Fitzgerald, Robert W.

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Southern Illinois University CarbondaleOpenSIUCArticles and Preprints Department of Mathematics1997K-Regular Witt RingsRobert W. FitzgeraldSouthern Illinois University Carbondale, rfitzg@math.siu.eduFollow this and additional works at: http://opensiuc.lib.siu.edu/math_articlesFirst published in Proceedings of the American Mathematical Society,125, 1309-1313, published bythe American Mathematical Society. ⓒ Copyright 1997, American Mathematical SocietyThis Article is brought to you for free and open access by the Department of Mathematics at OpenSIUC. It has been accepted for inclusion in Articlesand Preprints by an authorized administrator of OpenSIUC. For more information, please contact opensiuc@lib.siu.edu.Recommended CitationFitzgerald, Robert W. "K-Regular Witt Rings." ( Jan 1997).K-REGULAR WITT RINGSRobert W. Fitzgerald(R, G, q) will denote an abstract Witt ring in the sense of [4]. Nearly all examples ofinterest are Witt rings of non-singular quadratic forms over a field of characteristic nottwo,however using abstract Witt rings does simplify some proofs. The Witt ring is k-regularif there exists a 2-power k such that for all 1 6= x ∈ G we have |D〈1,−x〉| = k. Such Wittrings were first studied in [1] primarily because the block design counting arguments therewere perfectly suited to k-regular rings. However they remain unclassified.We will always assume that G is finite; set g = |G|. If k = g then R is totally degenerateand so classified by [4]. If k = g/2 then R is of local type [2] which are again classified in[4]. If k = 2 then R is a group ring extension of Z2 or Z4. If 2 < k < g/2 then R is notof elementary type and no examples are known or even expected. We will always assumethat 2 < k < g/2 and call such k-regular Witt rings exceptional.It was shown in [1] that exceptional k-regular Witt rings satisfy 8 ≤ k and 2k2 ≤ g.Kula [3] improved both bounds and added an upper bound, showing:16 ≤ k8k2 ≤ g ≤ k4/4 if k ≡ 1 (mod 3)8k2 ≤ g ≤ k4/8 if k ≡ 2 (mod 3).Here we show that k3 ≤ g and that if k ≡ 1 (mod 3) then g ≡ 1 (mod 3).We fix some notation, which will agree with Kula’s. G∗ denotes G \ {1}. We set e =log2 k. For a ∈ G∗ and i ≥ 0 set:Xi(a) = {x ∈ G : x 6= 1, a and |Q(a) ∩Q(x)| = 2i},where Q(x) = {q(x, y) : y ∈ G}. Now for x 6= a, |Q(a) ∩ Q(x)| = |D〈1,−ax〉|/|D〈1,−a〉 ∩D〈1,−x〉| = k/|D〈1,−a〉 ∩D〈1,−x〉|. Thus we also have that:Xi(a) = {x ∈ G : x 6= 1, a and |D〈1,−a〉 ∩D〈1,−x〉| = 2e−i}.In particular, we may assume 0 ≤ i ≤ e. We further set ni(a) = |Xi(a)| and write X(a)for Xe(a). For a 2-fold Pfister form ρ we let ρ′ denote the pure part of ρ.Typeset by AMS-TEX12 ROBERT W. FITZGERALDWe will use various equations derived by Kula:e−1∑i=0(2e−i − 1)ni(a) = k2 − 3k + 2(1)g +∑1 6=ρ∈Q(a)|D(ρ′)| = 1 + gk+e∑i=02ini(a)(2)|X(a) ∩X(b)| ≥ g − 2k2 + 6k − 7 ≥ g − 2k2,(3)where a 6= b in G∗ for (3). Equation (1) is [3,4.3b], (2) is equation (4.5.2) on [3,p.45]and the first inequality of (3) is equation (4.3.1) on [3,p.43]. The second inequality of (3)follows from our assumption that k > 2.We will also use two simple equations:(4)e∑i=0ni(a) = g − 2(5) |D(ρ′)| < k2 (if ρ 6= 1).Both (4) and (5) appear in [3] but direct proofs are quick. (4) follows from G\{1, a} beingthe union of the Xi(a). For (5), suppose ρ′ = 〈a, b, ab〉. ThenD(ρ′) = a · ∪x∈D〈1,a〉D〈1, bx〉.Since 1 occurs in each D〈1, bx〉 we have that |D(ρ′)| < |D〈1, a〉| · k = k2.Using equation (4) to find ne(a) and equation (1) to find ne−1(a), equation (2) may bere-written (see [3, pp. 45-46]) as:(6) g+∑1 6=ρ∈Q(a)|D(ρ′)| = 1+ gk+ gk− k32+3k22−3k+e−2∑i=02i(2e−i−1−1)(2e−i−1)ni(a).Proposition 1. If k ≡ 1 (mod 3) then g ≡ 1 (mod 3).Proof. We may pick an a ∈ G∗ with 〈〈1, 1〉〉 /∈ Q(a) \ {1} (otherwise −G∗ ⊂ D〈1, 1, 1〉while |D〈1, 1, 1〉| < k2 by (5) and |G∗| ≥ 8k2 − 1 by [3,4.4]. Then for each anisotropicρ ∈ Q(a) we have that |D(ρ′)| ≡ 0 (mod 3) by [3,2.9]. Also, since for each i, in equation(6) one of e− i− 1 or e− i is even, we have that one of 2e−i−1 − 1 or 2e−i − 1 is divisiblebt 3. Assuming k ≡ 1 (mod 3), equation (6) gives:g ≡ g + 1 + g − 2 (mod 3),and so 1 ≡ g (mod 3). ¤K-REGULAR WITT RINGS 3Theorem 1. g ≥ k3.Proof. Suppose there exists an exceptional k-regular Witt ring (R,G) with g < k3. Amongall such Witt rings, choose one with minimal h ≡ g/k2. Let a and b be distinct elements ofG∗. Choose x ∈ X(a) ∩X(b), which is possible by equation (3) and the fact that g ≥ 8k2[3,4.4]. We use the equation (4.3.2) from [3,p.43]:hk = g/k = |Q(x)| ≥ |(Q(x) ∩Q(a))(Q(x) ∩Q(b)|(7)=k2|Q(x) ∩Q(a) ∩Q(b)|≥ k2|Q(a) ∩Q(b)| = k|D〈1,−a〉 ∩D〈1,−b〉|.A simple consequence of (7) is that |D〈1,−a〉 ∩D〈1,−b〉| ≤ h. Pick minimal s ≥ 0 so thatthere exists distinct a and b in G∗ with |D〈1,−a〉∩D〈1,−b〉| = h/2s. Set 2t = |Q(a)∩Q(b)|.then we have:(8) g ≥ 2s+2k2 and t− s ≥ 1Namely, if the first inequality failed then h = g/k2 ≤ 2s+1. But then |D〈1,−a〉∩D〈1,−b〉| ≤2 for all distinct a and b in G∗. while as noted in the first sentence of [K,p.44] we can alwaysfind distinct a and b in G∗ with |D〈1,−a〉∩D〈1,−b〉| ≥ 4. For the second inequality of (8)note that:2t = |Q(a) ∩Q(b)| = k|D〈1,−a〉 ∩D〈1,−b〉| =2skh.Thus 2t−sh = k. By the assumption that g = hk2 < k3 we have 2h ≤ k and so t− s ≥ 1.For each x ∈ X(a) ∩X(b) we can rewrite (7) as:(9) hk ≥ k2|Q(x) ∩Q(a) ∩Q(b)| ≥k2|Q(a) ∩Q(b)| =hk2s.Then|Q(x) ∩Q(a) ∩Q(b)| ≥ 2t−ssince otherwise |Q(x)∩Q(a)∩Q(b)| < 2t−s = |Q(a)∩Q(b)|/2s and equation (9) becomes:hk ≥ k2|Q(x) ∩Q(a) ∩Q(b)| >2sk2|Q(a) ∩Q(b)| = hk.List the elements of Q(a)∩Q(b) as 1, ρ2, . . . , ρ2t .We have that for each x ∈ X(a)∩X(b)that 2t−s − 1 of the ρi’s lie in Q(x), or equivalently, satisfy −x ∈ D(ρ′i). Set Tx equal tothe number of i’s, 2 ≤ i ≤ 2t, such that −x ∈ D(ρ′i). Then:(10)∑x∈X(a)∩X(b)Tx ≥ (2t−s − 1)|X(a) ∩X(b)|4 ROBERT W. FITZGERALDNow this sum counts the number of pairs (i, x) with 2 ≤ i ≤ 2t, x ∈ X(a) ∩ X(b) and−x ∈ D(ρ′i). We can also count the number of such pairs by first fixing i. Namely:(11)∑x∈X(a)∩X(b)Tx =2t∑i=2|D(ρ′i) ∩ −(X(a) ∩X(b))|.Now (11) implies that there exists an i, 2 ≤ i ≤ 2t, such that:|D(ρ′i) ∩ −(X(a) ∩X(b))| ≥12t − 1∑x∈X(a)∩X(b)Txand hence when combined with (9):|D(ρ′i)| ≥2t−s − 12t − 1 |X(a) ∩X(b)|.Applying equations (5) and (3) yields:(12) k2 >2t−s − 12t − 1 (g − 2k2)If s = 0 then (12) becomes k2 > g− 2k2 which is impossible as g ≥ 8k2 [3,4.4]. Supposethen that s ≥ 1. (11) is then:(2t + 2t−s+1 − 3)k2 > (2t−s − 1)g.Use g ≥ 2s+2k2 from the first part of (8) to get:(2t + 2t−s+1 − 3)k2 > (2t+2 − 2s+2)k22s+2 + 2t−s+1 − 3 > 3 · 2t.Lastly, using t− 1 ≥ s from the second part of (8) gives:2t+1 + 2t−s+1 − 3 > 3 · 2t,2t−s+1 − 3 > 2t,which is impossible for s ≥ 1. This contradiction shows g ≥ k3. ¤We combine these results with Kula’s upper bound on g and bound on k.Corollary 1. For an exceptional k-regular Witt ring (R,G) with g = |G| : k ≥ 16 and(1) if k ≡ 1 (mod 3) then g ≡ 1 (mod 3) andk3 ≤ g ≤ 14k4,(2) if k ≡ 2 (mod 3) thenk3 ≤ g ≤ 18k4.¤We note that the first open case is k = 16 and g = 163 = 4096.Kula has shown that an exceptional k-regular Witt ring is non-formally real [3,Re-mark,p.41] so that InR = 0 for some n. We have:K-REGULAR WITT RINGS 5Corollary 2. If (R,G) is an exceptional k-regular Witt ring then I3R 6= 0. In fact, forany anisotropic 2-fold Pfister form ρ,D(ρ) 6= G.Proof. D(ρ) = ∪b∈D(ρ′)D〈1, b〉 so that |D(ρ)| ≤ k|D(ρ′)| < k3 by equation (5). ThusD(ρ) 6= G by Theorem 1. ¤References1. R. Fitzgerald and J. Yucas, Combinatorial techniques and abstract Witt rings I, J. Algebra 114 (1988),40–52.2. I. Kaplansky, Fro¨hlich’s local quadraric forms, J. Reine Angew. Math. 239 (1969), 74–77.3. M. Kula, Finitely Generated Witt Rings, Uniwersytet S´laski, Katowice, 1991.4. M. Marshall, Abstract Witt Rings, Queen’s Papers in Pure and Applied Math.,No. 57, Queen’s Uni-versity, Kingston, Ontario, 1980.Department of MathematicsSouthern Illinois UniversityCarbondale, IL 62901-4408e-mail address: rfitzg@math.siu.edu

K-Regular Witt Rings

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K-Regular Witt Rings

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