*269*
*72*
*436KB*

*English*
*Pages 25*
*Year 2018*

- Author / Uploaded
- Zhengyao Wu (吴正尧)

**Commentary**- Downloaded from https://wuzhengyao.gitee.io/homepage/u-char2.pdf

*Table of contents : 1. The square number vs u and u"0362u......Page 12. Henselian discrete valued fields......Page 63. Fields with prescribed u and u"0362u......Page 144. What is known......Page 22References......Page 25*

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2 ZHENGYAO WU (吴正尧)

Abstract. Notes for discussion with Yong Hu (胡勇) and Peng Sun (孙鹏) on 26 May 2018 at SUSTC. Main references: [Bae82b] and [MMW91].

Contents 1.

The square number vs u and u b

1

2.

Henselian discrete valued fields

6

3.

Fields with prescribed u and u b

14

4.

What is known

22

References

25

Throughout this note, let K be a field. For a quadratic form over K, let D(q) be the set of elements of K ∗ represented by q. We avoid h i because we reserve it for diagonal quadratic forms over fields of characteristic not 2. Also, [ ] is easier to type than \langle \rangle Let W (K) be the Witt ring of nonsingular symmetric bilinear forms over K. Let I = I(K) be the ideal of W (K) of even dimensional forms. Let Wq (K) be the Witt group of nonsingular quadratic forms over K. We have Wq (K) is a W (K)-module. The ideal I n Wq (K) is generated by n-fold Pfister forms hha1 , . . . , an−1 ii ⊗ [1, b] for a1 , . . . , an−1 ∈ F ∗ and b ∈ F . 1. The square number vs u and u b 1.1 Definition Let q : V → K be a quadratic from with associated symmetric bilinear form bq . We call q nonsingular if Rad(bq ) = 0; regular if Rad(q) = 0. 1.2 Lemma Suppose char K = 2. 1

2

ZHENGYAO WU (吴正尧)

(1) Every nonsingular form over K is regular. (2) Every anisotropic quadratic form over K is regular (but not necessarily nonsingular). An anisotrpic Pfister quadratic form over K is nonsingular. (3) The form [c], c ∈ K ∗ , i.e. q(x) = cx2 is anisotropic and singular. (4) The hyperbolic plane [0, 0], i.e. H(x, y) = xy is isotropic and nonsingular. (5) The form [a, b], a, b ∈ K, i.e. q(x, y) = ax2 + xy + by 2 is nonsingular. (6) Every regular quadratic form has shape [a1 , b1 ] ⊥ · · · ⊥ [an , bn ] ⊥ [c1 ] ⊥ · · · ⊥ [cm ], ai , bi ∈ K, cj ∈ K ∗ {z } | {z } | nonsingular part

singular part

In particular, every nonsingular quadratic form over K has even dimension and u(K) is even. 1.3 Definition u(K) = sup{dimK V | ∃ quadratic form q : V → K anisotropic and nonsingular.} u b(K) = sup{dimK V | ∃ quadratic form q : V → K anisotropic. } 1.4 Lemma (1) If char K 6= 2, then u(K) = u b(K). (2) If char K = 2, then u(K) ≤ u b(K). (3) There exists a field K such that char K = 2 and u(K) < u b(K). Proof. (3) Let K be a quadratically closed field of characteristic 2. Then u b(K) = 1. Also every 1-dimensional quadratic form is of the form [c], c ∈ K, which is singular by Lemma 1.2(3), so u(K) = 0.

1.5 Example This is [Bae82b, Rem. 1.2 b)]. Suppose char K = 2. If K is perfect, then either u(K) = 0 or u(K) = u b(K) = 2. Proof. Suppose u(K) > 0. By Lemma 1.2(5), 2 ≤ u(K). By Lemma 1.4(2), u(K) ≤ u b(K). Since x 7→ x2 is an automorphism of K, [a, b] is universal for all a, b ∈ K. If q is anisotropic of dimension > 1, then q ' [a, b] for some a, b ∈ K by Lemma 1.2(6) then u b(K) ≤ 2.

Let K 2 be the subfield of square elements of K. We call [K : K 2 ] the square number of K.

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

3

1.6 Lemma This is [Bae82b, Rem. 1.2 b)c)d)]. Suppose char K = 2. (0) If finite, [K : K 2 ] is a power of 2. (1) u b(K) ≥ [K : K 2 ]. (2) If u(K) = 0, then u b(K) = [K : K 2 ]. (3) There exists K such that u(K) > 0 and u b(K) > [K : K 2 ]. Proof. (0) Since K/K 2 is purely inseparable, we may decompose K/K 2 into a tower of simple extensions of degree 2. For details see [Bou81, Ch. V, § 5, no. 2, Prop. 4]. (1) The form [c1 ] ⊥ · · · ⊥ [cm ], cj ∈ K is anisotropic iff c1 , . . . , cm are linearly independent over K 2 . By Lemma 1.2(6), u b(K) ≥ [K : K 2 ]. (2) When u(K) = 0, by Lemma 1.2(2)(5)(6), every anisotropic quadratic form has shape [c1 ] ⊥ · · · ⊥ [cm ], cj ∈ K. Then u b(K) = [K : K 2 ]. (3) Since [1, 1] is anisotropic over F2 , u b(F2 ) = 2 > 1 = [F2 : F22 ].

1.7 Lemma This is [MMW91, Prop. 1, Cor. 1]. Suppose char K = 2. Suppose q = [a1 , b1 ] ⊥ · · · ⊥ [an , bn ] ⊥ [c1 ] ⊥ · · · ⊥ [cm ] is anisotropic over K. (1) n + m ≤ [K : K 2 ]. (2) u b(K) ≤ 2[K : K 2 ] (and hence u(K) ≤ 2[K : K 2 ]). (3) If n + m = [K : K 2 ] and n ≥ 1, then u(K) = u b(K) = 2[K : K 2 ]. Proof. (1) For all di ∈ D([ai , bi ]), [d1 ] ⊥ · · · ⊥ [dn ] ⊥ [c1 ] ⊥ · · · ⊥ [cm ] is anisotropic as a subform of q. Then d1 , . . . , dn , c1 , . . . , cm is linearly independent over K 2 . (2) dim(q) = 2n + m ≤ 2(n + m) ≤ 2[K : K 2 ] by (1). (3) Our goal is to find an anisotropic nonsingular form of dimension 2n + 2m. Then 2[K : K 2 ] ≤ u(K) and the rest follows from (2). Next, we construct such a form. We have a K 2 -basis d1 , d2 , . . . , dn , c1 , . . . , cm of K. Let t : K → K 2 be the K 2 -linear map defined by t(d1 ) = 1, t(d2 ) = · · · = t(dn ) = t(c1 ) = · · · = t(cm ) = 0. Then the composition t ◦ [a1 , b1 ] is a 2[K : K 2 ]-dimensional quadratic form over K 2 . Since [a1 , b1 ] is nonsingular, t ◦ [a1 , b1 ] is nonsingular. Finally, we show that t◦[a1 , b1 ] is anisotropic. If v 6= 0 such that t([a1 , b1 ](v)) = 0, then [a1 , b1 ](v) ∈ SpanK 2 {d2 , . . . , dn , c1 , . . . , cm }, a contradiction to the fact that q is anisotropic.

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ZHENGYAO WU (吴正尧)

1.8 Corollary This is [MMW91, Cor. 2]. Suppose char K = 2. Then u b(K) 6= 2n − 1 for all n ≥ 2. Proof. If not, then u b(K) = 2n − 1. By Lemma 1.6(1) and Lemma 1.7(2), we have 2n−1 − 0.5 ≤ [K : K 2 ] ≤ 2n − 1. By Lemma 1.6(0), [K : K 2 ] = 2n−1 . Suppose q = [a1 , b1 ] ⊥ · · · ⊥ [ar , br ] ⊥ [c1 ] ⊥ · · · ⊥ [cs ] is anisotropic over K such that 2r + s = 2n − 1. Then r + s ≤ 2n−1 . Hence s is odd and s+2n −1 = 2(r+s) ≤ 2n . Then s = 1, r = 2n−1 −1 and r+s = 2n−1 = [K : K 2 ]. It follows from Lemma 1.7(3) that u b(K) = 2[K : K 2 ] = 2n , a contradiction.

1.9 Lemma This is [Bae82b, p. 107, First paragraph]. Let k be a field of characteristic 2. If K/k is finite, then [K : K 2 ] = [k : k 2 ]. Proof. Let (wi )1≤i≤n be a k-basis of K. It follows from the identity (a1 w1 + · · · + an wn )2 = a21 w12 + · · · + a2n wn2 , (ai ∈ k) that (wi2 )1≤i≤n is a k 2 -basis of K 2 . Then [K : k] = [K 2 : k 2 ]. Thus [K : K 2 ] =

[K : k 2 ] [K : k][k : k 2 ] = = [k : k 2 ]. 2 2 [K : k ] [K 2 : k 2 ]

1.10 Corollary This is [MMW91, Prop. 4(iii)]. Let k be a field. Suppose char k = 2. Let K/k 1 b(k) ≤ u b(K) ≤ 2b u(k). be an extension such that [K : k] < ∞. Then u 2 Proof. u b(k) ≤ 2[k : k 2 ],

by Lemma 1.7(2).

2

= 2[K : K ], by Lemma 1.9. ≤ 2b u(K),

by Lemma 1.6(1).

u b(K) ≤ 2[K : K 2 ], by Lemma 1.7(2). = 2[k : k 2 ],

by Lemma 1.9.

≤ 2b u(k),

by Lemma 1.6(1).

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

5

1.11 Lemma This is [Bae82b, Lem. 1.3 A.1]. Let k be a field of characteristic 2. Suppose K/k is separable and algebraic. If (wi )1≤i≤n are linearly independent elements of K over k, then (wi2 )1≤i≤n are linearly independent over k. Proof. Suppose E = k(w1 , . . . , wn ). Let (wi )1≤i≤m (m ≥ n) be a k-basis of E m P containing (wi )1≤i≤n . Let F = Spank (wi2 )1≤i≤m . Suppose wi wj = cijl wl for cijl ∈ k. Suppose a =

m P

ai wi2

and b =

i=1 m P

• a±b=

(ai ±

m P l=1

• For a ∈

l=1

bi wi2 .

i=1

bi )wi2

∈ F; !

i=1

• ab =

m P

m P

ai bj c2ijl i,j=1 F ∗ , a1 = a( a1 )2

wl ∈ F ; where ( a1 )2 ∈ E 2 ⊂ F and hence

1 a

∈ F.

Thus F is a subfield of E. Finally, since E/k is separable, E/F is separable. Then E = F since E/E 2 is purely inseparable. Then (wi2 )1≤i≤m are linearly independent over k. Therefore its subset (wi2 )1≤i≤n are linearly independent over k.

1.12 Lemma This is [Bae82b, Lem. 1.3 A.2]. Let k be a field of characteristic 2. Suppose K/k is separable and algebraic. If (ei )1≤i≤n are linearly independent elements of k over k 2 , then they are also linearly independent over K 2 . Proof. Suppose a2i ∈ K 2 (ai ∈ K) such that a21 e1 + · · · + a2n en = 0. We want to show that a2i = 0 (i.e. ai = 0) for all 1 ≤ i ≤ n. Suppose E = k(a1 , . . . , an ). m P Let (wi )1≤i≤m be a k-basis of E. Then ai = bij wj for bij ∈ k. Then j=1

0=

n X

a2i ei

=

i=1

By Lemma 1.11, we have

n m X X i=1 n P

j=1

!2 bij wj

ei =

m n X X j=1

! b2ij ei

wj2

i=1

b2ij ei = 0 for all 1 ≤ j ≤ m and hence bij = 0 since

i=1

(ei )1≤i≤n are linearly independent over k 2 . Therefore ai = 0 for all 1 ≤ i ≤ n.

1.13 Lemma This is [Bae82b, Lem. 1.3]. Let k be a field of characteristic 2. If K/k is

6

ZHENGYAO WU (吴正尧)

separable and algebraic, then [K : K 2 ] = [k : k 2 ]. Proof. Let (wi )1≤i≤n be elements of K linearly independent over K 2 . Then (wi )1≤i≤n are linearly independent over E 2 where E = k(w1 , . . . , wn ) is subfield of K such that E/k is finite. Then [K : K 2 ] ≤ [E : E 2 ]. By Lemma 1.9, [E : E 2 ] = [k : k 2 ]. Hence [K : K 2 ] ≤ [k : k 2 ]. Conversely, let (ei )1≤i≤n be elements of k linearly independent over k 2 . By Lemma 1.12, (ei )1≤i≤n are linearly independent over K 2 . Hence [k : k 2 ] ≤ [K : K 2 ].

1.14 Example This is [Bae82b, p.108, two paragraphs before § 2]. Let k = F2 (X1 , X2 , . . . , Xn ) (resp. k = F2 (X1 , X2 , . . .)). Then [k : k 2 ] = 2n (resp. ∞). Let K be the quadratic separable closure of k, i.e. elements in K − k are in an algebraic closure of k whose minimal polynomial is a quadratic separable. It follows from Lemma 1.13 that [K : K 2 ] = [k : k 2 ] = 2n (resp. ∞). By Lemma 1.4(3), u(K) = 0. By Lemma 1.6(2), u b(K) = 2n (resp. ∞). 2. Henselian discrete valued fields Let (K, v) be a Henselian discrete valued field with valuation ring A, uniformizer π and residue field K = A/πA. 2.1 Lemma If f (T ) = a2 T 2 + a1 T + a0 is irreducible in K[T ], then 2v(a1 ) ≥ v(a0 ) + v(a2 ). Proof. Suppose ai = ui π ni for i ∈ {0, 1, 2}. If 2n1 < n0 + n2 , then g(T ) = π n2 −2n1 f (π n1 −n2 T ) = u2 T 2 + u1 T + u0 π n0 +n2 −2n1 . Hence g(T ) = T (u2 T + u1 ). By Hensel’s lemma, g(T ) is reducible and thus f (T ) is reducible, a contradiction.

2.2 Lemma Schwarz inequality. This is [MMW91, p. 342 (4)]. Let q : V → K be an anisotropic quadratic form with associated symmetric bilinear form bq . Then 2v(bq (x, y)) ≥ v(q(x)) + v(q(y)) for all x, y ∈ V . This is true regardless of char K and char K,

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

7

Proof. Suppose x, y are linearly independent. Since q is anisotropic, f (T ) = q(xT +y) = q(x)T 2 +bq (x, y)T +q(y) is irreducible. By Lemma 2.1, 2v(bq (x, y)) ≥ v(q(x)) + v(q(y)). Suppose x, y are linearly dependent. Then v(bq (x, y)) = v(0) = +∞.

2.3 Lemma This is [Bae82b, After Lem. 2.3]. Let q : V → K be an anisotropic quadratic form. Let Mi = {x ∈ V | v(q(x)) ≥ i}, i ≥ 0. Then Mi+1 ⊂ Mi and (1) Mi is an A-module. (2) Mi+2 = πMi for all i ≥ 0. (3) If x ∈ Mi and y ∈ Mi+1 , then v(bq (x, y)) ≥ i + 1. Proof. For all a ∈ K and x ∈ Mi , v(q(x)) ≥ i. We have v(q(ax)) = v(a2 q(x)) = 2v(a) + v(q(x)). (1) For all a ∈ A v(a) ≥ 0. Then v(q(ax)) ≥ i and hence ax ∈ Mi . (2) For all x ∈ Mi , v(q(πx)) = 2 + v(q(x)) ≥ i + 2 and hence πx ∈ Mi+2 . Conversely, for all y ∈ Mi+2 , then v(π −1 y) = −2 + v(q(y)) ≥ −2 + (i + 2) = i and hence y ∈ Mi . (3) By (2) symmetric, it suffices to assume i = 0. By Lemma 2.2, v(bq (x, y)) ≥ v(q(x)) + v(q(y)) 0+1 1 ≥ = and thus v(bq (x, y)) ≥ 1. 2 2 2 2.4 Lemma This is [Bae82b, After Lem. 2.3]. Let q : V → K be an anisotropic quadratic form. Let V0 = M0 /M1 and V1 = M1 /M2 . They are vector spaces over K. Let q(x) q0 (x) = q(x) for all x ∈ M0 . Let q1 (x) = for all x ∈ M1 . We call q0 π and q1 residue forms of q. Then (Vi , qi ) are anisotropic over K for i ∈ {0, 1}. Proof. Suppose q1 (x) = 0 in K for some x ∈ M0 . Then v(q(x)) ≥ 1, i.e. x ∈ M1 . Hence x = 0 in V0 and thus q0 is anisotropic.

q(x) Suppose q1 (x) = 0 in K for some x ∈ M1 . Then v π i.e. x ∈ M2 . Hence x = 0 in V1 and thus q1 is anisotropic.

≥ 1, v(q(x)) ≥ 2,

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ZHENGYAO WU (吴正尧)

2.5 Lemma This is [Bae82b, p. 109-111(2.6)-(2.9)]. Let ai , bi ∈ A for all 1 ≤ i ≤ n and uj ∈ A∗ for all 1 ≤ j ≤ m. Suppose q = [a1 , b1 ] ⊥ · · · ⊥ [an , bn ] ⊥ [u1 ] ⊥ · · · ⊥ [um ] is anisotropic over K. Let q0 = [a1 , b1 ] ⊥ · · · ⊥ [an , bn ]. Then q = q0 ⊥

q0 u1 um ⊥ [u1 , ] ⊥ · · · ⊥ [um , ] π π π

is anisotropic and nonsingular over K. q0 (t) Proof. By Lemma 1.4(5), q is nonsingular. If q is isotropic, then q0 (s) + + π m P uj uj x2j + xj yj + yj2 = 0. Suppose s = x0 s0 and t = y0 t0 such that x0 , y0 ∈ K π j=1 and s0 6≡ 0 mod π, t0 6≡ 0 mod π. Then ? πx20 q0 (s0 ) + y02 q0 (t0 ) +

m X

πx2j uj + πxj yj + yj2 uj = 0

j=1

Here xj , yj (0 ≤ j ≤ m) are not all 0 in K. Since q is anisotropic, we must have min(v(πx20 ), v(y02 ), v(πx2j ), v(yj2 ), v(πxj yj )(1 ≤ j ≤ m)) > 0. Three cases: (1) min = πx2k for some 0 ≤ k ≤ m; (2) min = yk2 for some 0 ≤ k ≤ m; (3) min = πxk yk for some 1 ≤ k ≤ m. (1) Dividing ? by min = πx2k , we have 2 2 2 2 m X 1 y0 xj xj yj 1 yj x0 0 0 q0 (s )+ q0 (t )+ uj + + uj = 0 xk π xk xk xk xk π xk j=1 2 ! v(πx2j ) − v(πx2k ) xj 1 yj Here v = v(yj2 ) − v(πx2k ) ≥ 0 = ≥ 0; v xk 2 π xk and thus ≥ 1 since v(yj2 ) is even and v(πx2k ) is odd.

Modulo π, we have

Since

x0 x0

x0 xk

2 q0

(s0 )

+

2 m X xj j=1

xk

uj = 0.

xk = 1 if 1 ≤ k ≤ m, we have xk ! x0 x x 1 m s0 , ,..., 6= 0, xk xk xk

s0 = s0 6= 0 if k = 0; or

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

9

a contradiction to the fact that q is anisotropic. (2) Dividing ? by min = yk2 , we have 2 2 2 2 m X x0 y0 xj xj yj yj 0 0 π q0 (s )+ q0 (t )+ π uj +π + uj = 0 yk yk yk yk yk yk j=1 2 ! v(yj2 ) − v(yk2 ) xj yj ≥ 0; v π = = v(πx2j ) − v(yk2 ) ≥ 0 Here v yk 2 yk xj yj and thus ≥ 1 since v(πx2j ) is odd and v(yk2 ) is even; v π = yk yk 1 + v(πx2j ) + v(yj2 ) 1 − v(yk2 ) ≥ and thus ≥ 1. 2 2 Modulo π, we have 2 2 m X y0 yj 0 q0 (t ) + uj = 0. yk yk j=1 Since

y0 y0

t0

=

t0

yk 6 0 if k = 0; or = = 1 if 1 ≤ k ≤ m, we have yk ! y0 0 y1 ym t, ,..., 6= 0, yk yk yk

a contradiction to the fact that q is anisotropic. (3) min = πxk yk . Then −1 ≥ −v(π) + v(πxk yk ) −

v(yk2 )

=v

xk yk

= v(πx2k ) − v(πxk yk ) ≥ 0

a contradiction.

2.6 Definition Let B be a valuation ring with field of fractions F = Frac(B). Let M be the maximal ideal of B with residue field F = B/M . Let v : F ∗ → Γ be the valuation to an ordered group. Let V be a finite dimensional vector space over F . Let P be a full B-lattice in V , i.e. a B-module such that P ⊂ V and F P = V . The residue defect of P is rd(P ) = dimF V − dimF (P/M P ). 2.7 Lemma This is [MMW91, Prop. 5(i)(ii)]. (1) rd(P ) ≥ 0. (2) rd(P ) = 0 iff P is a finitely generated B-module.

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ZHENGYAO WU (吴正尧)

Proof. (1) Suppose p1 , . . . , pk ∈ P such that p1 , . . . , pk ∈ P/M P are linearly independent over F . We show that p1 , . . . , pk are linearly independent over F and thus dimF V ≥ dimF (P/M P ). k P If not, then there exists ci ∈ F , 1 ≤ i ≤ k not all zero and ci pi = 0. Suppose i=1 P ci ci v(cj ) = min (v(ci )). Then cj 6= 0 and pi + pj = 0, where ∈ B. Modulo 1≤i≤k cj i6=j cj P ci M , we have pi + pj = 0, a contradiction to the fact that p1 , . . . , pk are i6=j cj linearly independent over F .

(2) If P is a finitely generated B-module, then P is a free over B of finite rank by [Bou64, Ch. VI, § 4, no. 1, Lem. 1]. Suppose P ' B n . Then F P ' F n n

and P/M P ' P ⊗B B/M ' B n ⊗B F ' F . Hence rd(P ) = dimF V − dimF (P/M P ) = n − n = 0. Conversely, suppose (p1 , . . . , pk ) is a F -basis of P/M P . By (1), (p1 , . . . , pk ) are linearly independent over B (then over F ). Since rd(P ) = 0, we have dimF V = k and hence (p1 , . . . , pk ) is a F -basis of V . It suffices to show that (p1 , . . . , pk ) span P . If not, then there exists

k P

ci pi ∈ P − (Bp1 ⊕ · · · ⊕ Bpk ) such that v(cj ) =

i=1

min (v(ci )) < 0. Then

1≤i≤k

k P ci 1 1 P ∈ M, pi + pj = ci pi ∈ M P . Modulo M , cj cj i=1 i6=j cj

P ci pi + pj = 0, a contradiction to the fact that p1 , . . . , pk are linearly i6=j cj independent over F . we have

2.8 Lemma This is [MMW91, Prop. 5(iii)]. Let W ⊂ V be an F -subspace. Let Q = P ∩W . Then rd(P ) = rd(Q)+rd(P/Q). Proof. Since P/Q = P/P ∩ W ' (P + W )/W ⊂ V /W , we may assume that P/Q ⊂ V /W . Since F Q = F (P ∩ W ) = F P ∩ W = V ∩ W = W and F is flat over B, we have F (P/Q) ' F P/F Q = V /W . Then P/Q is torsion-free over B. By [Bou64, Ch. VI, § 4, no. 1, Lem. 1] again, P/Q is flat over B. By [Bou07, Ch. X, § 4, no. 4, Cor. of Prop. 5], TorB 1 (B/M, P/Q) = 0. From the exact sequence 0 → Q → P → P/Q → 0, we have another exact sequence 0 → Q/M Q → P/M P →

P/Q → 0. M (P/Q)

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

11

Therefore rd(Q) + rd(P/Q) = dimF W − dimF (Q/M Q) + dimF (V /W ) − dimF

P/Q M (P/Q)

= dimF V − dimF (P/M P ) = rd(P ). 2.9 Definition Let (K, v) be a Henselian discrete valued field with valuation ring A, uniformizer π and residue field K = A/πA. Let q : V → K be an anisotropic quadratic form. Let q0 and q1 be residue forms of q as in Lemma 2.4. The residue defect of q is rd(q) = dimK q − dimK q0 − dimK q1 . 2.10 Lemma This is [MMW91, p. 342 (5)]. rd(q) = rd(M0 ). Proof. First KM0 = V . Then dimK V = dimK KM0 . Also, by Lemma 2.3, dimK q0 + dimK q1 = dimK (M0 /M1 ) + dimK (M1 /M2 ) = dimK (M0 /M1 ) + dimK (M1 /πM0 ) = dimK (M0 /πM0 ). Therefore rd(q) = dimK q−dimK q0 −dimK q1 = dimK KM0 −dimK (M0 /πM0 ) = rd(M0 ). 2.11 Lemma This is [MMW91, Th. 1(a)]. Let (K, v) be a Henselian discrete valued field with valuation ring A, uniformizer π and residue field K = A/πA. Let q : V → K be an anisotropic quadratic form. If char K = 2 and q is totally singular, then 2

rd(q) ≤ [K : K 2 ] − 2[K : K ].

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ZHENGYAO WU (吴正尧)

Proof. Since q(cy) = c2 q(y) for all c ∈ K and q is anisotropic, we have dimK V = dimK 2 D(q). Similarly, dimK (M0 /πM0 ) = dimK 2 (q(M0 )/π 2 q(M0 )). rd(q) = rd(M0 ),

by Lemma 2.10.

= dimK V − dimK (M0 /πM0 ) = dimK 2 D(q) − dimK 2 (q(M0 )/π 2 q(M0 ))

Since q is a totally singular.

= rd(q(M0 )) = rd(A) − rd(A/q(M0 )),

by Lemma 2.8.

2

2

≤ rd(A) = [K : K ] − dimK 2 (A/π A) 2

= [K : K 2 ] − 2[K : K ],

since dimK (A/π 2 A) = 2.

2.12 Theorem This is [MMW91, Th. 1(b)]. Let (K, v) be a Henselian discrete valued field with valuation ring A, uniformizer π and residue field K = A/πA. Let q : V → K be an anisotropic quadratic form. (1) If char K = 2, then 2

rd(q) = rd(Rad(bq )) ≤ [K : K 2 ] − 2[K : K ]. (2) If char K 6= 2, then rd(q) = 0. Proof. Let y1 , . . . , yk be a K-basis of V . Since KM0 = V , we may assume that yi ∈ M0 . It follows from Lemma 2.2 that bq (z, yi ) ∈ A for all 1 ≤ i ≤ k and z ∈ M0 . Let E = Ay1 ⊕ · · · ⊕ Ayk . Then γ : M0 → E ∗ = HomA (E, A) such that γ(z) = bq (z, •) is A-linear. Then ker(γ) = Rad(bq ) ∩ M0 . Since E ∗ is a free A-module of finite rank, its sub-A-module Im(γ) is also free and thus rd(Im(γ)) = 0. Then M0 ' ker(γ) ⊕ Im(γ) ' Rad(bq ) and rd(q) = rd(Rad(bq ) ∩ M0 ) + rd(Im(γ)) = rd(Rad(bq )). (1) Suppose char K = 2. Since q|Rad(bq ) is totally singular, by Lemma 2.11, 2

rd(q) = rd(Rad(bq )) ≤ [K : K 2 ] − 2[K : K ]. (2) Suppose char K 6= 2. Since q is anisotropic, it is nonsingular and hence Rad(bq ) = 0. Therefore rd(q) = rd(Rad(bq )) = 0.

2.13 Theorem This is [MMW91, Th. 2]. Let (K, v) be a Henselian discrete valued field with

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

13

valuation ring A, uniformizer π and residue field K = A/πA. Suppose char K = char K = 2. (1) u(K) = 2b u(K) (analogous to Springer’s theorem for char K 6= 2). 2

(2) max{2b u(K), [K : K 2 ]} ≤ u b(K) ≤ 2b u(K) + [K : K 2 ] − 2[K : K ]. Proof. First, it follows from Lemma 2.5 that there exists an anisotropic nonsingular quadratic form q over K such that q0 ' q1 ' q and dimK q = 2 dimK q. So 2b u(K) ≤ u(K). Next, let q be an anisotropic form over K. By Lemma 2.4, q0 and q1 are anisotropic over K. Then u(K) + rd(q). dimK q = dimK q0 + dimK q1 + rd(q) ≤ 2b (1) Suppose q is nonsingular and anisotropic. Then rd(q) = 0 and thus u(K) ≤ 2b u(K). 2

(2) By Theorem 2.12, u b(K) ≤ 2b u(K) + rd(q) ≤ 2b u(K) + [K : K 2 ] − 2[K : K ]. By Lemma 1.4(2), we have u(K) ≤ u b(K) and hence 2b u(K) ≤ u b(K). By Lemma 1.7(1), we have [K : K 2 ] ≤ u b(K).

2.14 Example Let k0 be an algebraically closed field of char k0 = 2. Let k1 = k0 (t1 , . . . , tn−1 ) be the rational function field over k0 with n−1 variables. Let L = k1 (s1 , . . . , sm−n+1 ) where si ∈ k1 ((X)) and (s1 , . . . , sm−n+1 ) are algebraically independent. Let v be the restriction of the X-adic valuation on k1 ((X)) to L. Let K be the Henselization of L relative to v. Then K = L = k1 . By Theorem 2.13(1), u(K) = 2b u(k1 ) = 2n . 2m = [L : L2 ], 2

since tr. deg.(L/k0 ) = m.

= [K : K ],

by Lemma 1.13.

≤ u b(K),

by Lemma 1.6(1).

m

≤ 2

by Tsen-Lang theorem and k0 is C0 .

2.15 Theorem This is [Bae82b, Th. 1.1] and [MMW91, Cor. 3]. Let (K, v) be a complete discrete valued field with residue field K. Suppose char K = 2. Then u(K) = u b(K) = 2b u(K).

14

ZHENGYAO WU (吴正尧)

Proof. By Theorem 2.13(1), 2b u(K) = u(K). By Lemma 1.4(2), u(K) ≤ u b(K). 2

Since K is complete, we have [K : K 2 ] = 2[K : K ]. By Theorem 2.13(2), u b(K) ≤ 2b u(K).

2.16 Corollary This is [Bae82b, Cor. 2.10]. Let k be a field of characteristic 2. Let K = k((T1 ))((T2 )) . . . ((Tn )). Then u(K) = u b(K) = 2n u b(k). Proof. For n = 1, use k((T1 )) = k and Theorem 2.15. For n ≥ 2, use induction.

3. Fields with prescribed u and u b 3.1 Review Cassels-Pfister theorem ([EKM08, Th. 17.3] without assumption on char K). Let q : V → K be a quadratic form. If a polynomial f ∈ K[T ] is represented by qK(T ) , then f is also represented by qK[T ] . 3.2 Review A field k is Cn if every homogeneous form of degree d in > dn variables has a nontrivial zero. In paticular, if a field k is Cn , then every quadratic form in > 2n variables is isotropic and hence u(k) ≤ 2n . Chevalley-Warning theorem: If k is finite, then it is C1 . Tsen-Lang theorem: If k is Cn , then k(T ) is Cn+1 . 3.3 Lemma This is [Bae82b, Eg. 2.11]. Let k be a field. Suppose char k = 2. Let K = k(T1 , . . . , Tn ) be the function field in n variables. (1) The n-fold Pfister form qn = hhT1 , . . . , Tn−1 ii ⊗ [1, Tn ] is anisotropic and nonsingular. (2) Suppose u(k) = 2. If a ∈ k − ℘(k), then the (n + 1)-fold Pfister form qn = hhT1 , . . . , Tn ii ⊗ [1, a] is anisotropic and nonsingular. (3) Suppose u(k) = 0 and u b(k) = 2. If a ∈ k − k 2 , then the (n + 1)-fold Pfister form qn = hha, T1 , . . . , Tn−1 ii ⊗ [1, Tn ] is anisotropic and nonsingular. Proof. The form qn is nonsingular because it is orthogonal sums of forms the shape [a, b] since it has a tensor factor [1, Tn ] (resp. [1, a], [1, Tn ]).

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

15

(1) We use induction to show that qn is anisotropic for all n ≥ 1. First, [1, T1 ] is anisotropic since T1 6∈ ℘(K) where ℘(x) = x2 + x for all x ∈ K. Suppose q 0 = hhT2 , . . . , Tn−1 ii[1, Tn ] is anisotropic over K. If qn is isotropic, then it is hyperbolic as it is Pfister. Since qn = q 0 ⊥ T1 q 0 , we have q 0 ' T1 q 0 . Then q 0 represents T1 over k(T2 , . . . , Tn )(T1 ). By the Cassels-Pfister theorem, Then q 0 represents T1 over k(T2 , . . . , Tn )[T1 ]. Suppose q 0 (f1 , . . . , f2n −1 ) = T1 where fi ∈ k(T2 , . . . , Tn )[T1 ] for all 1 ≤ i ≤ 2n − 1. If T1 |fi for all i, then T12 |q 0 (f1 , . . . , f2n −1 ) = T1 , a contradiction. Take T1 = 0, (fi (0))1≤i≤2n −1 are not all 0. Then q 0 is isotropic over k(T2 , . . . , Tn ). Hence q 0 is isotropic over K, a contradiction. (2) Similar proof for q0 = [1, a] and q 0 = hhX2 , . . . , Tn ii ⊗ [1, a]. (2) Similar proof for q1 = hhaii⊗[1, Tn ] and q 0 = hha, X2 , . . . , Tn−1 ii⊗[1, Tn ]. 3.4 Example This is [Bae82b, Eg. 2.11]. Let k be a field. Suppose char k = 2. Let K = k(T1 , . . . , Tn ) be the function field in n variables. (1) If k is algebraically closed, then u(K) = u b(K) = 2n . (2) If k is C1 such that u(k) = 2 or u(k) = 0 and u b(k) = 2, then u(K) = u b(K) = 2n+1 . Proof. (1) By Lemma 3.3(1), the n-fold Pfister form hhX1 , . . . , Xn−1 ii ⊗ [1, Xn ] is anisotropic and nonsingular over K, then 2n ≤ u(K). By Lemma 1.4(2), u(K) ≤ u b(K). Since k is a C0 -field, by the Tsen-Lang theorem, K is a Cn field. Then u b(K) ≤ 2n . (2) If u(k) = 2, then there exists some anisotropic nonsingular form [c, d], we take a =

d c

∈ k − ℘(k) (resp. u(k) = 0 and u b(k) = 2, by Lemma 1.4(3), k is

not quadratically closed, we take a ∈ k − k 2 ). By Lemma 3.3(2), the (n + 1)fold Pfister form hhX1 , . . . , Xn ii ⊗ [1, a] (resp. hha, T1 , . . . , Tn−1 ii ⊗ [1, Tn ] ) is anisotropic and nonsingular over K, then 2n+1 ≤ u(K). By Lemma 1.4(2), u(K) ≤ u b(K). Since k is a C1 -field, by the Tsen-Lang theorem, K is a Cn+1 field. Then u b(K) ≤ 2n+1 . 3.5 Definition (1) Suppose char K 6= 2. Let ν(K) = min{n | I n Wq (K) is torsion-free.}. (2) Suppose char K = 2. Let ν(K) = min{n | I n+1 Wq (K) = 0}

16

ZHENGYAO WU (吴正尧)

= sup{n | ∃n-fold anisotropic Pfister form over K.}. 3.6 Theorem (1) If L/K is finite separable, then ν(K) ≤ ν(L) ≤ ν(K) + 1. (2) There exists L/K finite separable such that ν(L) = ν(K) + 1. (3) If L/K is finite purely inseparable, then ν(K) = ν(L). We omit its proof since we care more about u and u b, see [AB89]. 3.7 Example This is [MMW91, Prop. 2]. Let k be a field. Suppose char k = 2. Let K/k be a finitely generated field extension such that tr. deg.(K/k) = n. (1) If k is algebraically closed, then u(K) = u b(K) = [K : K 2 ] = 2n . (2) If k is finite, then u(K) = u b(K) = 2[K : K 2 ] = 2n+1 . Proof. We first show that [K : K 2 ] = 2n . Let (X1 , . . . , Xn ) be a transcendence basis of K/k. Let L = k(X1 , . . . , Xn ). It follows from [Bou81, Ch. V, § 14, no. 7, Prop. 17] that K/L is finite. By Lemma 1.9, [K : K 2 ] = [L : L2 ] = 2n . (1) If k is algebraically closed, then by Lemma 3.3(1), hhX1 , . . . , Xn−1 , Xn ]] is anisotropic over L. Then 2n ≤ 2ν(L) , ν(K)

≤ 2

,

since ν(L) ≥ n. by Theorem 3.6 ν(L) ≤ ν(K).

≤ u(K),

by Definition 3.5.

≤ u b(K),

by Lemma 1.6(1).

n

≤ 2 ,

by Tsen-Lang theorem and k is C0 .

(2) If k is finite, then u(k) = 2. By Lemma 3.3(2), hhX1 , . . . , Xn−1 , Xn , a]] is anisotropic over L for a ∈ k − ℘(k). Then 2n+1 ≤ 2ν(L) , ν(K)

≤ 2

,

since ν(L) ≥ n + 1. by Theorem 3.6 ν(L) ≤ ν(K).

≤ u(K),

by Definition 3.5.

≤ u b(K),

by Lemma 1.6(1).

n+1

≤ 2

,

by Tsen-Lang, Chevalley-Warning theorems.

3.8 Definition A field extension K/k is separably generated if tr. deg.(K/k) = n and there

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

17

exists a transcendence basis (X1 , . . . , Xn ) of K/k such that K is separable and algebraic over k(X1 , . . . , Xn ). 3.9 Example This is [MMW91, Prop. 3]. Let k be an algebraically closed field. Suppose char k = 2. Let K/k be a separably generated extension with tr. deg.(K/k) = n. Let X be an indeterminate. Then u(K(X)) = u b(K(X)) = 2[K : K 2 ] = 2n+1 . Proof. Let (X1 , . . . , Xn ) be a transcendence basis of K/k. Let L = k(X1 , . . . , Xn ). By Lemma 1.13, [K : K 2 ] = [L : L2 ] = 2n . Let a1 , . . . , an be a 2-basis of K/k. Then hha1 , . . . , an ii is anisotropic over K. Since X is an indeterminate, hha1 , . . . , an , X]] is anisotropic over K(X). Then 2n+1 ≤ 2ν(K(X)) ,

since ν(K(X)) ≥ n.

≤ u(K(X)),

by Definition 3.5.

≤ u b(K(X)),

by Lemma 1.6(1).

n+1

≤ 2

,

by Tsen-Lang theorem and k is C0 .

3.10 Open Problem By Corollary 2.16, for all field K such that char K = 2 we have u(K((X))) = u b(K((X))) = 2b u(K). Does u(K(X)) = u b(K(X)) for all K? 3.11 Lemma This is [MMW91, p. 340, line 4]. Suppose char K = 2. Then for every simple algebraic extension L/K, u(K(X)) ≥ u b(L). Proof. Suppose π ∈ K[X] is irreducible such that L ' K[X]/(π). Let E be the (π)-adic completion of K(X). Then u(K(X)) ≥ u(E). By Theorem 2.15, u(E) = u b(E) = 2b u(L).

We have 2b u(K) ≤ u(K(X)),

by Lemma 3.11.

≤ u b(K(X)),

by Lemma 1.4(2).

≤ 2[K(X) : K(X)2 ],

by Lemma 1.7(2).

2

= 4[K : K ] ≤ 4b u(K),

by Lemma 1.6(1).

18

ZHENGYAO WU (吴正尧)

3.12 Lemma This is [Mor84, Lem. 1]. Suppose char K = 2. Let L/K be a cyclic extension and Gal(L/K) has generator s. Take β ∈ L − ℘(L). Let M = L(α) where ℘(α) = α2 + α = β. If there exists γ ∈ L such that β s + β = γ 2 + γ and TrL/K (γ) 6= 0, then M/K is cyclic. (The converse is also true [Alb37, p.197, Th. 3], but is irrelavant to this note.) Proof. Extend s to t : M → M such that t|L = s and t(α) = α + γ. The map t is well-defined since ℘(t(α)) = (α + γ)2 + (α + γ) = (α2 + α) + (γ 2 + γ) = β + (β s + β) = β s . Suppose [L : K] = 2e . Then the order of t in Gal(M/K) is a multiple of 2e . t[L:K] (α) = α + (γ + s(γ) + · · · + s2

e

−1

(γ)) = α + TrL/K (γ) 6= α.

Then the order of t is > 2e and a factor of 2e+1 = | Gal(M/K)|, so it can only be 2e+1 . Thus Gal(M/K) is generated by t and hence cyclic.

3.13 Lemma This is [Mor84, Lem. 2]. Suppose char K = 2 and [K : ℘(K)] = 2. Take β ∈ K−℘(K). Let M = K(α) where ℘(α) = α2 +α = β. Then [M : ℘(M )] = 2. Proof. First, ℘(M ) = K ⊕ ℘(K)α. For all x + yα ∈ M , x, y ∈ K, ℘(x + yα) = x2 + y 2 α2 + x + yα = (x2 + x + y 2 β) + (y 2 + y)α ∈ K ⊕ ℘(K)α. Conversely, ℘(K) ( ℘(K) + F2 β ⊂ ℘(K)(β) = K. We have ℘(K) + F2 β = K since [K : ℘(K) + F2 β] < [K : ℘(K)] = 2. Suppose z + ℘(y)α ∈ K ⊕ ℘(K)α where y, z ∈ K. Suppose z = ℘(w) + nβ for w ∈ K, n ∈ {0, 1}. Suppose y 2 β = ℘(v) + mβ for v ∈ K, m ∈ {0, 1}. If m = n, then ℘(v + w + yα) = ℘(v) + ℘(w) + ℘(y)α + y 2 β = z + ℘(y)α. If m 6= n, then ℘(v + w + (y + m + n)α) = ℘(v) + ℘(w) + ℘(y + m + n)α + (y 2 + m2 + n2 )β = ℘(v) + ℘(w) + ℘(y)α + (y 2 + m + n)β = z + ℘(y)α. Hence ℘(M ) = K ⊕ ℘(K)α.

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

19

Next, the map f : E = K(α) → K/℘(K) x + yα 7→ y + ℘(K), (x, y ∈ K). is a sujective additive group homomorphism with ker(f ) = K ⊕℘(K)α = ℘(M ). Therefore M/℘(M ) ' K/℘(K) and [M : ℘(M )] = [K : ℘(K)] = 2.

3.14 Definition Suppose char K = 2. Let Ks be the separable closure of K. The 2-separable closure K2,s of K is the direct limit of finite separable extensions of K in Ks with 2-power degree. 3.15 Theorem This is [Mor84, Th. 1]. Suppose char K = 2 and α ∈ K − ℘(K). The following are equivalent: (1) If L is an extension of K such that α 6∈ ℘(L), then L = K. (2) [K : ℘(K)] = 2. (3) Every finite separable extension of K with a 2-power degree is cyclic. (4) Gal(K2,s /K) ' Z2 . Proof. (1) implies (2). Since α 6∈ ℘(K), we have [K : ℘(K)] ≥ 2. For all β ∈ K − ℘(K), let L = K(b) where ℘(b) = β. If α + ℘(K) 6= β + ℘(K), then α 6∈ ℘(L). By (1), L = K and hence β ∈ ℘(K), a contradiction. Therefore α + ℘(K) = β + ℘(K) and [K : ℘(K)] = 2. (2) implies (3). Let L/K be a finite separable subextension of K2,s /K with a 2power degree. By the primitive element theorem, L = K(λ). Then the splitting field of λ, i.e. the Galois closure L is finite over L. Suppose [L : K] = 2m . We have a tower of fields K = L0 ⊂ L1 ⊂ · · · ⊂ Lm = L such that [Li : Li−1 ] = 2 for all 1 ≤ i ≤ m. Next, we use induction to show that Li /K is cyclic and [Li : ℘(Li )] = 2. For i = 0, L0 = K and [K : ℘(K)] is given by (2). Suppose it is true for i − 1. Take β ∈ Li−1 − ℘(Li−1 ) such that Li = Li−1 (δ) and ℘(δ) = β. Here (β, δ) can be replaced by (β + ℘(x), δ + x) for all x ∈ Li−1 . It follows from Lemma 3.13 that [Li : ℘(Li )] = 2.

20

ZHENGYAO WU (吴正尧)

We have Li−1 = ℘(Li−1 ) + F2 β by [Li−1 : ℘(Li−1 )] = 2. Suppose Gal(Li−1 /K) is generated by s. Then β s = ℘(γ) + nβ for some γ ∈ Li−1 and n ∈ {0, 1}. Since β s

2i−1

= β, we have n = 1 and hence β s + β = γ 2 + γ. Since Li−1 /K is

separable, TrLi−1 /K 6= 0 (see [Bou81, Ch. V, § 8, no. 2, Cor. to Prop. 1]). Since (β, δ) can be replaced by (β + ℘(x), δ + x) for all x ∈ Li−1 , we may assume that TrLi−1 /K (γ) 6= 0. By Lemma 3.12, we have Li /K is cyclic for all 1 ≤ i ≤ m. In particular, L = Lm is cyclic over K and thus abelian. Then the subgroup Gal(L/L) of Gal(L/K) is normal, or equivalently, L/K is Galois. Therefore L = L and L/K is cyclic. (3) implies (4). K2,s = lim L where L runs through finite separable extensions −→

of K with 2-power degrees. By (3), Gal(L/K) ' Z/2n Z for [L : K] = 2n . Gal(K2,s /K) = lim Gal(L/K) ' lim Z/2n Z ' Z2 . ←−

←−

(4) implies (2) and (1). First, we show (2). If [K : ℘(K)] > 2, then take α1 , α2 ∈ K −℘(K) such that α1 −α2 6∈ ℘(K). Suppose ℘(δi ) = αi for i ∈ {1, 2}. Then Gal(K(δ1 , δ2 )/K) ' Z/2Z × Z/2Z, a contradiction to (4). Next, we show (1). Suppose L/K and α ∈ K − ℘(L). If K ( L, then there exists δ ∈ L − K such that ℘(δ) = α. Let L1 = K(δ). Then [L1 : K] = 2 and by the proof of Lemma 3.13, ℘(L1 ) = K ⊕ ℘(K)δ. Then K ⊂ ℘(L1 ) ⊂ ℘(L), a contradiction. Therefore K = L.

3.16 Lemma This is [Sal77, Th. 3]. Let K be a field of char K = 2. If [K : ℘(K)] < ∞, then Br2 (K) = 0. Proof. It suffices to show that every quarternin algebra (a, b] splits over K. Here (a, b] is generated by 1, i, j, ij + 1 with relations i2 = a, j 2 + j = b, ij + ji = i. We will use the fact that (a, b] is split over K iff its norm form hha, b]] ' hhaii ⊗ hhb]] is isotropic over K If K is perfect, then a ∈ K 2 and hence (a, b] is split. Now suppose K is not perfect and a 6∈ K 2 . Since finite fields are perfect, K is infinite and thus K 2 is infinite. Then K(i)2 /K 2 is an infinite dimensional F2 -vector space. Let (ak )k∈I be representatives of a F2 -basis of K(i)2 /K 2 . Let c1 , . . . , cn be representatives of a F2 -basis of K/℘(K). We have cl +℘(K) = n P 2 c2l + ℘(K) for all 1 ≤ l ≤ n. For ak ∈ K, suppose ak c2l = am kl cm + zk for j=1

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

21

aikl ∈ F2 and zk ∈ ℘(K). Since Mn (F2 ) is finite and (am kl )ml ∈ M2 (F2 ) for all k ∈ I. By the pigeonhole principle, there are two representatives as , at with the same matrix. Suppose a∗ = 1 + as + at . Then a∗ c2m + c2m = zs + zt ∈ ℘(K) for all 1 ≤ m ≤ n and a∗ ∈ K(i)2 − K 2 since as + K 2 6= at + K 2 . Then (a, b] ' (a∗ , b∗ ] for some n P b∗ ∈ K. Suppose (a∗ )2 = a∗ and b∗ = b∗m c2m +z for b∗m ∈ F2 and z ∈ ℘(K). m=1

Finally, since (b∗m )2 c2m a∗ + b∗m c2m = b∗m (c2m a∗ + c2m ) ∈ ℘(K), we have ∗

∗

∗

∗

(a , b ] ' (a , b +

n X

(b∗m )2 c2m a∗ ]

∗

∗

' (a , b +

n X

(b∗m )2 c2m ] = (a∗ , z] ' M2 (K).

m=1

m=1 1

Using [EKM08, Fact 98.14(1)(4)] .

3.17 Theorem This is [Mor84, Th. 3]. For all integer n > 0, there exists a field K, char K = 2 such that u(K) = 2 and u b(K) = 2n (resp. ∞). Proof. Let k0 be an algebraically closed field in characteristic 2. Let k = k0 (X1 , X2 , . . . , Xn ) (resp. k = k0 (X1 , X2 , . . .)). Take α ∈ k − ℘(k). Suppose K/k is a finite separable extension with a 2-power degree such that α 6∈ ℘(K) and is maximal with repect to this property. (1) First, we show that u(K) = 2. Since α ∈ k − ℘(K), the form [1, α] is anisotropic. Then u(K) ≥ 2. Also, by Theorem 3.15, [K : ℘(K)] = 2. Then Br2 (K) = 0 since Lemma 3.16. Then the norm form of (a, ab], i.e. the 2-fold Pfister form hha, ab]] ' [a, b] ⊥ [1, ab] is isotropic and hence hyperbolic. Let [a, b] ⊥ [c, d] be any nonsingular form of dimension 4. Then [a, b] ⊥ [c, d] ⊥ [1, ab] ⊥ [1, cd] ' H4 . Since [1, ab] ⊥ [1, cd] ' [1, ab + cd] ⊥ H (see [EKM08, e.g. 7.23]). Cancel2 H = [0, 0], we have [a, b] ⊥ [c, d] ⊥ [1, ab + cd] ' H3 Since [a, b] ⊥ [c, d] has dimension 4 > 3, it is isotropic. Therefore u(K) ≤ 2. (2) If k = k0 (X1 , X2 , . . . , Xn ) we show that u b(K) = 2n . By Tsen-Lang’s theorem, k is a Cn field. Since K/k is finite, K is also a Cn -field. Then u b(K) ≤ 2n . " 1The quaternion algebra

a, b F

# in [EKM08] corresponds the notation (a, ab].

2Althrough Witt cancellation does not hold in general, we can cancel nonsingular forms

[EKM08, Th. 8.4].

22

ZHENGYAO WU (吴正尧)

Also, [k : k 2 ] = 2n and ℘(k) ( k. By Lemma 1.6(1), u b(K) ≥ [K : K 2 ]. By Lemma 1.13, [K : K 2 ] = [k : k 2 ]. Therefore u(K) ≥ 2n . (3) When k = k0 (X1 , X2 , . . .), similarly u b(K) ≥ [K : K 2 ] = [k : k 2 ] = ∞. 4. What is known See tables after this page.

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

Assumptions for field K,

u(K)

u b(K)

23

References

char K = 2 u(K) = 0

0

[K : K 2 ] [Bae82b,

Rem. 1.2

c)],

Lemma 1.6 Quadratically closed (e.g.

0

1

[Bae82b,

alg. closed)

Rem. 1.2 a)],

Lemma 1.4(3)

Quadratic separable clo2

0

n

2

[Bae82b, Rem. 1.2 e) ], Ex-

n

sure of k, [k : k ] = 2 (e.g.

ample 1.14

k = F2 (X1 , X2 , · · · , Xn )) Quadratic separable clo-

0

∞

[Bae82b, p. 108], Exam-

sure of F2 (X1 , X2 , · · · )

ple 1.14

Prefect and u(K) > 0 (e.g.

2

2

[Bae82b, Rem. 1.2 b)], Ex-

finite)

ample 1.5

Complete discrete valued

2b u(K)

2b u(K)

2b u(K)

3

field

[Bae82b, Th. 1.1], Theorem 2.15

Henselian discrete valued

below

field

[MMW91, Th. 2], Theorem 2.13(2)

k((T1 ))((T2 )) . . . ((Tn ))

n

2 u b(k)

n

2 u b(k)

[Bae82b, Cor. 2.10], Corollary 2.16

k(T1 , . . . , Tn ),

k

alge-

n

2

n

2

braically closed k(T1 , . . . , Tn ),

[Bae82b, Eg. 2.11], Example 3.4(1)

k

is C1 ,

2n+1

2n+1

u(k) = 2 or u(k) = 0 and

[Bae82b, Eg. 2.11], Example 3.4(2)

u b(k) = 2 K/k

finitely

generated

2n

2n

with tr. deg.(K/k) = n, k

[MMW91, Prop. 2], Example 3.7(1)

alg. closed K/k

finitely

generated

2n+1

2n+1

with tr. deg.(K/k) = n, k

[MMW91, Prop. 2], Example 3.7(2)

finite K/k separably generated with tr. deg.(K/k) = n, k

2n

2n

[MMW91, Prop. 3], Example 3.9

alg. closed 2 3max{2b u(K), [K : K 2 ]} ≤ u b(K) ≤ 2b u(K) + [K : K 2 ] − 2[K : K ]

24

ZHENGYAO WU (吴正尧)

Assumptions for field K,

u(K)

u b(K)

References

TBA

6

∞

[MTW91, Th. 4.1]

TBA

6

6

[MTW91, 5.1]

2n

2

m

[MTW91, 5.2]

2n

∞

[MTW91, 5.3]

2

2n

[Mor84,

char K = 2

m

2

≥ 2n ≥ 4, m ≥ n − 1

TBA n ≥ 2 TBA k = k0 (T1 , . . . , Tn ), k0 alg. closed, K/k maximal rela-

Th. 3],

Theo-

rem 3.17

tive to α ∈ k − ℘(K) n ≤ m, k0 alg. closed, k1

=

2n

2m

k0 (t1 , . . . , tn−1 ),

[MMW91, p. 344, Eg.], Example 2.14

L = k1 (s1 , . . . , sm−n−1 ) ⊂ k1 ((X)), K a Henselization of L rel. v such that K = L = k1 , K

= k((X)), u(k) =

12

12

2

u b(k) = 6, [k : k ] = 4 2

n

[K : K ] = 2 TBA

[MMW91, p. 344, Rem. (i)]

12

n

n

[2 , 2 + 4] [MMW91, p. 344, Rem. (i)]

n ≥ 2 TBA

n

2 ·3

[MMW91, p. 344, Rem. (i)] claimed in [Lag15, p. 808]

n≥2

6= 2n − 1

[MMW91, Cor. 2], Corollary 1.8

TBA

6= 5

[MMW91, Cor. 2] need and

could

[Bae82a]

not

obtain

NOTES TO u-INVARIANTS IN CHARACTERISTIC 2

25

References [AB89]

R. Aravire and R. Baeza, The behavior of the ν-invariant of a field of characteristic 2 under finite extensions, Rocky Mountain J. Math. 19 (1989), no. 3, 589–600, Quadratic forms and real algebraic geometry (Corvallis, OR, 1986). MR 1043232 →16

[Alb37]

A. Adrian Albert, Mordern higher algebra, The University of Chicago Press, Chicago Illinois, 1937. →18

[Bae82a]

R. Baeza, Algebras de cuaterniones, formas cuadraticas y un resultado de c. arf., Notas. Soc. Mat. Chile (1982), no. 2, 1–19. →24

[Bae82b]

, Comparing u-invariants of fields of characteristic 2, Bol. Soc. Brasil. Mat. 13 (1982), no. 1, 105–114. MR 692281 →1, →2, →3, →4, →5, →6, →7, →8, →13, →14, →15, →23

[Bou64]

´ ements de math´ N. Bourbaki, El´ ematique. Fasc. XXX. Alg` ebre commutative. Chapitre 5: Entiers. Chapitre 6: Valuations, Actualit´ es Scientifiques et Industrielles, No. 1308, Hermann, Paris, 1964. MR 0194450 →10 ´ ements de math´ , El´ ematique, Masson, Paris, 1981, Alg` ebre. Chapitres 4 ` a 7.

[Bou81]

MR 643362 →3, →16, →20 ´ ements de math´ , El´ ematique. Alg` ebre. Chapitre 10. Alg` ebre homologique,

[Bou07]

Springer-Verlag,

Berlin,

2007,

Reprint of the 1980 original [Masson,

Paris;

MR0610795]. MR 2327161 →10 [EKM08]

Richard Elman, Nikita Karpenko, and Alexander Merkurjev, The algebraic and geometric theory of quadratic forms, American Mathematical Society Colloquium Publications, vol. 56, American Mathematical Society, Providence, RI, 2008. MR 2427530 →14, →21

[Lag15]

Ahmed Laghribi, Quelques invariants de corps de caract´ eristique 2 li´ es au u ˆinvariant, Bull. Sci. Math. 139 (2015), no. 7, 806–828. MR 3407516 →24

[MMW91] P. Mammone, R. Moresi, and A. R. Wadsworth, u-invariants of fields of characteristic 2, Math. Z. 208 (1991), no. 3, 335–347. MR 1134579 →1, →3, →4, →6, →9, →10, →11, →12, →13, →16, →17, →23, →24 [Mor84]

Remo Moresi, On a class of fields admitting only cyclic extensions of prime power degree, Bol. Soc. Brasil. Mat. 15 (1984), no. 1-2, 101–107. MR 794732 →18, →19, →21, →24

[MTW91] P. Mammone, J.-P. Tignol, and A. Wadsworth, Fields of characteristic 2 with prescribed u-invariants, Math. Ann. 290 (1991), no. 1, 109–128. MR 1107665 →24 [Sal77]

David J. Saltman, Splittings of cyclic p-algebras, Proc. Amer. Math. Soc. 62 (1977), no. 2, 223–228. MR 0435044 →20

Department of Mathematics, Shantou University, 243 Daxue Road, Shantou, Guangdong, China 515063 E-mail address: [email protected]