contents u bu · 2020. 4. 30. · notes to u-invariants in characteristic 2 zhengyao wu (吴正尧)...

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 1

2. Henselian discrete valued fields 6

3. Fields with prescribed u and u 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 〈〉 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 InWq(K) is generated by n-fold Pfister forms

〈〈a1, . . . , an−1〉〉 ⊗ [1, b] for a1, . . . , an−1 ∈ F ∗ and b ∈ F .

1. The square number vs u and u

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 charK = 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 + by2 is nonsingular.

(6) Every regular quadratic form has shape

[a1, b1] ⊥ · · · ⊥ [an, bn]︸︷︷︸nonsingular part

⊥ [c1] ⊥ · · · ⊥ [cm]︸︷︷︸singular part

, ai, bi ∈ K, cj ∈ K∗

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(K) = sup{dimK V | ∃ quadratic form q : V → K anisotropic. }

1.4 Lemma

(1) If charK 6= 2, then u(K) = u(K).

(2) If charK = 2, then u(K) ≤ u(K).

(3) There exists a field K such that charK = 2 and u(K) < u(K).

Proof. (3) Let K be a quadratically closed field of characteristic 2. Then

u(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 charK = 2. If K is perfect, then either

u(K) = 0 or u(K) = u(K) = 2.

Proof. Suppose u(K) > 0. By Lemma 1.2(5), 2 ≤ u(K). By Lemma 1.4(2),

u(K) ≤ u(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 ∈ Kby Lemma 1.2(6) then u(K) ≤ 2. �

Let K2 be the subfield of square elements of K. We call [K : K2] 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 charK = 2.

(0) If finite, [K : K2] is a power of 2.

(1) u(K) ≥ [K : K2].

(2) If u(K) = 0, then u(K) = [K : K2].

(3) There exists K such that u(K) > 0 and u(K) > [K : K2].

Proof. (0) Since K/K2 is purely inseparable, we may decompose K/K2 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 K2. By Lemma 1.2(6), u(K) ≥ [K : K2].

(2) When u(K) = 0, by Lemma 1.2(2)(5)(6), every anisotropic quadratic form

has shape [c1] ⊥ · · · ⊥ [cm], cj ∈ K. Then u(K) = [K : K2].

(3) Since [1, 1] is anisotropic over F2, u(F2) = 2 > 1 = [F2 : F22]. �

1.7 Lemma

This is [MMW91, Prop. 1, Cor. 1]. Suppose charK = 2. Suppose q = [a1, b1] ⊥· · · ⊥ [an, bn] ⊥ [c1] ⊥ · · · ⊥ [cm] is anisotropic over K.

(1) n+m ≤ [K : K2].

(2) u(K) ≤ 2[K : K2] (and hence u(K) ≤ 2[K : K2]).

(3) If n+m = [K : K2] and n ≥ 1, then u(K) = u(K) = 2[K : K2].

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 K2.

(2) dim(q) = 2n+m ≤ 2(n+m) ≤ 2[K : K2] by (1).

(3) Our goal is to find an anisotropic nonsingular form of dimension 2n+ 2m.

Then 2[K : K2] ≤ u(K) and the rest follows from (2).

Next, we construct such a form. We have a K2-basis d1, d2, . . . , dn, c1, . . . , cm

of K. Let t : K → K2 be the K2-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 : K2]-dimensional quadratic form over K2. 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) ∈ SpanK2{d2, . . . , dn, c1, . . . , cm}, a contradiction to the fact

that q is anisotropic. �


1.8 Corollary

This is [MMW91, Cor. 2]. Suppose charK = 2. Then u(K) 6= 2n − 1 for all

n ≥ 2.

Proof. If not, then u(K) = 2n − 1. By Lemma 1.6(1) and Lemma 1.7(2), we

have 2n−1 − 0.5 ≤ [K : K2] ≤ 2n − 1. By Lemma 1.6(0), [K : K2] = 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 : K2].

It follows from Lemma 1.7(3) that u(K) = 2[K : K2] = 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 : K2] = [k : k2].

Proof. Let (wi)1≤i≤n be a k-basis of K. It follows from the identity (a1w1 +

· · ·+ anwn)2 = a21w

21 + · · ·+ a2

nw2n, (ai ∈ k) that (w2

i )1≤i≤n is a k2-basis of K2.

Then [K : k] = [K2 : k2]. Thus

[K : K2] =[K : k2]

[K2 : k2]=

[K : k][k : k2]

[K2 : k2]= [k : k2].

�

1.10 Corollary

This is [MMW91, Prop. 4(iii)]. Let k be a field. Suppose char k = 2. Let K/k

be an extension such that [K : k] <∞. Then1

2u(k) ≤ u(K) ≤ 2u(k).

Proof.

u(k) ≤ 2[k : k2], by Lemma 1.7(2).

= 2[K : K2], by Lemma 1.9.

≤ 2u(K), by Lemma 1.6(1).

u(K) ≤ 2[K : K2], by Lemma 1.7(2).

= 2[k : k2], by Lemma 1.9.

≤ 2u(k), by Lemma 1.6(1).

�


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 (w2i )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

containing (wi)1≤i≤n. Let F = Spank(w2i )1≤i≤m. Suppose wiwj =

m∑l=1

cijlwl for

cijl ∈ k. Suppose a =m∑i=1

aiw2i and b =

m∑i=1

biw2i .

• a± b =m∑i=1

(ai ± bi)w2i ∈ F ;

• ab =m∑l=1

(m∑

i,j=1

aibjc2ijl

)wl ∈ F ;

• For a ∈ F ∗, 1a

= a( 1a)2 where ( 1

a)2 ∈ E2 ⊂ 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/E2 is

purely inseparable. Then (w2i )1≤i≤m are linearly independent over k. Therefore

its subset (w2i )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 k2, then they are also linearly independent over K2.

Proof. Suppose a2i ∈ K2 (ai ∈ K) such that a2

1e1 + · · · + a2nen = 0. We want

to show that a2i = 0 (i.e. ai = 0) for all 1 ≤ i ≤ n. Suppose E = k(a1, . . . , an).

Let (wi)1≤i≤m be a k-basis of E. Then ai =m∑j=1

bijwj for bij ∈ k. Then

0 =n∑i=1

a2i ei =

n∑i=1

(m∑j=1

bijwj

)2

ei =m∑j=1

(n∑i=1

b2ijei

)w2j

By Lemma 1.11, we haven∑i=1

b2ijei = 0 for all 1 ≤ j ≤ m and hence bij = 0 since

(ei)1≤i≤n are linearly independent over k2. 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


separable and algebraic, then

[K : K2] = [k : k2].

Proof. Let (wi)1≤i≤n be elements of K linearly independent over K2. Then

(wi)1≤i≤n are linearly independent over E2 where E = k(w1, . . . , wn) is subfield

of K such that E/k is finite. Then [K : K2] ≤ [E : E2]. By Lemma 1.9,

[E : E2] = [k : k2]. Hence [K : K2] ≤ [k : k2].

Conversely, let (ei)1≤i≤n be elements of k linearly independent over k2. By

Lemma 1.12, (ei)1≤i≤n are linearly independent over K2. Hence [k : k2] ≤ [K :

K2]. �

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 : k2] = 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 : K2] = [k : k2] = 2n (resp.∞). By Lemma 1.4(3),

u(K) = 0. By Lemma 1.6(2), u(K) = 2n (resp.∞).

2. Henselian discrete valued fields

Let (K, v) be a Henselian discrete valued field with valuation ring A, uni-

formizer π and residue field K = A/πA.

2.1 Lemma

If f(T ) = a2T2 + a1T + a0 is irreducible in K[T ], then 2v(a1) ≥ v(a0) + v(a2).

Proof. Suppose ai = uiπni for i ∈ {0, 1, 2}. If 2n1 < n0 + n2, then

g(T ) = πn2−2n1f(πn1−n2T ) = u2T2 + u1T + u0π

n0+n2−2n1 .

Hence g(T ) = T (u2T + 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

charK and charK,


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(a2q(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−1y) = −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))

2≥ 0 + 1

2=

1

2and thus v(bq(x, y)) ≥ 1. �

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

q0(x) = q(x) for all x ∈ M0. Let q1(x) =

(q(x)

π

)for all x ∈ M1. We call q0

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.

Suppose q1(x) = 0 in K for some x ∈ M1. Then v

(q(x)

π

)≥ 1, v(q(x)) ≥ 2,

i.e. x ∈M2. Hence x = 0 in V1 and thus q1 is anisotropic. �


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

π⊥ [u1,

u1

π] ⊥ · · · ⊥ [um,

umπ

]

is anisotropic and nonsingular over K.

Proof. By Lemma 1.4(5), q is nonsingular. If q is isotropic, then q0(s)+q0(t)

π+

m∑j=1

ujx2j +xjyj +

ujπy2j = 0. Suppose s = x0s

′ and t = y0t′ such that x0, y0 ∈ K

and s′ 6≡ 0 modπ, t′ 6≡ 0 modπ. Then ?

πx20q0(s′) + y2

0q0(t′) +m∑j=1

πx2juj + πxjyj + y2

juj = 0

Here xj , yj (0 ≤ j ≤ m) are not all 0 in K. Since q is anisotropic, we must have

min(v(πx20), v(y2

0), v(πx2j), v(y2

j ), v(πxjyj)(1 ≤ j ≤ m)) > 0. Three cases:

(1) min = πx2k for some 0 ≤ k ≤ m;

(2) min = y2k for some 0 ≤ k ≤ m;

(3) min = πxkyk for some 1 ≤ k ≤ m.

(1) Dividing ? by min = πx2k, we have(

x0

xk

)2

q0(s′)+1

π

(y0

xk

)2

q0(t′)+m∑j=1

(xjxk

)2

uj+

(xjxk

)(yjxk

)+

1

π

(yjxk

)2

uj = 0

Here v

(xjxk

)=v(πx2

j)− v(πx2k)

2≥ 0; v

(1

π

(yjxk

)2)

= v(y2j ) − v(πx2

k) ≥ 0

and thus ≥ 1 since v(y2j ) is even and v(πx2

k) is odd.

Modulo π, we have (x0

xk

)2

q0(s′) +m∑j=1

(xjxk

)2

uj = 0.

Since

(x0

x0

)s′ = s′ 6= 0 if k = 0; or

(xkxk

)= 1 if 1 ≤ k ≤ m, we have((

x0

xk

)s′,

(x1

xk

), . . . ,

(xmxk

))6= 0,


a contradiction to the fact that q is anisotropic.

(2) Dividing ? by min = y2k, we have

π

(x0

yk

)2

q0(s′)+

(y0

yk

)2

q0(t′)+m∑j=1

π

(xjyk

)2

uj+π

(xjyk

)(yjyk

)+

(yjyk

)2

uj = 0

Here v

(yjyk

)=

v(y2j )− v(y2

k)

2≥ 0; v

(π

(xjyk

)2)

= v(πx2j) − v(y2

k) ≥ 0

and thus ≥ 1 since v(πx2j) is odd and v(y2

k) is even; v

(π

(xjyk

)(yjyk

))=

1 + v(πx2j) + v(y2

j )

2− v(y2

k) ≥1

2and thus ≥ 1.

Modulo π, we have (y0

yk

)2

q0(t′) +m∑j=1

(yjyk

)2

uj = 0.

Since

(y0

y0

)t′ = t′ 6= 0 if k = 0; or

(ykyk

)= 1 if 1 ≤ k ≤ m, we have((

y0

yk

)t′,

(y1

yk

), . . . ,

(ymyk

))6= 0,

a contradiction to the fact that q is anisotropic.

(3) min = πxkyk. Then

−1 ≥ −v(π) + v(πxkyk)− v(y2k) = v

(xkyk

)= v(πx2

k)− v(πxkyk) ≥ 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

FP = V . The residue defect of P is

rd(P ) = dimF V − dimF (P/MP ).

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.


Proof. (1) Suppose p1, . . . , pk ∈ P such that p1, . . . , pk ∈ P/MP are linearly

independent over F . We show that p1, . . . , pk are linearly independent over F

and thus dimF V ≥ dimF (P/MP ).

If not, then there exists ci ∈ F , 1 ≤ i ≤ k not all zero andk∑i=1

cipi = 0. Suppose

v(cj) = min1≤i≤k

(v(ci)). Then cj 6= 0 and∑i 6=j

cicjpi+pj = 0, where

cicj∈ B. Modulo

M , we have∑i6=j

cicjpi + pj = 0, a contradiction to the fact that p1, . . . , pk are

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 ' Bn. Then FP ' Fn

and P/MP ' P ⊗B B/M ' Bn ⊗B F ' Fn. Hence rd(P ) = dimF V −

dimF (P/MP ) = n− n = 0.

Conversely, suppose (p1, . . . , pk) is a F -basis of P/MP . 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 existsk∑i=1

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

min1≤i≤k

(v(ci)) < 0. Then1

cj∈ M ,

∑i 6=j

cicjpi + pj =

1

cj

k∑i=1

cipi ∈ MP . Modulo M ,

we have∑i 6=j

cicjpi+pj = 0, a contradiction to the fact that p1, . . . , pk are linearly

independent over F . �

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 FQ = F (P ∩W ) = FP ∩W = V ∩W = W and F is

flat over B, we have F (P/Q) ' FP/FQ = 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], TorB1 (B/M,P/Q) = 0. From the

exact sequence 0→ Q→ P → P/Q→ 0, we have another exact sequence

0→ Q/MQ→ P/MP → P/Q

M(P/Q)→ 0.


Therefore

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

P/Q

M(P/Q)

= dimF V − dimF (P/MP ) = rd(P ).

�

2.9 Definition

Let (K, v) be a Henselian discrete valued field with valuation ring A, uni-

formizer 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 charK = 2 and q is totally singular, then

rd(q) ≤ [K : K2]− 2[K : K2].


Proof. Since q(cy) = c2q(y) for all c ∈ K and q is anisotropic, we have

dimK V = dimK2 D(q). Similarly, dimK(M0/πM0) = dimK

2(q(M0)/π2q(M0)).

rd(q) = rd(M0), by Lemma 2.10.

= dimK V − dimK(M0/πM0)

= dimK2 D(q)− dimK

2(q(M0)/π2q(M0)) Since q is a totally singular.

= rd(q(M0)) = rd(A)− rd(A/q(M0)), by Lemma 2.8.

≤ rd(A) = [K : K2]− dimK

2(A/π2A)

= [K : K2]− 2[K : K2], since dimK(A/π2A) = 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 charK = 2, then

rd(q) = rd(Rad(bq)) ≤ [K : K2]− 2[K : K2].

(2) If charK 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 charK = 2. Since q|Rad(bq) is totally singular, by Lemma 2.11,

rd(q) = rd(Rad(bq)) ≤ [K : K2]− 2[K : K2].

(2) Suppose charK 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


valuation ring A, uniformizer π and residue field K = A/πA. Suppose charK =

charK = 2.

(1) u(K) = 2u(K) (analogous to Springer’s theorem for charK 6= 2).

(2) max{2u(K), [K : K2]} ≤ u(K) ≤ 2u(K) + [K : K2]− 2[K : K2].

Proof. First, it follows from Lemma 2.5 that there exists an anisotropic nonsin-

gular quadratic form q over K such that q0 ' q1 ' q and dimK q = 2 dimK q.

So 2u(K) ≤ u(K).

Next, let q be an anisotropic form over K. By Lemma 2.4, q0 and q1 are

anisotropic over K. Then

dimK q = dimK q0 + dimK q1 + rd(q) ≤ 2u(K) + rd(q).

(1) Suppose q is nonsingular and anisotropic. Then rd(q) = 0 and thus u(K) ≤2u(K).

(2) By Theorem 2.12, u(K) ≤ 2u(K) + rd(q) ≤ 2u(K) + [K : K2]− 2[K : K2].

By Lemma 1.4(2), we have u(K) ≤ u(K) and hence 2u(K) ≤ u(K).

By Lemma 1.7(1), we have [K : K2] ≤ u(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) = 2u(k1) = 2n.

2m = [L : L2], since tr.deg.(L/k0) = m.

= [K : K2], by Lemma 1.13.

≤ u(K), by Lemma 1.6(1).

≤ 2m 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 charK = 2. Then u(K) =

u(K) = 2u(K).


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

Since K is complete, we have [K : K2] = 2[K : K2]. By Theorem 2.13(2),

u(K) ≤ 2u(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(K) = 2nu(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

3.1 Review

Cassels-Pfister theorem ([EKM08, Th. 17.3] without assumption on charK).

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 = 〈〈T1, . . . , Tn−1〉〉 ⊗ [1, Tn] is anisotropic and

nonsingular.

(2) Suppose u(k) = 2. If a ∈ k − ℘(k), then the (n + 1)-fold Pfister form

qn = 〈〈T1, . . . , Tn〉〉 ⊗ [1, a] is anisotropic and nonsingular.

(3) Suppose u(k) = 0 and u(k) = 2. If a ∈ k− k2, then the (n+ 1)-fold Pfister

form qn = 〈〈a, T1, . . . , Tn−1〉〉 ⊗ [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]).


(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′ = 〈〈T2, . . . , Tn−1〉〉[1, Tn] is anisotropic over K. If qn is isotropic,

then it is hyperbolic as it is Pfister. Since qn = q′ ⊥ T1q′, we have q′ ' T1q

′.

Then q′ represents T1 over k(T2, . . . , Tn)(T1). By the Cassels-Pfister theorem,

Then q′ represents T1 over k(T2, . . . , Tn)[T1]. Suppose q′(f1, . . . , f2n−1) = T1

where fi ∈ k(T2, . . . , Tn)[T1] for all 1 ≤ i ≤ 2n − 1. If T1|fi for all i, then

T 21 |q′(f1, . . . , f2n−1) = T1, a contradiction. Take T1 = 0, (fi(0))1≤i≤2n−1 are

not all 0. Then q′ is isotropic over k(T2, . . . , Tn). Hence q′ is isotropic over K,

a contradiction.

(2) Similar proof for q0 = [1, a] and q′ = 〈〈X2, . . . , Tn〉〉 ⊗ [1, a].

(2) Similar proof for q1 = 〈〈a〉〉⊗[1, Tn] and q′ = 〈〈a,X2, . . . , Tn−1〉〉⊗[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(K) = 2n.

(2) If k is C1 such that u(k) = 2 or u(k) = 0 and u(k) = 2, then u(K) =

u(K) = 2n+1.

Proof. (1) By Lemma 3.3(1), the n-fold Pfister form 〈〈X1, . . . , Xn−1〉〉⊗ [1, Xn]

is anisotropic and nonsingular over K, then 2n ≤ u(K). By Lemma 1.4(2),

u(K) ≤ u(K). Since k is a C0-field, by the Tsen-Lang theorem, K is a Cn-

field. Then u(K) ≤ 2n.

(2) If u(k) = 2, then there exists some anisotropic nonsingular form [c, d], we

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

not quadratically closed, we take a ∈ k − k2). By Lemma 3.3(2), the (n + 1)-

fold Pfister form 〈〈X1, . . . , Xn〉〉 ⊗ [1, a] (resp. 〈〈a, T1, . . . , Tn−1〉〉 ⊗ [1, Tn] ) is

anisotropic and nonsingular over K, then 2n+1 ≤ u(K). By Lemma 1.4(2),

u(K) ≤ u(K). Since k is a C1-field, by the Tsen-Lang theorem, K is a Cn+1-

field. Then u(K) ≤ 2n+1. �

3.5 Definition

(1) Suppose charK 6= 2. Let ν(K) = min{n | InWq(K) is torsion-free.}.(2) Suppose charK = 2. Let

ν(K) = min{n | In+1Wq(K) = 0}


= 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, 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(K) = [K : K2] = 2n.

(2) If k is finite, then u(K) = u(K) = 2[K : K2] = 2n+1.

Proof. We first show that [K : K2] = 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 : K2] = [L : L2] = 2n.

(1) If k is algebraically closed, then by Lemma 3.3(1), 〈〈X1, . . . , Xn−1, Xn]] is

anisotropic over L. Then

2n ≤ 2ν(L), since ν(L) ≥ n.≤ 2ν(K), by Theorem 3.6 ν(L) ≤ ν(K).

≤ u(K), by Definition 3.5.

≤ u(K), by Lemma 1.6(1).

≤ 2n, by Tsen-Lang theorem and k is C0.

(2) If k is finite, then u(k) = 2. By Lemma 3.3(2), 〈〈X1, . . . , Xn−1, Xn, a]] is

anisotropic over L for a ∈ k − ℘(k). Then

2n+1 ≤ 2ν(L), since ν(L) ≥ n+ 1.

≤ 2ν(K), by Theorem 3.6 ν(L) ≤ ν(K).

≤ u(K), by Definition 3.5.

≤ u(K), by Lemma 1.6(1).

≤ 2n+1, 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


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. LetK/k be a separably generated extension with tr.deg.(K/k) = n.

Let X be an indeterminate. Then u(K(X)) = u(K(X)) = 2[K : K2] = 2n+1.

Proof. Let (X1, . . . , Xn) be a transcendence basis ofK/k. Let L = k(X1, . . . , Xn).

By Lemma 1.13, [K : K2] = [L : L2] = 2n. Let a1, . . . , an be a 2-basis of

K/k. Then 〈〈a1, . . . , an〉〉 is anisotropic over K. Since X is an indeterminate,

〈〈a1, . . . , 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(K(X)), by Lemma 1.6(1).

≤ 2n+1, by Tsen-Lang theorem and k is C0.

�

3.10 Open Problem

By Corollary 2.16, for all field K such that charK = 2 we have

u(K((X))) = u(K((X))) = 2u(K).

Does u(K(X)) = u(K(X)) for all K?

3.11 Lemma

This is [MMW91, p. 340, line 4]. Suppose charK = 2. Then for every simple

algebraic extension L/K, u(K(X)) ≥ u(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(E) = 2u(L). �

We have2u(K) ≤ u(K(X)), by Lemma 3.11.

≤ u(K(X)), by Lemma 1.4(2).

≤ 2[K(X) : K(X)2], by Lemma 1.7(2).

= 4[K : K2]

≤ 4u(K), by Lemma 1.6(1).


3.12 Lemma

This is [Mor84, Lem. 1]. Suppose charK = 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 βs + β = γ2 + 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) + (γ2 + γ) = β + (βs + β) = βs.

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) + · · ·+ s2e−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 charK = 2 and [K : ℘(K)] = 2. Take

β ∈ K−℘(K). LetM = 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 + y2α2 + x+ yα = (x2 + x+ y2β) + (y2 + y)α ∈ K ⊕ ℘(K)α.

Conversely, ℘(K) ( ℘(K) + F2β ⊂ ℘(K)(β) = K. We have ℘(K) + F2β = K

since [K : ℘(K) + F2β] < [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 y2β = ℘(v) +mβ for v ∈ K, m ∈ {0, 1}.If m = n, then

℘(v + w + yα) = ℘(v) + ℘(w) + ℘(y)α+ y2β = z + ℘(y)α.

If m 6= n, then

℘(v + w + (y +m+ n)α) = ℘(v) + ℘(w) + ℘(y +m+ n)α+ (y2 +m2 + n2)β

= ℘(v) + ℘(w) + ℘(y)α+ (y2 +m+ n)β = z + ℘(y)α.

Hence ℘(M) = K ⊕ ℘(K)α.


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 charK = 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 charK = 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 2-

power 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.


We have Li−1 = ℘(Li−1) + F2β by [Li−1 : ℘(Li−1)] = 2. Suppose Gal(Li−1/K)

is generated by s. Then βs = ℘(γ) + nβ for some γ ∈ Li−1 and n ∈ {0, 1}.Since βs

2i−1

= β, we have n = 1 and hence βs + β = γ2 + 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/2nZ for [L : K] = 2n.

Gal(K2,s/K) = lim←−

Gal(L/K) ' lim←−

Z/2nZ ' 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 charK = 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, j2 + j = b,

ij + ji = i. We will use the fact that (a, b] is split over K iff its norm form

〈〈a, b]] ' 〈〈a〉〉 ⊗ 〈〈b]] is isotropic over K

If K is perfect, then a ∈ K2 and hence (a, b] is split.

Now suppose K is not perfect and a 6∈ K2. Since finite fields are perfect, K

is infinite and thus K2 is infinite. Then K(i)2/K2 is an infinite dimensional

F2-vector space. Let (ak)k∈I be representatives of a F2-basis of K(i)2/K2.

Let c1, . . . , cn be representatives of a F2-basis of K/℘(K). We have cl+℘(K) =

c2l + ℘(K) for all 1 ≤ l ≤ n. For ak ∈ K, suppose akc

2l =

n∑j=1

amklc2m + zk for


aikl ∈ F2 and zk ∈ ℘(K). Since Mn(F2) is finite and (amkl)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 + c2

m = zs + zt ∈ ℘(K) for all 1 ≤ m ≤ nand a∗ ∈ K(i)2 −K2 since as +K2 6= at +K2. Then (a, b] ' (a∗, b∗] for some

b∗ ∈ K. Suppose (a∗)2 = a∗ and b∗ =n∑

m=1

b∗mc2m+z for b∗m ∈ F2 and z ∈ ℘(K).

Finally, since (b∗m)2c2ma∗ + b∗mc2

m = b∗m(c2ma∗ + c2

m) ∈ ℘(K), we have

(a∗, b∗] ' (a∗, b∗+n∑

m=1

(b∗m)2c2ma∗] ' (a∗, b∗+

n∑m=1

(b∗m)2c2m] = (a∗, z] 'M2(K).

Using [EKM08, Fact 98.14(1)(4)]1. �

3.17 Theorem

This is [Mor84, Th. 3]. For all integer n > 0, there exists a field K, charK = 2

such that u(K) = 2 and u(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 〈〈a, 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(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(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].


Also, [k : k2] = 2n and ℘(k) ( k. By Lemma 1.6(1), u(K) ≥ [K : K2]. By

Lemma 1.13, [K : K2] = [k : k2]. Therefore u(K) ≥ 2n.

(3) When k = k0(X1, X2, . . .), similarly u(K) ≥ [K : K2] = [k : k2] =∞. �

4. What is known

See tables after this page.


Assumptions for field K,

charK = 2

u(K) u(K) References

u(K) = 0 0 [K : K2] [Bae82b, Rem. 1.2 c)],

Lemma 1.6

Quadratically closed (e.g.

alg. closed)

0 1 [Bae82b, Rem. 1.2 a)],

Lemma 1.4(3)

Quadratic separable clo-

sure of k, [k : k2] = 2n (e.g.

k = F2(X1, X2, · · · , Xn))

0 2n [Bae82b, Rem. 1.2 e) ], Ex-

ample 1.14

Quadratic separable clo-

sure of F2(X1, X2, · · · )0 ∞ [Bae82b, p. 108], Exam-

ple 1.14

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

finite)

2 2 [Bae82b, Rem. 1.2 b)], Ex-

ample 1.5

Complete discrete valued

field

2u(K) 2u(K) [Bae82b, Th. 1.1], Theo-

rem 2.15

Henselian discrete valued

field

2u(K) below3 [MMW91, Th. 2], Theo-

rem 2.13(2)

k((T1))((T2)) . . . ((Tn)) 2nu(k) 2nu(k) [Bae82b, Cor. 2.10], Corol-

lary 2.16

k(T1, . . . , Tn), k alge-

braically closed

2n 2n [Bae82b, Eg. 2.11], Exam-

ple 3.4(1)

k(T1, . . . , Tn), k is C1,

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

u(k) = 2

2n+1 2n+1 [Bae82b, Eg. 2.11], Exam-

ple 3.4(2)

K/k finitely generated

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

alg. closed

2n 2n [MMW91, Prop. 2], Exam-

ple 3.7(1)

K/k finitely generated


finite

2n+1 2n+1 [MMW91, Prop. 2], Exam-

ple 3.7(2)

K/k separably generated


alg. closed

2n 2n [MMW91, Prop. 3], Exam-

ple 3.9

3max{2u(K), [K : K2]} ≤ u(K) ≤ 2u(K) + [K : K2]− 2[K : K2]


Assumptions for field K,

charK = 2

u(K) u(K) References

TBA 6 ∞ [MTW91, Th. 4.1]

TBA 6 6 [MTW91, 5.1]

2m ≥ 2n ≥ 4, m ≥ n − 1

TBA

2n 2m [MTW91, 5.2]

n ≥ 2 TBA 2n ∞ [MTW91, 5.3]

k = k0(T1, . . . , Tn), k0 alg.

closed, K/k maximal rela-

tive to α ∈ k − ℘(K)

2 2n [Mor84, Th. 3], Theo-

rem 3.17

n ≤ m, k0 alg. closed,

k1 = k0(t1, . . . , tn−1),

L = k1(s1, . . . , sm−n−1) ⊂k1((X)), K a Henseliza-

tion of L rel. v such that

K = L = k1,

2n 2m [MMW91, p. 344, Eg.],

Example 2.14

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

u(k) = 6, [k : k2] = 4

12 12 [MMW91, p. 344, Rem.

(i)]

[K : K2] = 2n TBA 12 [2n, 2n + 4] [MMW91, p. 344, Rem.

(i)]

n ≥ 2 TBA 2n · 3 [MMW91, p. 344, Rem.

(i)] claimed in [Lag15, p.

808]

n ≥ 2 6= 2n − 1 [MMW91, Cor. 2], Corol-

lary 1.8

TBA 6= 5 [MMW91, Cor. 2] need

and could not obtain

[Bae82a]


References

[AB89] R. Aravire and R. Baeza, The behavior of the ν-invariant of a field of characteris-

tic 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] N. Bourbaki, Elements de mathematique. Fasc. XXX. Algebre commutative.

Chapitre 5: Entiers. Chapitre 6: Valuations, Actualites Scientifiques et Industrielles,

No. 1308, Hermann, Paris, 1964. MR 0194450 →10

[Bou81] , Elements de mathematique, Masson, Paris, 1981, Algebre. Chapitres 4 a 7.

MR 643362 →3, →16, →20

[Bou07] , Elements de mathematique. Algebre. Chapitre 10. Algebre homologique,

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 geo-

metric theory of quadratic forms, American Mathematical Society Colloquium Pub-

lications, vol. 56, American Mathematical Society, Providence, RI, 2008. MR 2427530

→14, →21

[Lag15] Ahmed Laghribi, Quelques invariants de corps de caracteristique 2 lies au u-

invariant, 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 character-

istic 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 pre-

scribed 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]

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