automorphism; similarly for $s \circ s'$. Hence $s$ is an automorphism.

$g(0, \ldots, 0, X_n) = f(X_n^{u(1)}, \ldots, X_n^{u(n-1)}, X_n)$. Lemma 3 is therefore a consequence of Lemma 2.

(a) We show that $f \mid B \cap M = (0)$. Suppose that there is a relation:

We show that the $b_i$ (and hence the $r_j$) are all zero, which will prove in particular that $f$ is not a divisor of zero in $B$. Since $A$ is Hausdorff, it suffices to show that $b_i \in m^n$ for all $i \geq 0$ and all $n \geq 0$. It is obvious for $n = 0$. We shall argue by double induction: we shall assume that $b_i \in m^{n-1}$ for all $i$ and $b_i \in m^n$ for $i < k$ and show that this implies that $b_k \in m^n$. For this, we write $f = \sum_{i=1}^{\infty} a_i X^i$ and compare the coefficients of $X^{s+k}$ in (3); then: The terms in the first bracket belong to $m^n$ since the $b_i \in m^n$ for $i < k$; similarly for those in the second, since the $b_i \in m^{n-1}$ for all $i$ and the $a_i \in m$ for $i \leq s-1$. Hence $b_k a_s \in m^n$ and, as $a_s$ is an invertible element of $A$, $b_k \in m^n$.

(b) We show that $f \mid B + M = B$. We write

it is an invertible element of $B$. Then if therefore we write $fg^{-1} - X^s = (f - X^s g) g^{-1} = -h$, the coefficients of $h$ belong to $m$. Then let $r$ be an element of $B$. By induction on $n$ we define a sequence $(q^{(n)})$ of elements of B: we take $q^{(0)}$ to be the unique series satisfying: writing $h = \sum_{i=0}^m h_i X^i$ and $q^{(n)} = \sum_{i=0}^\infty q_i^{(n)} X^i$, the $q_i^{(n)}$ are defined by: It follows immediately from (6) that: As $h_j \in m$ for all $j$, it also follows from (6), by induction on $n$, that $q_i^{(n)} \in m^n$ for all $i$ and all $n$. As A is complete, it follows that the series converges to an element $q$ of B. By (5) and (7), As M is closed, at the limit (8) gives $r \equiv (X^s - h)q \pmod{M}$, that is We may also use the results of Chapter III, § 2 to show the relation $B = fB + M$ (cf. Exercise 12). The method followed here has the advantage of being applicable to convergent series.

Let the $B'$-module B be given the (T)-adic filtration, which consists of the $f^n B$ for $n \geq 0$ (Chapter 111, § 2, no. 1). Then $B/fB$ is a free module over the ring $A = B'/TB'$ and the images of the $X^i$ ($0 \leq i \leq s-1$) in this A-module form a basis of it (Proposition 5); as moreover $f$ is not a divisor of zero in B (Proposition 5), $Bf^n/Bf^{n+1}$ is also a free $(B'/TB')$-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.

We write $F = X^s + G$, where $G = g_0 + \cdots g_{s-1} X^{s-1}$ ($g_i \in \mathbf{A}$). The relation $f = uF$ is equivalent to $F = u^{-1}f$, that is to $X^s = u^{-1}f - G$. Hence Proposition 5 shows the uniqueness of $G$ and $u^{-1}$ and therefore of $F$ and $u$. It also shows that there exist $v \in B$ and a polynomial $G = g_0 + \cdots + g_{s-1} X^{s-1}$ ($g_i \in \mathbf{A}$) such that $X^s = v - G$; it remains to show that $v$ is invertible in $B$ and that $g_i \in \mathfrak{m}$ for all $i$. Now, writing $\bar{g}_i$ for the canonical image of $g_i$ in $k$ and $\bar{f}, \bar{v}$ for the reduced series of $f, g$,

since $\bar{f}$ is of order $s$, $\bar{g}_i = 0$ for all $i$ and $\bar{v}$ is of order 0, hence $v$ is invertible.

In fact, the reduced series of $g$ and $h$ are $\neq 0$; hence, by Proposition 6, there exist invertible elements $u, v$ of $B$ such that $ug$ and $vh$ are distinguished polynomials. Then $uvF = (ug)(vh)$ is a distinguished polynomial and $uv$ is invertible. Passing to the reduced series, it is seen immediately that $F$ and $uvF$ have the same reduced order, that is the same degree. The uniqueness assertion in Proposition 6 therefore shows that $F = uvF$, whence $uv = 1$.

Suppose that $F$ is not extremal in $\mathbf{A}[X]$, so that $F = f_1 f_2$, where $f_1$ and $f_2$ are non-invertible elements of $\mathbf{A}[X]$; the product of the dominant coefficients of $f_1$ and $f_2$ being equal to 1, these coefficients are invertible in $\mathbf{A}$ and the hypothesis implies that $f_1$ and $f_2$ are of degrees $> 0$ and $< s$; as the reduced polynomials $\bar{f}_1, \bar{f}_2$ satisfy $\bar{f}_1 \bar{f}_2 = X^s$, neither $\bar{f}_1$ nor $\bar{f}_2$ can be invertible in $k[[X]]$, for, if $\bar{f}_1$ were invertible, $\bar{f}_2$ would be of order $s$, which is absurd. A fortiori, neither $f_1$ nor $f_2$ is invertible in $B$ and $F$ is not extremal in $B$.

Conversely, if F is not extremal in $A[[X]]$, then $F = gh$, where neither g nor his invertible in B; their reduced orders are therefore $\geq 1$; then the distinguished polynomials $ug$ and $u^{-1}h$ of Proposition 7 are not constant, which shows that F is not extremal in $A[X]$.

Let $p$ be the maximal ideal of C and $x$ a generator of $p$ (if C is a field, then $\pi = 0$). Let C be given the $p$-adic topology, which is Hausdorff. As C is a Noetherian local ring, $B = C[[X_1, \ldots, X_n]]$ is a Noetherian local ring and its completion is $\hat{C}[[X_1, \ldots, X_n]]$ (Chapter III, §2, no. 6, Proposition 6). By the Corollary to Proposition 4 (no. 6), it suffices to prove that $\hat{C}[[X_1, \ldots, X_n]]$ is factorial. Now, if C is a field, then $\hat{C} = C$; if C is a discrete valuation ring, the same is true of $\hat{C}$ (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 $n = 0$, we shall assume that it has been proved that $A = C[[X_1, \ldots, X_{n-1}]]$ is factorial. We shall identify B with $A[[X_n]]$ and denote by m the maximal ideal of A (generated by $\pi, X_1, \ldots, X_{n-1}$). 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 $C/C\pi$; as $B/B\pi$ is identified with $K[[X_1, \ldots, X_n]]$, the ideal $Bx$ is prime and $x$ is extremal. If $x \neq 0$, $B_{B\pi}$ 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 $g = \pi^{w(g)} f$, where $f \in B$ and f is not a multiple of $\pi$. It will therefore suffice to show that $f$ is an essentially unique product of extremal elements. Now the canonical image of $f$ in $K[[X_1, \ldots, X_n]]$ is not zero; Lemma 3 (no. 7) therefore shows that there exists an automorphism of B which maps $f$ to an element $f'$ such that the coefficients of $f'(0, \ldots, 0, X_n)$ are not all in $Cx$; this means that the coefficients of the series $f'$, considered as a formal power series in $X_n$, are not all in m. It will suffice to prove our assertion for $f'$.

In what follows, all the elements of B will be considered as formal power series in $X_n$ 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 $\neq 0$), $f'$ is associated, in B, with a unique distinguished polynomial F. By Proposition 7 of no. 8, every series which divides $f'$ (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 $f' = u F_1 \cdots F_q$, where $u$ is invertible and the $F_i$ are extremal distinguished polynomials (in B) such that $F = F_1 \cdots F_q$. By the Corollary to Proposition 7 of no. 8, the $F_i$ are also extremal in $\mathbf{A}[X_n]$. Now, as $\mathbf{A}$ is factorial by the induction hypothesis, so is $\mathbf{A}[X_n]$ (Theorem 2, no. 5); hence, since they are monic, the $F_i$ are uniquely determined by $F$ (up to a permutation). This shows the uniqueness of the decomposition $f' = u F_1 \ldots F_r$; its existence follows from the fact that $B$ is Noetherian, which completes the proof.

Remarks (1) There exist factorial rings $\mathbf{A}$ such that the ring $\mathbf{A}[[X]]$ is not factorial (Exercise 8). However, if $\mathbf{A}$ is a principal ideal domain, $\mathbf{A}[[X_1, \ldots, X_n]]$ 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. \*