Isotropy and Factorization in Reduced Witt Rings

Robert W. Fitzgerald

Received: April 5, 2001

Communicated by Ulf Rehmann

Abstract. We consider reduced Witt rings of finite chain length. We show there is a bound, in terms of the chain length and maximal signature, on the dimension of anisotropic, totally indefinite forms. From this we get the ascending chain condition on principal ideals and hence factorization of forms into products of irreducible forms.

2000 Mathematics Subject Classification: 11E81, 12D15

Keywords and Phrases: isotropy, quadratic forms, Witt ring

R will denote a (real) reduced Witt ring. A form q ∈ R is totally indefinite if |sgn_{α}q| < dim q for all orderings α of R. It is well-known that such a form need not be isotropic. However, when R has finite chain length, cl(R), we show there are restrictions on the possible dimensions of anisotropic, totally indefinite forms. To be specific,

dim q ≤ \frac{1}{2}cl(R) max_{α}\{| sgn_{α}q|^{2}\},

unless R = ℤ and q is one-dimensional. The proof depends on Marshall’s classification of reduced Witt rings of finite chain length. This bound allows us to show that R, of finite chain length, satisfies the ascending chain condition on principal ideals. One consequence of this result is that chains of basic clopen sets H(a_{1}, …, a_{n}), for fixed n, stabilize. Another consequence is that non-zero, non-units of R factor into a finite product of irreducible elements (in the sense of Anderson and Valdes-Leon). This had been previously known only for odd dimensional forms in rings with only finitely many orderings. Conversely, we show, for a wide class of reduced Witt rings R, that the ascending chain condition on principal ideals implies R has finite chain length. The proof relies on Marshall’s notion of a sheaf product. We close with examples of factorization into irreducible elements. These illustrate how the factorization of even dimensional forms is less well behaved than the factorization of odd dimensional forms studied in [8].

We set some of the notation. R will be an abstract Witt ring, in the sense of Marshall [11], and reduced. The main case of interest is the Witt ring of a Pythagorean field. X_{R}, or just X if the ring is understood, denotes the set of orderings (equivalently, signatures) on R. We always assume X is non-empty. For a form q ∈ R and ordering α ∈ X, the signature of q at α will be denoted by either sgn_{α}q or q̂(α). We let G_{R}, or just G when R is understood, denote the group of onedimensional forms of R. When R is the Witt ring of a field, G = F^{∗}/F^{∗2}. Forms in R are written as ⟨a_{1}, …, a_{n}⟩, with each a_{i} ∈ G. An n-fold Pfister form is a product ⟨1, a_{1}⟩⟨1, a_{2}⟩ ⋯ ⟨1, a_{n}⟩, denoted by ⟨⟨a_{1}, a_{2}, …, a_{n}⟩⟩. The set of orderings X has a topology with basic clopen sets

H(a_{1}, …, a_{n}) = {α ∈ X : a_{i} >_{α} 0 for all i},

where each a_{i} ∈ G. The chain length of R, denoted by cl(R), is the supremum of the set of integers k for which there is a chain

H(a_{0}) ⊊ H(a_{1}) ⊊ ⋯ ⊊ H(a_{k})

of length k (each a_{i} ∈ G). A subgroup F ⊂ G is a fan if it satisfies : any subgroup P ⊃ F such that −1 ∉ P and P has index 2 in G is an ordering. The index of the fan is [G : F]. The set of orderings P that contain F is denoted X/F. Note that |X/F | = 2^{n−1} if F has index 2^{n}. The stability index of R, denoted by st(R), is the supremum of log_{2} |X/F | over all fans in G. If R_{1} and R_{2} are reduced Witt rings then so is the product

R_{1} ⊓ R_{2} = {(r_{1}, r_{2}) : r_{1} ∈ R_{1}, r_{2} ∈ R_{2} and dim r_{1} ≡ dim r_{2} (mod 2)}.

E will always denote a group of exponent 2. If R is a reduced Witt ring then so is the group ring generated by E, denoted by R[E]. E_{k} will denote the group of exponent 2 and order 2^{k}. We will always take t_{1}, …, t_{k} as generators of E_{k} (except when k = 1 when we use just t). For an arbitrary E we use t_{1}, t_{2}, … as generators. When E is uncountable we are assuming the use of infinite ordinals as indices. Lastly, if S ⊂ G we write sp(S) for the subgroup generated by S.

1. Isotropy. Over ℝ a form q is hyperbolic iff sgn q = 0 and isotropic iff |sgn q| < dim q. The first statement holds for any reduced Witt ring but not the second. Our goal is to find a limit on the difference between |sgn q| and dim q for anisotropic forms. We restrict ourselves to reduced Witt rings with a finite chain length. Recall [12, 4.4.2] ([5] in the field case) that such rings are built up from copies of ℤ by finite products and arbitrary group ring extensions. The decomposition is unique except that ℤ ⊓ ℤ = ℤ[E_{1}]. We introduce some notation. Recall that E_{k} is generated by t_{1}, …, t_{k}. We fix a listing x_{1}, …, x_{2^{k}} of the elements of E_{k} as follows. The list for E_{1} is 1, t_{1}. The list for E_{k+1} is the list of E_{k} followed by t_{k+1} times the list for E_{k}. We also fix a listing α_{1}, …, α_{2^{n}} of the orderings on ℤ[E_{k}]. For the k = 1 we take α_{1} to be the ordering with t_{1} positive and α_{2} to be the ordering with t_{1} negative. The list for ℤ[E_{k+1}] consists of the orderings on ℤ[E_{k}] extended by taking t_{k+1} positive, followed by the extensions with t_{k+1} negative. Lastly, we define P_{k} to be the 2^{k} × 2^{k}-matrix whose (i, j) entry is the sign of x_{j} at the α_{i} ordering. Thus P_{1} = ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ).

Lemma 1.1. For each k ≥ 1 (1) P_{k} is symmetric.

(2) P_{k}^{2} = 2^{k}I.

(3) For q = ∑ n_{i}x_{i} ∈ ℤ[E_{k}] let s_{i} = q̂(α_{i}). Set n̄ = (n_{1}, … n_{2^{k}})^{T}, where T denotes the transpose, and s̄ = (s_{1}, …, s_{2^{k}})^{T}. Then P_{k}n̄ = s̄.

Proof. We use induction on k to prove (1) and (2). Both are clear for k = 1. By our construction,

P_{k+1} = ( \begin{matrix}P_{k} & P_{k} \\ P_{k} & −P_{k} \end{matrix} ) .

Thus P_{k} symmetric implies P_{k+1} is also. And

P_{k+1}^{2} = ( \begin{matrix}2P_{k}^{2} & 0 \\ 0 & 2P_{k}^{2} \end{matrix} ) = 2^{k+1}I.

Statement (3) is simple to check. □

The reader may notice that each P_{k} is a Hadamard matrix, indeed the simplest examples of Hadamard matrices, namely Kronecker products of copies of

( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ).

Notation. Let M(q) = max{|q̂(α)| : α ∈ X}.

Proposition 1.2. Let R = ℤ[E], where E is an arbitrary group of exponent two. Suppose q ∈ R is anisotropic. Then dim q ≤ M(q)^{2}.

Proof. We may assume q ∈ ℤ[E_{k}] for some k. Write q = ∑ n_{i}x_{i} where n_{i} ∈ ℤ and the x_{i} form the list of the elements of E_{k} described above. Let n̄ and s̄ be as in (1.1). Then:

P_{k}n̄ = s̄
2^{k}n̄ = P_{k}^{2}n̄ = P_{k}s̄
∑ n_{i}^{2} = n̄^{T} n̄ = \frac{1}{2^{2k}} s̄^{T} P_{k}^{T} P_{k}s̄

= \frac{1}{2^{k}} s̄^{T} s̄ = \frac{1}{2^{k}} ∑ s_{i}^{2}.

Now for each i we have s_{i}^{2} ≤ M(q)^{2}. So ∑ n_{i}^{2} ≤ M(q)^{2}. Further, |n_{i}| ≤ n_{i}^{2} so dim q = ∑ |n_{i}| ≤ M(q)^{2}. □

Remarks. (1) The bound in (1.2) is sharp infinitely often. Let ϵ = (ϵ_{1}, …, ϵ_{k}) be a choice of signs, that is, each ϵ_{i} = ±1. Pick a one-to-one correspondence between the 2^{k} many sign choices and the elements of sp{t_{k+1}, …, t_{2k}}, say ϵ ↦ x_{ϵ}. Then consider

q = ∑_{ϵ} x_{ϵ}⟨⟨ϵ_{1}t_{1}, …, ϵ_{k}t_{k}⟩⟩ ∈ ℤ[E_{2k}],

where the sum is over all possible sign choices. At each ordering of ℤ[E_{k}] exactly one of the Pfister forms has signature 2^{k}, the others having signature zero. In any extension of this ordering to ℤ[E_{2k}] we get sgn q = ±2^{k}. Thus q is anisotropic, dim q = 2^{2k} and M(q) = 2^{k}. Hence dim q = M(q)^{2}.

(2) The bound of (1.2) is not sharp for M’s that are not 2-powers. For instance, suppose q is anisotropic and M(q) = 3. We may assume (see (2.6)) that q has signature 3 or −1 at each ordering. Let q_{0} = (q − 1)_{an}, the anisotropic part. Then M(q_{0}) = 2 and so dim q_{0} ≤ 4. Thus dim q ≤ 5 < M(q)^{2}. The bound of (1.2) can also be improved if k is fixed. For instance, one can show for anisotropic q ∈ ℤ[E_{3}] that dim q ≤ \frac{5}{2}M(q).

Theorem 1.3. Suppose R is a reduced Witt ring of finite chain length. Let q ∈ R be anisotropic. Then dim q ≤ \frac{1}{2}cl(R)M(q)^{2}, unless R = ℤ and q is one-dimensional.

Proof. The result is clear if dim q = 1 so assume dim q ≥ 2. We may thus ignore the exceptional case. We will prove the result for R = S[E], any E, by induction on the chain length of S. Say cl(S) = 1 so that S = ℤ. If E = 1 then dim q = M(q) ≤ \frac{1}{2}M(q)^{2} as dim q ≥ 2. If E ≠ 1 then we are done by (1.2) as cl(ℤ[E]) = 2. In the general case we may assume S = S_{1} ⊓ S_{2}, with at least one of S_{1} or S_{2} not ℤ. Then both S_{1} and S_{2} have smaller chain length than S and so we are assuming the result holds for S_{i}[E], i = 1, 2 and any E. First suppose E = 1. Write q = (a, b) with a ∈ S_{1} and b ∈ S_{2}. We may assume that dim a ≥ dim b. Then dim q = dim a. We have by induction

dim q = dim a ≤ \frac{1}{2}cl(S_{1})M(a)^{2}
≤ \frac{1}{2}cl(R)M(a)^{2}, since cl(R) = cl(S_{1}) + cl(S_{2})
≤ \frac{1}{2}cl(R)M(q)^{2},

as q̂(α) = â(α) or b̂(α) for every α ∈ X so that M(a) ≤ M(q). Next suppose E ≠ 1. Since q has only finitely many entries we may assume that q ∈ (S_{1} ⊓ S_{2})[E_{k}], for some k. Write q = ∑(a_{i}, b_{i})x_{i}, where each a_{i} ∈ S_{1} and b_{i} ∈ S_{2} and the x_{i}’s are our listing of the elements of E_{k}. Set

φ = (∑ a_{i}x_{i}, 0) + r(0, ∑ b_{i}x_{i}) ∈ (S_{1}[E_{k}] ⊓ S_{2}[E_{k}])[E_{1}].

Now

dim q = ∑_{i} max{dim a_{i}, dim b_{i}}
dim φ = ∑ dim a_{i} + ∑ dim b_{i} ≥ dim q.

We check the signatures. If α ∈ X_{S_{1}} and α^{ϵ} is an extension of α to R = (S_{1} ⊓ S_{2})[E_{k}] then q̂(α^{ϵ}) = ∑ â_{i}(α)ϵ_{i} (here ϵ_{i} = ±1 depending on the sign of x_{i} in the extension). Similarly, if β ∈ X_{S_{2}} and β^{ϵ} is an extension to R then q̂(β^{ϵ}) = ∑ b̂_{i}(β)ϵ_{i}. We may also view α^{ϵ} as an extension of α to S_{1}[E_{k}] and hence to S_{1}[E_{k}]⊓S_{2}[E_{k}]. Let α^{ϵ+} denote the further extension to (S_{1}[E_{k}] ⊓ S_{2}[E_{k}])[E_{1}] with r positive. We also have the other extensions α^{ϵ−}, β^{ϵ+} and β^{ϵ−}. Then:

φ̂(α^{ϵ+}) = ∑ â_{i}(α)ϵ_{i}
φ̂(α^{ϵ−}) = ∑ â_{i}(α)ϵ_{i}
φ̂(β^{ϵ+}) = ∑ b̂_{i}(β)ϵ_{i}
φ̂(β^{ϵ−}) = − ∑ b̂_{i}(β)ϵ_{i}.

Thus M(φ) = M(q). Set φ_{1} = ∑ a_{i}x_{i} ∈ S_{1}[E_{k}] and φ_{2} = ∑ b_{i}x_{i} ∈ S_{2}[E_{k}]. Then by induction we have:

dim φ_{1} ≤ \frac{1}{2}cl(S_{1})M(φ_{1})^{2}
dim φ_{2} ≤ \frac{1}{2}cl(S_{2})M(φ_{2})^{2}.

The previous computation shows that for any ordering γ of (S_{1}[E_{k}]⊓S_{2}[E_{k}])[E_{1}] that φ̂(γ) equals φ̂_{1}(α) or ±φ̂_{2}(β) where γ restricts to either α on S_{1}[E_{k}] or β on S_{2}[E_{k}]. Thus M(φ_{i}) ≤ M(φ) for i = 1, 2. We obtain

dim φ = dim φ_{1} + dim φ_{2} ≤ \frac{1}{2}(cl(S_{1}) + cl(S_{2}))M(φ)^{2}
= \frac{1}{2}cl(R)M(φ)^{2},

using [12, 4.2.1]. Lastly, we have already checked that dim q ≤ dim φ and M(q) = M(φ), giving the desired bound. □

Remarks. (1) The bound of (1.3) is sometimes achieved. For example, in

R = (ℤ[E_{2}] ⊓ ℤ[E_{2}] ⊓ ℤ[E_{2}])[E_{2}],

where the last E_{2} is generated by s_{1}, s_{2}, let φ = ⟨1, t_{1}, t_{2}, −t_{1}t_{2}⟩ and set q = (φ, 0, 0) + s_{1}(0, φ, 0) + s_{2}(0, 0, φ). Then q is anisotropic, dim q = 12, M(q) = 2 and cl(R) = 6. Thus dim q = \frac{1}{2}cl(R)M(q)^{2}.

(2) Bröcker [3] has a result that looks similar to (1.3) but is apparently unrelated. There, in the version of [12, 7.7.3], if q is anisotropic, q̂(α) = ±2^{k} for all α and Y = {α : q̂(α) = 2^{k}} is the union of basic open sets each of stability index at most k + 1, then dim q ≤ 2^{2k} = M(q)^{2}.

(3) Bonnard [2] also has a result that looks like (1.3), which in fact uses Bröcker’s result in the proof. In our notation, her result is: if R has finite stability index s and q ∈ R is anisotropic then dim q ≤ 2^{s−1}M(q). Her bound is slightly better than this. Chain length and stability index are independent invariants so again there is no apparent connection between (1.3) and Bonnard’s result. Recall that a form q is weakly isotropic if mq is isotropic for some m ∈ ℕ.

Corollary 1.4. Let R be a real Witt ring (not necessarily reduced) of finite chain length. Let q ∈ R be a form of dimension at least 2. If dim q > \frac{1}{2}cl(R)M(q)^{2} then q is weakly isotropic.

Proof. Let q_{r} = q + R_{t} ∈ R_{red}, the reduced Witt ring. Then q_{r} is isotropic by (1.3). Hence q_{r} ≃ ⟨1, −1⟩ + φ_{r}, for some form φ_{r} = φ + R_{t} ∈ R_{red}. Then 2^{k}q ≃ 2^{k}⟨1, −1⟩ + 2^{k}φ, for some k, and so q is weakly isotropic. □

2. Chains of principal ideals. We use the standard abbreviation ACC for ascending chain condition.

Proposition 2.1. If ACC holds for the principal ideals of R then R has finite chain length.

Proof. Suppose we have a tower

H(a_{1}) ⊇ H(a_{2}) ⊇ ⋯ ⊇ H(a_{n}) ⊇ · · · .

Set q_{n} = ⟨1, 1, a_{n}⟩. Then q̂_{n}(α) is 1 or 3, with q̂_{n}(α) = 3 iff α ∈ H(a_{n}). In particular, for every n we have q̂_{n+1}(α) divides q̂_{n}(α), for every α ∈ X. Then q_{n+1} divides q_{n} by [7, 1.7]. Thus we have a tower of principal ideals :

(q_{1}) ⊆ (q_{2}) ⊆ ⋯ ⊆ (q_{n}) ⊆ · · · .

The ACC implies there exists a N such that (q_{N}) = (q_{m}) for all m > N. Then q̂_{N}(α) divides q̂_{m}(α) for all α ∈ X and so H(a_{N}) = H(a_{m}), for all m > N. □

We need some technical terms for the next result.

Definitions. A fan tower is a strictly decreasing tower of fans F_{1} > F_{2} > ⋯ > F_{n} > ⋯, each of finite index plus a fixed choice of complements C_{n} where G = C_{n} × F_{n}. We set F_{∞} = ∩F_{n}. A separating set of fan towers is a finite set of fan towers s_{1}, …, s_{ℓ}, with s_{i} = {F_{in}} such that

(1) Given any q ∈ R there exists m, possibly depending on q, such that all entries of q are in C_{im}F_{i∞}, for each i between 1 and ℓ.

(2) Given K ⊂ ℤ and forms q_{1}, q_{2} ∈ R, there exists N, depending on q_{1} and q_{2} but not K, such that if for some n > N

q̂_{1}^{−1}(K) ∩ (X/F_{in}) = q̂_{2}^{−1}(K) ∩ (X/F_{in})

for all i then q̂_{1}^{−1}(K) = q̂_{2}^{−1}(K).

Example. For a simple example, let R = ℤ[E] with E countably infinite. Let F_{i} = sp{t_{i+1}, t_{i+2}, … } and C_{i} = sp{−1, t_{1}, …, t_{i}}. Then each F_{i} is a fan of finite index, each C_{i} is a complement and the F_{i} are strictly decreasing. Hence {F_{i}} is a fan tower. Note that here F_{∞} = 1. This fan tower is a separating (singleton) set of fan towers. A given form q has entries involving only a finite number of t_{i}’s and so its entries lie in some C_{m}; this is the first condition. If we are given two forms q_{1} and q_{2} then again all of their entries lie in some C_{N}. So the signatures of the q_{i} depend only on the signs of t_{1}, … t_{N} in that ordering. Hence if q̂_{1} and q̂_{2} agree on X/F_{N} then they agree at every ordering. This is the second condition. Roughly, our fan towers will look like this example. When there is a product we will need one tower in each coordinate, hence a separating set.

Lemma 2.2. If R has finite chain length then R has a separating set of fan towers.

Proof. We prove this by induction on the chain length. When cl(R) = 1 then R = ℤ and the result is clear. We first consider the case R = S_{1} ⊓ S_{2}. Write G_{1} and X_{1} for G_{S_{1}} and X_{S_{1}} and similarly for G_{2} and X_{2}. Let {s_{1}^{1}, …, s_{ℓ_{1}}^{1}} be a separating set of fan towers for S_{1}. Here s_{k}^{1} = {F_{ki}^{1}} with complements C_{ki}^{1}. Set F_{ki} = F_{ki}^{1} × G_{2}, which is a fan in G = G_{1} × G_{2} with complement C_{ki} = C_{ki}^{1} × 1. Then for 1 ≤ k ≤ ℓ_{1}, r_{k} = {F_{ki}} is a fan tower. Note that F_{k∞} = F_{k∞}^{1} × G_{2}. Similarly, let {s_{1}^{2}, …, s_{ℓ_{2}}^{2}} be a separating set of fan towers for S_{2}, with s_{k}^{2} = {F_{ki}^{2}} and complements C_{ki}^{2}. Set F_{ℓ_{1}+k i} = G_{1} ×F_{ki}^{2} and C_{ℓ_{1}+k i} = 1×C_{ki}^{2}. Then for 1 ≤ k ≤ ℓ_{2}, r_{ℓ_{1}+k} = {F_{ℓ_{1}+k i}} is a fan tower. We check that r_{1}, …, r_{ℓ_{1}}, r_{ℓ_{1}+1}, …, r_{ℓ_{1}+ℓ_{2}} is a separating set of fan towers for R. We check the first condition. We are given a form q = ⟨(a_{1}, b_{1}), …, (a_{n}, b_{n})⟩ ∈ R. By induction, there exists a m_{1} such that a_{1}, …, a_{n} ∈ C_{km_{1}}^{1}F_{k∞}^{1} for all k. So

(a_{1}, b_{1}), …, (a_{n}, b_{n}) ∈ C_{km_{1}}F_{k∞} = C_{km_{1}}^{1}F_{k∞}^{1}G_{2},

for all k with 1 ≤ k ≤ ℓ_{1}. Similarly, there exists a m_{2} such that b_{1}, …, b_{n} ∈ C_{km_{2}}^{2}F_{k∞}^{2}, for all 1 ≤ k ≤ ℓ_{2}. Hence (a_{1}, b_{1}), …, (a_{n}, b_{n}) ∈ C_{km_{2}}F_{k∞} for all k with ℓ_{1} < k ≤ ℓ_{1} + ℓ_{2}. So take m to be the maximum of m_{1} and m_{2}. We next check the second condition. We are given K ⊂ ℤ and forms q_{1} = (u_{1}, v_{1}) and q_{2} = (u_{2}, v_{2}). Note that q̂_{1}^{−1}(K) = û_{1}^{−1}(K) ∪ v̂_{1}^{−1}(K) ⊂ X_{1} ∪ X_{2}, a disjoint union. By induction there exists a N_{1} satisfying the second condition for K, u_{1} and u_{2} and a N_{2} satisfying the second condition for K, v_{1} and v_{2}. Let N be the maximum of N_{1} and N_{2}. Suppose for some n > N we have

q̂_{1}^{−1}(K) ∩ (X/F_{kn}) = q̂_{2}^{−1}(K) ∩ (X/F_{kn}),

for all 1 ≤ k ≤ ℓ_{1} + ℓ_{2}. For 1 ≤ k ≤ ℓ_{1} we have:

q̂_{1}^{−1}(K) ∩ (X/F_{kn}) = q̂_{1}^{−1}(K) ∩ (X_{1}/F_{kn}^{1})
= û_{1}^{−1}(K) ∩ (X_{1}/F_{kn}^{1}).

We thus obtain

û_{1}^{−1}(K) ∩ (X_{1}/F_{kn}^{1}) = û_{2}^{−1}(K) ∩ (X_{1}/F_{kn}^{1}),

for all 1 ≤ k ≤ ℓ_{1}. By the second condition on S_{1} we have û_{1}^{−1}(K) = û_{2}^{−1}(K). Similarly, v̂_{1}^{−1}(K) = v̂_{2}^{−1}(K) and so q̂_{1}^{−1}(K) = q̂_{2}^{−1}(K). Now suppose R = S[E]. Set T_{i} = sp{t_{i+1}, t_{i+2}, … }. Let {s_{1}, …, s_{ℓ}} be a separating set of fan towers for S where s_{k} = {F′_{ki}} and the complements are C′_{ki}. Then F_{ki} = F′_{ki}T_{i} is a fan of finite index in R with complement C_{ki} = C′_{ki}sp{t_{1}, …, t_{i}}. Then r_{k} = {F_{ki}} is a fan tower. Note that F_{k∞} = F′_{k∞}. We show that {r_{1}, …, r_{ℓ}} is a separating set of fan towers for R. For the first condition we are given a form q ∈ R = S[E]. There exists a p such that q ∈ S[E_{p}]. Write q = ∑ a_{i}x_{i} where each a_{i} ∈ S and the x_{i}’s are some list of the elements of E_{p}. By induction, for each i there exists a m(i) such that every entry of a_{i} is in C′_{km(i)}F′_{k∞} for all k, 1 ≤ k ≤ ℓ. Let m be the maximum of the m(i) and p. Then every entry of every a_{i} lies in C′_{km}F′_{k∞} ⊂ C_{km}F_{k∞} and each x_{i} lies in sp{t_{1}, …, t_{p}} ⊂ C_{km}. So every entry of q lies in C_{km}F_{k∞}, for all k. For the second condition we are given K ⊂ ℤ and two forms q_{1}, q_{2} ∈ R. Again there exists a p such that q_{1}, q_{2} ∈ S[E_{p}]. Write q_{1} = ∑ a_{i}x_{i} and q_{2} = ∑ b_{i}x_{i} with a_{i}, b_{i} ∈ S and the x_{i} as before. Let ϵ ∈ {±1}^{p} be a choice of sign for t_{1}, …, t_{p}. Let ϵ(x_{i}) be the resulting sign of x_{i}. Set:

q_{1}^{ϵ} = ∑ a_{i}ϵ(x_{i}) q_{2}^{ϵ} = ∑ b_{i}ϵ(x_{i}),

both forms in S. For each ϵ there exists a N_{ϵ} so that condition 2 holds for q_{1}^{ϵ} and q_{2}^{ϵ}. Let N be the maximum of the N_{ϵ} and p. If α ∈ X_{S} we let α^{ϵ} be the extension of α to S[E_{p}] with t_{i} > 0 iff ϵ(t_{i}) = 1. Then we claim that:

q̂_{1}^{−1}(K) ∩ X_{S[E_{p}]} = ⋃_{ϵ}[(q̂_{1}^{ϵ})^{−1}(K)]^{ϵ}.

Namely if α^{ϵ} ∈ X_{S[E_{p}]} and q̂_{1}(α^{ϵ}) ∈ K then

q̂_{1}(α^{ϵ}) = ∑ â_{i}(α)ϵ(x_{i}) = q̂_{1}^{ϵ}(α).

Hence α^{ϵ} ∈ (q̂_{1}^{ϵ})^{−1}(K)^{ϵ}. The reverse inclusion is similar.

Now let α^{ϵe} denote any extension of α^{ϵ} to R = S[E]. Then by the claim we have:

(2.3)  q̂_{1}^{−1}(K) = ⋃_{e} (⋃_{ϵ}[(q̂_{1}^{ϵ})^{−1}(K)]^{ϵ})^{e}

So q̂_{1}^{−1}(K) ∩ (X/F_{kn}) = q̂_{2}^{−1}(K) ∩ (X/F_{kn}) implies that

(q̂_{1}^{ϵ})^{−1}(K) ∩ (X_{s}/F′_{kn}) = (q̂_{2}^{ϵ})^{−1}(K) ∩ (X_{s}/F′_{kn}),

for all sign choices ϵ. Hence by condition 2 applied to S we obtain (q̂_{1}^{ϵ})^{−1}(K) = (q̂_{2}^{ϵ})^{−1}(K) for all ϵ. Then (2.3) gives q̂_{1}^{−1}(K) = q̂_{2}^{−1}(K). □

Lemma 2.4. Suppose R has a separating set of fan towers {s_{1}, …, s_{ℓ}}. Let q ∈ R and K ⊂ ℤ. Let m be the index such that every entry of q lies in C_{km}F_{k∞}, for all 1 ≤ k ≤ ℓ. Let n > m. Then for each k we have:

|q̂^{−1}(K) ∩ (X/F_{kn})| = \frac{|X/F_{kn}|}{|X/F_{km}|}|q̂^{−1}(K) ∩ (X/F_{km})|.

Proof. Pick a k with 1 ≤ k ≤ ℓ. F_{kn} ⊂ F_{km} are both fans of finite index so we can write F_{km} = H × F_{kn} with H spanned by h_{1}, …, h_{p}, where 2^{p} = |X/F_{kn}|/|X/F_{km}|. Every α ∈ X/F_{km} has 2^{p} extensions to X/F_{kn}, one for each choice of signs (±1) for the h_{i}. Specifically, if ϵ is a sign choice for the h_{i} and h ∈ H, let ϵ(h) be the resulting sign of h. Since G = C_{km}HF_{kn}, the extension of α ∈ X/F_{km} to X/F_{kn} via ϵ is: α^{ϵ}(chf) = α(c)ϵ(h), where c ∈ C_{km}, h ∈ H and f ∈ F_{kn}. We thus have

X/F_{kn} = ⋃_{ϵ}(X/F_{km})^{ϵ}.

Write q = ⟨a_{1}, a_{2}, … ⟩. By assumption, each a_{i} is in C_{km}F_{k∞} ⊂ C_{km}F_{kn}. Hence α^{ϵ}(a_{i}) = α(a_{i}). Thus :

q̂^{−1}(K) ∩ (X/F_{kn}) = ⋃_{ϵ} (q̂^{−1}(K) ∩ (X/F_{km}))^{ϵ} .

So |q̂^{−1}(K) ∩ (X/F_{kn}) = 2^{p}|q̂^{−1}(K) ∩ (X/F_{km})|, and the result follows. □

Lemma 2.5. Let q ∈ R be a form of dimension n. Let F be a fan of finite index and let K ⊂ ℤ. Then :

|q̂^{−1}(K) ∩ (X/F)| = \frac{k}{2^{n}} |X/F |,

for some integer k, 0 ≤ k ≤ 2^{n}.

Proof. Write q = ⟨a_{1}, …, a_{n}⟩. Then q̂^{−1}(K) is a disjoint union of H(ϵ_{1}a_{1}, …, ϵ_{n}a_{n}) for various choices of ϵ = (ϵ_{1}, …, ϵ_{n}) ∈ {±1}^{n}. Set ρ_{ϵ} = ⟨⟨ϵ_{1}a_{1}, …, ϵ_{n}a_{n}⟩⟩. Then by the easy half of the representation theorem

∑_{α∈X/F} ρ̂_{ϵ}(α) ≡ 0 (mod |X/F |)
2^{n}|H(ϵ_{1}a_{1}, …, ϵ_{n}a_{n}) ∩ (X/F)| = k_{ϵ}|X/F |,

for some non-negative integer k_{ϵ}. Then :

|q̂^{−1}(K) ∩ (X/F)| = ∑_{ϵ} \frac{k_{ϵ}}{2^{n}} |X/F | = \frac{k}{2^{n}} |X/F |,

for some non-negative integer k. □

The following is essentially from [9]. For a form q = ⟨a_{1}, …, a_{n}⟩ the discriminant is dis q = (−1)^{n(n−1)/2}a_{1} ⋯ a_{n}. This is sometimes called the signed discriminant.

Lemma 2.6. Let q be an odd dimensional form.

(1) dis q >_{α} 0 iff q̂(α) ≡ 1 (mod 4).

(2) sgn_{α}dis(q)q ≡ 1 (mod 4) for all α ∈ X.

(3) If 0 ≠ a = bc and â(α) = ±b̂(α) for all α ∈ X with â(α) ≠ 0 then there exists d ∈ G such that ⟨d⟩a = b.

Proof. (1) Suppose n = dim q. Let s = q̂(α). If r is the number of α-negative entries in q then

sgn_{α}dis q = (−1)^{n(n−1)/2}(−1)^{r} = (−1)^{\frac{n(n−1)}{2} + \frac{n−s}{2}} = (−1)^{(n^{2}−s)/2}.

This is positive iff n^{2} −s ≡ 0 (mod 4). As n is odd we get that the discriminant is positive iff q̂(α) = s ≡ n^{2} ≡ 1 (mod 4).

(2) is easy to check. For (3), let A = {α ∈ X : â(α) ≠ 0}. Then ĉ(α) = ±1 for all α ∈ A. In particular c is odd dimensional and â(α) = 0 iff b̂(α) = 0.Let d = dis c. Then ⟨d⟩c has signature 1 for all α ∈ A by (2). Hence ⟨d⟩bc and b have the same signature at each α ∈ B, and also at each α ∉ A (as both have signature 0 there). Thus ⟨d⟩a = ⟨d⟩bc = b. □

Theorem 2.7. Let R be a reduced Witt ring. Then ACC holds for principal ideals iff the chain length of R is finite.

Proof. (2.1) gives (⟶). For the converse, let (q) ⊂ (q_{1}) ⊂ (q_{2}) ⊂ ⋯ be an ascending chain of principal ideals in R. Note that as each q_{i} divides q we have M(q_{i}) ≤ M(q). Let M = M(q). Then (1.3) gives dim q_{i} ≤ \frac{1}{2}cl(R)M^{2} for all i (note q is not one-dimensional else all (q_{i}) = R).

We begin with some simple reductions. If all q_{i} are 0 then the result is clear. If some q_{i} is not zero then all the later q_{i}’s are not zero. We may start our tower there, that is, we may assume q ≠ 0. For a non-zero form φ define deg φ to be the largest d such that 2^{d} divides φ̂(α) for all α ∈ X. Since q_{i+1} divides q_{i} we have deg q_{i+1} ≤ deg q_{i}. Let d_{0} be the minimum of the degrees of the q_{i}. We may start our tower at a q_{j} of minimal degree, that is, we may assume that deg q = deg q_{i} for all i. Now we may write q = q_{i}φ_{i} for some form φ_{i}. We check that φ_{i} is odd dimensional. If instead φ_{i} is even dimensional then 2 divides φ̂_{i}(α) for all α and so 2^{d+1} divides q̂(α) for all α, contradicting our reduction to a tower of uniform degree. Hence φ_{i} is odd dimensional. In particular, q̂(α) = 0 iff q̂_{i}(α) = 0. Let D be the set of integers d > 1 that divide some non-zero q̂(α), α ∈ X. Write D = {d_{1}, …, d_{z}} with d_{1} < d_{2} < ⋯ < d_{z}. Set A(i, d_{j}) = q̂_{i}^{−1}(±d_{j}). Let d_{k} be the largest element of D (if any) for which {A(i, d_{k}) : i ≥ 1} is not finite. Our goal is to show that there is in fact no such d_{k}. Our assumption on d_{k} means that for each j > k we have a t_{j} such that A(t, d_{j}) = A(t_{j}, d_{j}) for all t ≥ t_{j}. Let T be the maximum of the t_{j}, j > k. Then by starting our tower of ideals with q_{T}, we may assume A(i, d_{j}) = A(1, d_{j}) for all j > k and all i ≥ 1. We first check that A(i + 1, d_{k}) ⊂ A(i, d_{k}) for any i. Namely, q_{i} = q_{i+1}φ for some form φ. So if α ∈ A(i + 1, d_{k}) then ±d_{k} divides q̂(α). Also |q̂_{i}(α)| is not of the d_{j} with j > k else α ∈ A(i, d_{j}) = A(i + 1, d_{j}), which is impossible as α ∈ A(i + 1, d_{k}). Thus |q̂_{i}(α)| ≤ d_{k} and is divisible by d_{k}. Hence q̂_{i}(α) = ±d_{k} and α ∈ A(i, d_{k}) as desired. Let s = {F_{m}} be one fan tower in a separating set of fan towers for R (which exists by (2.2)). The first condition for a separating set, plus a simple induction argument, shows that for each i there exists a least m(i) with every entry of q_{1}, …, q_{i} in C_{m(i)}F_{∞}. Note that m(i + 1) ≥ m(i). Let p(i) be the number of distinct values of

\frac{|A(j, d_{k}) ∩ (X/F_{m(i)})|}{|X/F_{m(i)}|} ≡ γ(i, j),

over j with 1 ≤ j ≤ i. Now, by (2.4)

γ(i + 1, j) = \frac{|A(j, d_{k}) ∩ (X/F_{m(i+1)})|}{|X/F_{m(i+1)}|}

= \frac{|X/F_{m(i+1)}|}{|X/F_{m(i)}|} \frac{|A(j, d_{k}) ∩ (X/F_{m(i)})|}{|X/F_{m(i+1)}|}

= γ(i, j).

Hence p(i + 1) ≥ p(i), with only γ(i + 1, i + 1) possibly being a new value. Since every dim q_{i} ≤ \frac{1}{2}cl(R)M^{2}, (2.5) implies each p(i) ≤ 2^{cl(R)M^{2}/2} +1. Hence there is a t_{0} such that p(t) = p(t_{0}) for all t ≥ t_{0}. Let p = p(t_{0}) and m = m(t_{0}). Say γ(t_{0}, j_{1}), …, γ(t_{0}, j_{p}) are the distinct γ-values over 1 ≤ j ≤ t_{0}. Let t > t_{0} and set n = m(t). Then γ(t, t) = γ(t_{0}, j_{s}) for some j_{s}. That is,

\frac{|A(t, d_{k}) ∩ (X/F_{n})|}{|X/F_{n}|} = \frac{|A(j_{s}, d_{k}) ∩ (X/F_{m})|}{|X/F_{m}|}

= \frac{|A(j_{s}, d_{k}) ∩ (X/F_{n})|}{|X/F_{n}|},

using (2.4) again. Further, A(t, d_{k}) ⊂ A(t_{0}, d_{k}) ⊂ A(j_{s}, d_{k}) so that we have

|A(t, d_{k}) ∩ (X/F_{n})| = |A(t_{0}, d_{k}) ∩ (X/F_{n})|,

and this holds for all t ≥ t_{0}. We can repeat this argument for each fan tower in the separating set. Let {s_{1}, …, s_{ℓ}} be the separating set and let s_{i} = {F_{in}}. Hence there exist an N and a T such that |A(t, d_{k}) ∩ (X/F_{in})| = |A(T, d_{k}) ∩ (X/F_{in})| for all 1 ≤ i ≤ ℓ and all t ≥ T. By the second property of a separating set we have A(t, d_{k}) = A(T, d_{k}) for all t ≥ T. This contradicts our choice of d_{k}. Hence we have a T such that A(t, d_{j}) = A(T, d_{j}) for all t ≥ T and all d_{j} ∈ D. Thus q̂_{t}(α) = ±q̂_{T} (α) for all α in the union of the A(T, d_{j}), that is, for all α with q̂(α) ≠ 0. By our early reduction, q̂(α) ≠ 0 iff q̂_{T} (α) ≠ 0. Thus q̂_{t}(α) = ±q̂_{T} (α) for all α with q̂_{T} (α) ≠ 0 and also q_{t} divides q_{T}. By (2.6) we obtain (q_{t}) = (q_{T}), for all t ≥ T. □

Corollary 2.8. Let R be a real (but necessarily reduced) Witt ring. If R has finite chain length then ACC holds for principal ideals generated by odd dimensional forms.

Proof. Every ideal containing an odd dimensional form contains the torsion ideal R_{t} by [7, 1.5]. Hence passing to the reduced Witt ring maintains a tower of principal ideals generated by odd dimensional forms. This reduced tower stabilizes by (2.7). Hence the original tower stabilizes. □

Corollary 2.9. Let (G, X) be a space of orderings. Let S denote the collection of subsets of G of order n. If X has finite chain length then any tower

H(S_{1}) ⊂ H(S_{2}) ⊂ ⋯ H(S_{k}) ⊂ · · · .

with each S ∈ S, stabilizes.

Proof. Suppose S_{i} = {a_{i1}, …, a_{in}}. Set q_{i} = ⟨⟨a_{i1}, …, a_{in}⟩⟩ + 1. Then q̂_{i}(X) = {1, 2^{n} + 1} and q̂_{i}^{−1}(2^{n} + 1) = H(S_{i}). Thus q̂_{i+1}(α) divides q̂_{i}(α) for all α ∈ X. So q_{i+1} divides q_{i} by [7, 1.7]. We thus have a tower of principal ideals (q_{1}) ⊂ (q_{2}) ⊂ · · ·. This stabilizes by (2.7) and so the tower of H(S_{i})’s also stabilizes. □

3. Factorization. Anderson and Valdes-Leon [1] have several notions of an associate in a commutative ring R. We need three of these. Two elements a and b are associates if their principal ideals are equal, (a) = (b). They are strong associates if a = bu, for some unit u ∈ R. Lastly, a and b are very strong associates if (a) = (b) and either a = b = 0 or a ≠ 0 and a = br implies r is a unit. An non-unit a is irreducible if a = bc implies either b or c is an associate of a. Similarly, a is strongly irreducible (very strongly irreducible) if a = bc implies either b or c is a strong associate (respectively, very strong associate) of a. Lastly, R is atomic if every non-zero non-unit of R can be written as a finite product of irreducible elements. Define strongly atomic and very strongly atomic similarly.

Proposition 3.1. Let R be a reduced Witt ring and let a, b ∈ R. Then a, b are associates iff a, b are strong associates. In particular, R is atomic iff R is strongly atomic.

Proof. Strong associates are always associates so we check the converse. Suppose (a) = (b). Write a = bx and b = ay. Then a = axy and a(1 − xy) = 0. Let Z = {α ∈ X : â(α) = 0}. Then for all α ∉ Z we have x̂(α) = ±1. From a = bx and (2.6) we get ⟨d⟩a = b for some d ∈ G. Clearly ⟨d⟩ is a unit. □

Strong associates need not be very strong associates in a reduced Witt ring. If ±1 ≠ g ∈ G then ⟨1, g⟩ is not even a very strong associate of itself. Namely, ⟨1, g⟩ = ⟨1, g⟩⟨1, 1, −g⟩ and ⟨1, g⟩ ≠ 0 and ⟨1, 1, −g⟩ is not a unit. So, except for R = ℤ, R will not be very strongly atomic.

Corollary 3.2. Let R be a real Witt ring (not necessarily reduced) and suppose R has finite chain length.

(1) Every odd dimensional form can be written as a finite product of irreducible forms.

(2) If R is reduced then R is atomic.

Proof. These are standard consequences of (2.8) and (2.7), see [1, 3.2]. □

We are unable to prove the converse to (3.2)(2) for all reduced Witt rings R. However, we can prove the converse for a wide class of rings. For this we need Marshall’s notion of a sheaf product [11]. Start with a non-empty Boolean space I, a collection of reduced Witt rings R_{C}, one for each clopen C ⊂ I and a collection of ring homomorphisms res_{C.D} : R_{C} → R_{D}, defined whenever D ⊂ C are clopen in I. We assume the usual sheaf properties, namely,

(1) R_{∅} = ℤ/2ℤ and R_{C} ≠ ℤ/2ℤ if C ≠ ∅.

(2) res_{C,C} is the identity map on C.

(3) If E ⊂ D ⊂ C then res_{C,E} = res_{D,E}res_{C,D}.

(4) If C = ∪_{j}C_{j} and if r_{j} ∈ R_{j} are given such that

res_{C_{j},C_{j}∩C_{k}}(r_{j}) = res_{C_{k},C_{j}∩C_{k}}(r_{k}),

for all j, k, then there exists a unique r ∈ R_{C} such that res_{C,C_{j}}(r) = r_{j},

for all j.

For fixed i ∈ I we form the stalk

R_{i} = \underrightarrow{lim}_{i∈C} R_{C}.

Each R_{i} is a reduced Witt ring. We call the reduced Witt ring R_{I} the sheaf product of the R_{i}’s and write R_{I} = ∏_{i∈I} R_{i}. When I is finite and discrete this is the usual product of Witt rings. We next define a sequence of classes of reduced Witt rings (which is slightly different from the sequence of Marshall [11, p. 219]). Let C_{1} denote the class of finitely generated reduced Witt rings. Inductively define C_{n} to be sheaf products of R_{i}[E^{i}], where E^{i} is a group of exponent two (not necessarily finite) and R_{i} ∈ C_{m} for some m < n. Lastly, let C_{ω} be the union of all C_{n}. This is a large class. Already C_{2} contains all SAP reduced Witt rings and C_{ω} contains all reduced Witt rings where X has only a finite number of accumulation points [11, 8.17]. We will prove that R ∈ C_{ω} atomic implies R has finite chain length. We begin with a lemma.

Lemma 3.3. Let S = R[E] and let T ⊂ G_{S} be a fan of finite index. Set T_{0} = T ∩ G_{R}.

(1) T_{0} is a fan in G_{R}.

(2) Suppose X_{R}/T_{0} = {P, Q}. Then X_{S}/T consists of extensions of P, Q to S. If x ∈ G_{S} \G_{R} then either none, exactly half or all of the extensions of P that lie in X_{S}/T make x positive.

Proof. (1) Write T = T_{0}H for some subgroup H of G_{S} with H ∩ G_{R} = 1. Extend H to subgroup L of G_{S} such that G_{S} = G_{R} × L. Suppose P ⊂ G_{R} is a subgroup of index 2, containing T_{0} but not −1. Then P L is a subgroup of index at most 2 containing T. If −1 ∈ P L then for some p ∈ P and y ∈ L we have −p = y ∈ P ∩ L = 1. But then −1 = p ∈ P, a contradiction. Thus P L is an ordering in G_{S}. It is easy to check that P is then an ordering in G_{R}. This shows T_{0} is a fan.

(2) The first statement is clear. Suppose P_{1}, … P_{m}, Q_{1}, …, Q_{m} are the extensions of P, Q that lie in X_{S}/T. Pick a ∈ G_{R} with â(P) = 1 and â(Q) = −1. Let k be the number of P_{i} for which x is positive. From the easy half of the Representation Theorem [11, 7.13]

∑_{α∈X_{S}/T} sgn_{α}⟨⟨a, x⟩⟩ ≡ 0 (mod 2m)
4k ≡ 0 (mod 2m).

So m divides 2k and clearly k ≤ m. Hence k = 0, \frac{1}{2}m or m. □

Our proof that R ∈ C_{ω} atomic implies finite chain length is not the usual induction argument since we are unable to show R[E] atomic implies R atomic. Instead we explicitly construct a form which does not factor into a finite product of irreducibles. Unfortunately, the construction requires considerable notation. We introduce this notation by first looking at a special case. Let ∗ denote a group ring extension. A ring in C_{n} looks like

R = ∏_{α∈A_{1}} W (α)^{∗}

= ∏_{α∈A_{1}}( ∏_{β∈A_{2}(α)} W (α, β)^{∗})^{∗}

= ∏_{α∈A_{1}}( ∏_{β∈A_{2}(α)}( ∏_{γ∈A_{3}(α,β)} W (α, β, γ)^{∗})^{∗})^{∗},

where each A_{1}, A_{2}(α) and A_{3}(α, β) is a Boolean space and each W (α, β, γ) is in C_{m}, for some m ≤ n − 3. Suppose we want to single out the product over A_{3}(α_{0}, β_{0}), for some particular α_{0} and β_{0}. We set :

R_{1} = ∏_{γ∈A_{3}(α_{0},β_{0})} W (α_{0}, β_{0}, γ)^{∗}

R_{2} = ∏_{\begin{matrix}β∈A_{2}(α_{0}) \\ β≠β_{0} \end{matrix}}( ∏_{γ∈A_{3}(α_{0},β)} W (α_{0}, β, γ)^{∗})^{∗}

R_{3} = ∏_{\begin{matrix}α∈A_{1} \\ α≠α_{0} \end{matrix}}( ∏_{β∈A_{2}(α)}( ∏_{γ∈A_{3}(α,β)} W (α, β, γ)^{∗})^{∗})^{∗}.

Then R = ((R_{1}^{∗} ⊓ R_{2}^{∗})^{∗} ⊓ R_{3}^{∗})^{∗}. We will want to single out the first infinite sheaf product. We have:

R = ((… ((R_{1}^{∗} ⊓ R_{2}^{∗})^{∗} ⊓ R_{3}^{∗})^{∗} ⊓ …)^{∗} ⊓ R_{s}^{∗})^{∗},

with R_{1} an infinite sheaf product, say

R_{1} = ∏_{δ∈A} W (δ)^{∗},

and each W (δ) in some C_{m}, m ≤ n − s. We will need explicit extension groups. We use the notation

R = (… ((R_{1}[E^{1}] ⊓ R_{2}[F^{1}])[E^{2}] ⊓ R_{3}[F^{2}])[E^{3}] ⊓ … ⊓ R_{s}[F^{s−1}])[E^{s}].

We further take {t_{j}^{i}} as generators of E^{i}.

Lastly, we need notation to express the orderings on R. Let X_{i} denote X_{R_{i}}. Let X_{1}(ϵ_{1}) denote the extensions of X_{1} to R_{1}[E^{1}]. Here ϵ_{1} is an arbitrary choice of signs. The extension is determined by the values ϵ_{1}(t_{j}^{1}) ∈ {±1}. To save on indices we will write ϵ_{1}(j) for ϵ_{1}(t_{j}^{1}). Next, X_{2}(η_{1}) denotes the extensions from R_{2} to R_{2}[F^{1}]. X_{1}(ϵ_{1}, ϵ_{2}) denotes the extensions from R_{1} to (R_{1}[E^{1}]⊓R_{1}[F^{1}])[E^{2}], with ϵ_{2} a sign choice for E^{2}. Continue with this pattern. We obtain for X_{R}

⋃_{ϵ,η}[X_{1}(ϵ_{1}, …, ϵ_{s}) ∪ X_{2}(η_{1}, ϵ_{2}, …, ϵ_{s}) ∪ X_{3}(η_{2}, ϵ_{3}, …, ϵ_{s}) ∪ … ∪ X_{s}(η_{s−1}, ϵ_{s})].

Theorem 3.4. Suppose R ∈ C_{ω}. The following are equivalent:

(1) R has finite chain length.

(2) R has ACC on principal ideals.

(3) R is atomic.

Proof. We need only show R atomic implies R has finite chain length, by (2.7) and (3.2). Suppose R ∈ C_{n} and let s be the first level (if any) with an infinite sheaf product. We follow the above notation. Fix some δ_{0} ∈ A and define a ∈ G_{R_{1}} with −1 in the δ_{0} coordinate and 1 in the other coordinates. Set

b = ((… (a, −1), −1), …), −1) ∈ G_{R},

and set q = ⟨b, t_{1}^{1}, bt_{1}^{1}⟩. Let X_{δ} be the orderings on W (δ)^{∗} so that X_{1} = ∪X_{δ}. Set C = q̂^{−1}(3). Then:

C = ⋃_{\begin{matrix}ϵ,η \\ ϵ_{1}(1)=1 \end{matrix}} [( ⋃_{δ≠δ_{0}} X_{δ})(ϵ_{1}, …, ϵ_{s}) ∪ X_{2}(η_{1}, ϵ_{2}, …, ϵ_{s}) ∪ … ∪ X_{s}(η_{s−1}, ϵ_{s}))] .

We are assuming R is atomic, so let q = φ_{1} ⋯ φ_{r} with each φ_{i} irreducible. We may assume φ̂_{i}(X) = {3, −1} by (2.6). Set D_{i} = φ̂_{i}^{−1}(3). Note D_{i} ⊂ C. We will show that in fact one of the φ_{i} factors and hence that no sheaf product in R is infinite. Our first goal is to show that each D_{i} consists of all extensions, with t_{1}^{1} positive, of some subset of X_{1}. Pick P ∈ X_{δ_{0}} and Q ∈ X_{δ} with δ ≠ δ_{0}. Fix some k and j. Let

e^{k} = sp{t_{1}^{k}, …, t_{j−1}^{k}, t_{j+1}^{k}, … }
e^{1} = sp{t_{2}^{1}, t_{3}^{1}, … }.

Let T be the fan

(… (((P ∩ Q)[e^{1}] ⊓ G_{R_{2}}[F^{1}])[E^{2}] …)[e^{k}] ⊓ … ⊓ G_{R_{s}}[F^{s−1}])[E^{s}].

Then X/T has 8 orderings, namely the extensions of P and Q with all t_{ℓ}^{i} positive except possibly t_{1}^{1} and t_{k}^{j}. Write these orderings as P (±1, ±1) and Q(±1, ±1), where the first coordinate gives the sign of t_{1}^{1} and the second gives the sign of t_{j}^{k}. C ∩ (X/T) = {Q(1, ±1)} so that |C ∩ (X/T)| = 2. To ease notation slightly, write D for one of the D_{i}. Let w = |D ∩ (X/T)|. Then by the easy part of the Representation Theorem we have:

∑_{γ∈X/T} φ̂(γ) ≡ 0 (mod |X/T |)
3w − (8 − w) ≡ 0 (mod 8)
w ≡ 0 (mod 2).

As D ∩ (X/T) ⊂ C ∩ (X/T) we have D ∩ (X/T) is either empty or all of C ∩ (X/T). Suppose for some k and j we are in the second case, D ∩ (X/T) = C ∩ (X/T). Choose another pair g, h. Pick the fan T′ generated over P ∩ Q by E^{i} for i ≠ 1, k, g, the same e^{1} as before and

e^{k′} = sp{t_{1}^{k}, …, t_{j−1}^{k}, −t_{j}^{k}, t_{j+1}^{k} … }
e^{g′} = sp{t_{1}^{g}, …, t_{h−1}^{g}, t_{h+1}^{g}, … }.

Then X/T′ has 8 orderings, namely the extensions of P and Q with all t_{ℓ}^{i} positive except t_{j}^{k} negative and t_{1}^{1}, t_{h}^{g} arbitrary. Write these as P (±1, −1, ±1) and Q(±1, −1, ±1) with the first coordinate the sign of t_{1}^{1}, the second coordinate indicating that t_{j}^{k} is negative and the third coordinate the sign of t_{h}^{g}. Again C ∩ (X/T′) consists of two orderings, Q(1, −1, ±1). And as before we get that D ∩ (X/T′) is either empty or all of C ∩ (X/T′). But Q(1, −1, 1) is the same ordering that was denoted by Q(1, −1) before (that is, with t_{1}^{1} positive, t_{j}^{k} negative and all other t’s positive). Hence we have D ∩ (X/T′) = C ∩ (X/T′). We continue to assume D ∩ (X/T) = C ∩ (X/T). If we repeat this argument ( first with a fan having t_{j}^{k} and t_{h}^{g} negative) we get that any extension Q with t_{1}^{1} positive and only a finite number of t_{ℓ}^{i} negative is in D. Now D = φ̂^{−1}(3) and the entries of φ involve only a finite number of t_{ℓ}^{i}. Hence we have that any extension of Q with t_{1}^{1} positive is in D. The assumption that D ∩ (X/T) ≠ ∅ means we are assuming some extension of Q with t_{1}^{1} positive is in D. From this we conclude that all such extensions are in D. Let X_{1}^{∗} denote the orderings on R_{1}[E^{1}], namely the extensions ϵ_{1} of X_{1}. Write D|X_{1}^{∗} for the orderings in D restricted to R_{1}[E^{1}]. We have shown that D|X_{1}^{∗} consists of all extensions, with t_{1}^{1} positive, of some subset (call it D|X_{1}) of X_{1}. Each factor φ_{i} of q has its set D_{i}. We have C = ∪D_{i} and

∪(D_{i}|X_{1}) = C|X_{1} = ⋃_{\begin{matrix}δ∈A \\ δ≠δ_{0} \end{matrix}} X_{δ}.

A is infinite so some D_{i}|X_{1} meets at least two X_{δ}’s. For simplicity, call this D_{i} simply D and the corresponding form φ. Suppose D|X_{1} meets X_{δ_{1}} and X_{δ_{2}}, δ_{1} ≠ δ_{2}. Set

D_{0} = ⋃_{ϵ(1)=1}[(D|X_{1}) ∩ X_{δ_{1}}](ϵ_{1}) ⊂ X_{1}^{∗}.

In words, D_{0} consists of the extensions for X_{δ_{1}} that lie in D|X_{1}^{∗}. We will use D_{0} to construct a factor of φ. Let f : X_{1}^{∗} → ℤ by f(P) = 3 if P ∈ D_{0} and f(P) = −1 if P ∉ D_{0}. We want to use the Representation Theorem [11,7.13] to show f is represented by a form in R_{1}[E^{1}]. Let T ⊂ G_{R_{1}}E^{1} be a fan of finite index. Then T_{1} = T ∩ G_{R_{1}} is a fan in G_{R_{1}} by (3.3) Case 1 : (X_{1}/T_{1}) ⊂ X_{δ} for some δ ∈ A. Here X_{1}^{∗}/T = (X_{δ}/T_{1})(ϵ), over some set of extensions ϵ to E^{1}. If δ ≠ δ_{1} then f(P) = −1 for all P ∈ (X_{1}^{∗}/T) since D_{0} only has extensions from X_{δ_{1}}. Thus

∑_{P ∈X_{1}^{∗}/T} f(P) = −|X_{1}^{∗}/T | ≡ 0 (mod |X_{1}^{∗}/T |).

If δ = δ_{1} then P ∈ D_{0} iff P ∈ D|X_{1}^{∗} iff some (equivalently, every) extension, with t_{1}^{1} positive, of P to X_{R} lies in D iff φ̂(P) = 3. So f(P) = φ̂(P) for all P ∈ X_{1}^{∗}/T. We obtain

∑_{P ∈X_{1}^{∗}/T} f(P) = ∑_{P ∈X_{1}^{∗}/T} φ̂(P) ≡ 0 (mod |X_{1}^{∗}/T |).

Case 2 : (X_{1}/T_{1}) ⊄ X_{δ} for some δ ∈ A. Here we must have |X_{1}/T_{1}| = 2 by [11, 8.12] Write X_{1}/T_{1} = {P_{α}, P_{β}} where α, β are distinct elements of A and P_{α} ∈ X_{α} and P_{β} ∈ X_{β}. Then X_{1}^{∗}/T consists of some set of extensions, to E^{1}, applied to P_{α} and P_{β}. Again, if neither α nor β are δ_{1} then all f(P) = −1 and we are done. So say α = δ_{1} (and so β ≠ δ_{1}). If P_{α} ∉ D|X_{1} then no extension is in D_{0} and all f(P) = −1 again. So suppose P_{α} ∈ (D|X_{1}) ∩ X_{δ_{1}}. Since P_{β} ∉ X_{δ_{1}} no extension of P_{β} in X_{1}^{∗}/T is in D_{0}. This is half of X_{1}^{∗}/T. The other half consists of extensions of P_{α} and by (3.3) either none, exactly half or all of these extensions make t_{1}^{1} positive, and hence lie in D_{0}. Thus |D_{0} ∩ (X_{1}^{∗})| = d|X_{1}^{∗}/T |, where d is either (i) 0, or (ii) \frac{1}{4} or (iii) \frac{1}{2}. In case (i) we have

∑_{P ∈X_{1}^{∗}} f(P) = −|X_{1}^{∗}/T | ≡ 0 (mod |X_{1}^{∗}/T |).

In case (ii) we have

∑_{P ∈X_{1}^{∗}} f(P) = \frac{1}{4}|X_{1}^{∗}/T | · 3 + \frac{3}{4}|X_{1}^{∗}/T | · (−1) ≡ 0 (mod |X_{1}^{∗}/T |).

In case (iii) we have

∑_{P ∈X_{1}^{∗}} f(P) = \frac{1}{2}|X_{1}^{∗}/T | · 3 + \frac{1}{2}|X_{1}^{∗}/T | · (−1) ≡ 0 (mod |X_{1}^{∗}/T |).

Thus in all cases we have ∑ f(P) ≡ 0 (mod |X_{1}^{∗}/T |). By the non-trivial half of the Representation Theorem we have f = ψ̂ for some form ψ ∈ R_{1}[E^{1}]. By construction ψ̂(X_{1}^{∗}) = {3, −1} and ψ̂^{−1}(3) = D_{0} < D. Hence by [7, 1.7] ψ is a proper divisor of φ. Hence φ is not irreducible, a contradiction. We thus have if R ∈ C_{n} is atomic then all sheaf products are finite. Hence cl(R) < ∞, using [12, 4.2.1]. □

Corollary 3.5. Let R ∈ C_{ω}. If R[E] is atomic then so is R.

Proof. R[E] atomic implies R[E] has finite chain length by (3.4). Then , as cl(R[E]) = cl(R), R has finite chain length and so is atomic by (3.2). □

It is unknown if the reduced Witt rings of finite stability index lie in C_{ω} so the following may improve (3.4), although (3.4) includes many atomic Witt rings with X infinite.

Proposition 3.6. Suppose R has finite stability index. The following are equivalent:

(1) R has finite chain length.

(2) R has ACC on principal ideals.

(3) R is atomic.

(4) X is finite.

Proof. (1) and (4) are equivalent by [10] (first shown, in the field case in [4]). As in the proof of (3.4) we need only show (3) implies (1). Suppose the stability index of R is n. We can find a prime p congruent to 1 mod 2^{n} by Dirichlet’s Theorem. R is atomic so p = φ_{1} ⋯ φ_{t} for some irreducible elements φ_{i}. Note that for each i we have |φ̂_{i}(X)| = {p, 1}. Let A_{i} = φ̂_{i}^{−1}(±p). The A_{i}’s form a clopen cover of X. We wish to show R has finite chain length. So suppose we have a tower

H(a_{1}) > H(a_{2}) > H(a_{3}) > · · · .

First suppose there is an s, 1 ≤ s ≤ t and a k such that A_{s} ∩ H(a_{k}) is a non-empty, proper subset of A_{s}. Define f : X → ℤ by

f(α) = \{ \begin{matrix}p, & if α ∈ A_{s} ∩ H(a_{k}) \\ 1, & if α ∉ A_{s} ∩ H(a_{k}). \end{matrix}

Let T be a fan, |X/T | = 2^{m}, where m ≤ n by definition of the stability index. Set w = |A_{s} ∩ H(a_{k}) ∩ (X/T)|. Then

∑_{α∈X/T} f(α) = wp + (2^{m} − w) = w(p − 1) ≡ 0 (mod 2^{m}),

since p − 1 is a multiple of 2^{n}. By the Representation Theorem, f = ψ̂ for some form ψ. Then ψ̂(α) divides φ̂_{s}(α) for all α and for α ∈ A_{s} \ H(a_{k}), ψ̂(α) ≠ ±φ̂_{s}(α). So, using [7, 1.7], we have ψ is proper divisor of φ_{s}, which is impossible. Thus there does not exists a pair s, k such that H(a_{k}) ∩ A_{s} is a non-empty, proper subset of A_{s}. That is, for all i, j we have H(a_{i}) ∩ A_{j} ≠ ∅ implies A_{j} ⊂ H(a_{i}). The A_{j}’s cover X so each H(a_{i}) is a union of A_{j}’s. Let n(i) be the number of A_{j}’s required to cover H(a_{i}). Then 1 ≤ n(i + 1) < n(i) ≤ t for all i. Thus the tower is finite and we are done. □

4. Irreducible elements. We look at some examples to illustrate factorization in reduced Witt rings.

Proposition 4.1. If 1 ≠ a ∈ G then ⟨1, −a⟩ is irreducible in R.

Proof. Suppose ⟨1, −a⟩ = qφ in R. We may assume q is even dimensional and φ is odd dimensional. If a <_{α} 0 then 2 = q̂(α)φ̂(α). Thus q̂(α) = ±2 = ±sgn_{α}⟨1, −a⟩, for all α with sgn_{α}⟨1, −a⟩ ≠ 0. By (2.6) there exists a d ∈ G such that ⟨d⟩⟨1, −a⟩ = q and so q is an associate of ⟨1, −a⟩. □

Example. If R ≠ ℤ then factorization into irreducible elements is not unique. Namely, if a ≠ ±1 then ⟨1, −a⟩⟨1, −a⟩ = ⟨1, 1⟩⟨1, −a⟩ gives two different factorizations of the Pfister form. This is quite different from the case of factoring odd dimensional forms. When X is finite there is unique factorization of odd dimensional forms if the ideal class group of R is trivial or, equivalently, the stability index is at most 2, by [6, 2.7] and [7, 1.17].

We next find the irreducible elements in ℤ[E_{1}]. Note that any form q in this ring is associate to some n + mt with n ≥ |m|.

Proposition 4.2. Let q = n + mt ∈ ℤ[E_{1}] with n ≥ |m|. Then q is irreducible iff (n, m) or (n, −m) equals one of the following:

(1) (1, 1)

(2) (2^{k} + 1, 2^{k} − 1), for some k ≥ 0

(3) (\frac{1}{2}(p + 1), \frac{1}{2}(p − 1)), for some odd prime p.

Proof. Let q be irreducible. First suppose q is even dimensional. If both n and m are even then 2 is a factor of q. So we have n and m odd. If n = ±m then n is a factor of q and we must have n = 1. Thus (n, m) = (1, ±1). We may thus suppose n + m and n − m are non-zero. Write n + m = 2^{g}h and n − m = 2^{k}ℓ with h and ℓ odd and g, k ≥ 1. Set

φ_{1} = \frac{1}{2}(2^{g} + 2^{k}) + \frac{1}{2}(2^{g} − 2^{k})t
φ_{2} = \frac{1}{2}(h + ℓ) + \frac{1}{2}(h − ℓ)t.

Then q = φ_{1}φ_{2} and φ_{2} is odd dimensional and so not an associate of q. Thus φ_{1} is an associate of q. If α is the ordering with t positive then n + m = q̂(α) = ±φ̂_{1}(α) = ±2^{g}. Since n ≥ −m we obtain n + m = 2^{g} and h = 1. Similarly, taking signatures at the ordering β with t negative gives ℓ = 1. If both g and k are at least 2 then n and m are even which is not possible. Suppose n+m = 2^{g} and n − m = 2. Then we get case (2). The reverse , n + m = 2 and n − m = 2^{k} gives case (2) for the pair (n, −m). Now suppose q is odd dimensional. If n + m is composite, say n + m = ab with a, b > 1, then set

φ_{1} = \frac{1}{2}(a + 1) + \frac{1}{2}(a − 1)t
φ_{2} = \frac{1}{2}(b + n − m) + \frac{1}{2}(b − n + m)t.

Then q = φ_{1}φ_{2}. Neither φ_{1} nor φ_{2} is an associate of q as q̂(α) = ab while φ̂_{1}(α) = a and φ̂_{2}(α) = b. Hence n + m is not composite. Similarly, n − m is not composite. If both n + m and n − m are prime then set

φ_{1} = \frac{1}{2}(n + m + 1) + \frac{1}{2}(n + m − 1)t
φ_{2} = \frac{1}{2}(n − m + 1) + \frac{1}{2}(1 − n + m)t.

We have q = φ_{1}φ_{2}. Neither φ_{1} nor φ_{2} is an associate of q as q̂(α) = n+m while φ̂_{2}(α) = 1 and q̂(β) = n − m while φ̂_{1}(β) = 1. Thus we must have n + m = p, p an odd prime, and n − m = 1 (or the reverse). This gives case (3). It is straightforward to check the forms in cases (1) - (3) are irreducible. □

Example. Already for ℤ[E_{1}], and in fact for any R ≠ ℤ, the number of irreducible factors in factorization of a given element can be arbitrarily large. For instance, ⟨1, 1, t⟩ is irreducible (take p = 3 in (4.2)(3)) and ⟨1, −t⟩⟨1, 1, t⟩ = ⟨1, −t⟩. Hence

⟨⟨1, −t⟩⟩ = ⟨1, 1⟩⟨1, 1, t⟩^{n}⟨1, −t⟩

is a factorization into irreducible elements for any n. Again the situation is quite different if we consider only factorizations of odd dimensional forms. When X is finite, the number of irreducible factors in a factorization is uniquely determined iff the stability index is at most 3 and R has no factor of the type (ℤ^{s})[E_{2}], with s ≥ 3, see [7].

Notice that the even prime of ℤ remains irreducible in ℤ[E_{1}] while the odd primes of ℤ all factor in ℤ[E_{1}]. This holds more generally.

Proposition 4.3. Let q ∈ R be irreducible.

(1) If q is even dimensional then q remains irreducible in R[E_{1}].

(2) If q is odd dimensional then q remains irreducible in R[E_{1}] iff q is not associate to 1 + 2q_{0}, for some q_{0} ∈ R.

Proof. First say q = 1 + 2q_{0}, for some q_{0} ∈ R. Since q is not a unit, there exists an α ∈ X_{R} with q̂(α) ≠ ±1. Let α^{+} and α^{−} denote the extensions of α to R[E_{1}] with, respectively, t positive and t negative. Now

q = (1 + q_{0}⟨1, t⟩)(1 + q_{0}⟨1, −t⟩).

Neither factor is an associate of q as the first has signature 1 at α^{−} and the second has signature 1 at α^{+}. Thus q is not irreducible in R[E_{1}]. Now suppose we have an irreducible q that factors in R[E_{1}]. We want to show q is odd dimensional and associate to some 1 + 2q_{0}. Write q = (a + b⟨1, t⟩)(c + d⟨1, −t⟩), with a, b, c, d ∈ R and neither factor an associate of q. The coefficient of t, namely bc − ad, must be zero and so q = ac + ad + bc. Then

(4.4)  q = ac + 2bc = c(a + 2b)
(4.5)  = ac + 2ad = a(c + 2d).

As q is irreducible in R, (4.4) shows that either c or a + 2b is an associate of q. We may assume c is the associate of q. Namely, if a + 2b is the associate then rewrite q as

q = ((c + 2d) + (−d)⟨1, t⟩)((a + 2b) + (−b)⟨1, −t⟩)
≡ (a′ + b′⟨1, t⟩)(c′ + d′⟨1, −t⟩).

Then c′ = a + 2b is associate to q. Write uq = c for some unit c ∈ R. Equation (4.5) shows that either a or c + 2d is an associate of q. Assume by way of contradiction that vq = c + 2d for some unit v ∈ R. Note (v − u)q = 2d; set χ = v − u. Let Z = {α ∈ X_{R} : q̂(α) ≠ 0}. From (4.4), q = qu(a + 2b) so that û = â + 2b̂ on Z. Similarly, from (4.5) q = qva so that v̂ = â on Z. Thus, on Z, χ̂ = v̂ − û = −2b̂. Now u and v are units and so have signatures ±1 at all orderings. Thus χ̂(X_{R}) ⊂ {2, 0, −2}. If b is even dimensional then we must have b̂ = 0 on Z. Then χ̂ = 0 on Z and 0 = qχ = 2d. But then d = 0 and the second factor of q, c + d⟨1, −t⟩ = c = uq is an associate of q, a contradiction. Hence b is odd dimensional. In particular, b̂ is never zero. So v̂ − û is not zero on Z. We must have v̂ = −û (as û and v̂ are always ±1). So χ̂ = 2v̂ on Z. Then 2vq = qχ = 2d and vq = d. But then the second factor of q is c + d⟨1, −t⟩ = uq + vq⟨1, −t⟩ = q(u + v − vt) = −vtq, an associate of q. This is impossible. Hence we must have that q is an associate of a as well as c. Write uq = c and vq = a for units u, v ∈ R. Equation (4.4) gives q = uq(a + 2b). If q is even dimensional then a + 2b is odd dimensional and so a is odd dimensional. But a is an associate of the even dimensional q so a must be even dimensional, a contradiction. We have then that q is odd dimensional. Then q = uq(a+2b) implies u(a+2b) = 1. So uvq = ua = 1 − 2ub, as desired. □

It can be shown that a + bt ∈ R[E_{1}] is irreducible if a + b is irreducible in R and a − b is a unit. Thus in the factorization of (4.3) 1 + 2q_{0} = (1 + q_{0} + q_{0}t)(1 + q_{0} − q_{0}t), both factors are irreducible. However, not every irreducible a + bt ∈ R[E_{1}] satisfies a + b irreducible and a − b a unit. For instance, one may easily check that q = ⟨1⟩ + ⟨⟨t_{1}, t_{2}, t_{3}⟩⟩ ∈ ℤ[E_{3}] is irreducible. As a form in R[E_{1}], where R = ℤ[E_{2}], we have q = a + bt_{3} with a = ⟨1⟩ + ⟨⟨t_{1}, t_{2}⟩⟩ and b = ⟨⟨t_{1}, t_{2}⟩⟩. Then a − b is a unit but a + b = 1 + 2⟨⟨t_{1}, t_{2}⟩⟩ = (1 − ⟨⟨t_{1}, t_{2}⟩⟩)^{2}. In fact, we have been unable to determine the irreducible elements of R[E_{1}] in terms of the irreducibles of R. For products, we can determine only the irreducible odd dimensional forms.

Proposition 4.6. If R = R_{1} ⊓ R_{2} and (a, b) ∈ R is odd dimensional then (a, b) is irreducible iff a is irreducible in R and b is a unit or the reverse, a is a unit and b is irreducible.

Proof. We have (a, b) = (a, 1)(1, b). So (a, b) irreducible implies either a or b is a unit. Say b is a unit. If a = xy then (a, b) = (x, b)(y, 1), so a must be irreducible in R. □

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Robert W. Fitzgerald

Southern Illinois University Carbondale, IL 62901-4408 USA rfitzg@math.siu.edu 164

Documenta Mathematica · Quadratic Forms LSU 2001