QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2

by

Lawrence Ervin Gosnell

Thesis submitted to the Graduate Faculty of the

Virginia Polytechnic Institute and State University

in partial fulfillment of the requirements for the degree of

APPROVED:

C. W. Patty

C. J. fa~ry

MASTER OF SCIENCE

in

Mathematics

E. A. Brown, Chairman

G. W. /rofts

R. J. tS'aigle

June, 1973

Blacksburg, Virginia

ACKNOWLEDGEMENTS

I would like to thank

under whose direction this thesis was written, for his

advice and encouragement.

ii

TABLE OF CONTENTS

INTRODUCTION . . . . . . . . . . . . . . . . . . . 1

I. QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC :; 2 • • • • • • • • • • • • • • • • • • • • • 4

II. QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2 • • • • • • • • • • • • • • • • 8

III. THE SYMPLECTIC GROUP . . . . . . . . . . . . IV. EQUIVALENCE OF FORMS . . . . . . . . . . . .

BIBLIOGRAPHY . . . . . . . APPENDIX

VITA

ABSTRACT

111

24 26

64

65

71

INTRODUCTION

By a form we shall mean a homogeneous polynomial

with coefficients in a given field. A form is linear if,

for each term of the polynomial, the sum of the powers

of the indeterminants is one. A form is quadratic if

the sum is two. In an n-ic form, the sum of the powers

is n. A form is unary if it involves only one

indeterminant, binary if it involves two indeterminants,

and n-ary if it involves n indeterminants. Thus 3x + y

- 4z is a ternary linear form over the integers.

Similarly, 4x2 + xy - 2y2 and 2xn + x2yn-2 + yn + zn are

respectively binary quadratic and ternary n-ic forms.

Quadratic forms (in fact forms in general) have

been extremely important in the study of number theory.

The two-square, three-square, and four-square problems

are merely questions of the representations of primes 2 2 2 2 2 2 2 2 2 by the forms x + y , x + y + z , and x + y + z + w

respectively. In addition, the properties of binary

quadratic forms over the reals play a large part in the

solutions of these problems. Fermat's Last Theorem

can be stated in terms of forms by saying that the

ternary n-ic form xn + yn ± zn never represents zero

with x, y, z ~ O, n > 2. For more on arithmetic use of

forms see Jones [4].

1

2

Quadratic forms over the integers have been

studied since the time of Gauss. However, as the

twentieth century began the idea of generalization· of many

mathematical concepts became popular. Among the concepts

generalized to arbitrary settings was that of quadratic

forms over the integers. But in generalization problems

arose. With standard definitions, forms over fields of

characteristic 2 had to be ruled out. An authoritative

and often cited source for quadratic forms over fields of

characteristic ~ 2 is the 1937 paper of Ernst Witt [7].

The characteristic 2 case, after some slight adjustments,

turned out to be non-trivial and some new concepts were

developed. One of the most important of these is the

Arf Invariant, first introduced in 1940 by Cahit Arf [l].

In this paper we shall investigate quadratic forms

over fields of characteristic 2. Some sources for back-

ground material on these forms are Artin [2], Dieudonne

[3], and Kaplansky [5]. In Section I we will outline the

generalization of quadratic forms to fields of

characteristic ~ 2. In Section II we complete the necessary

adjustments to make the characteristic 2 case non-trivial.

In passing, we shall mention in Section III some results

on the symplectic group. Finally, in Section IV we shall

investigate the properties of quadratic forms over fields of

3

characteristic 2; in particular, we study conditions for

equivalence of such forms.

I. QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC ~ 2

The extension of quadratic forms over the real

numbers to quadratic forms over arbitrary fields of

characteristic ~ 2 is standardly done in the following

manner. Let V be an n-dimensional vector space over a

field F of characteristic ~ 2. In our discussions we shall

limit ourselves to finite dimensional spaces. We then

define a map ( , ): VxV+F such that for x, y, z £ V and

a, S £ F:

1) (x,y) = (y,x)

2) (ax+ Sy,z) = a(x,z) + S(y,z).

This map is called the symmetric bilinear form on V (or

sometimes just the bilinear form on V). This map is very

much like an inner product on V and is called such by

Artin [2] and Kaplansky [5]. We shall follow their example

in referring to the map ( , ). Based upon this map we

define a second map, the quadratic form, Q:V+F by

Q(x) = (x,x).

It is immediate that for x, y, xi £ V and a, S, a1 £ F,

Q has the following properties:

1) Q(ax) = a2Q(x)

2) Q(ax + Sy) = a2Q(x) + s2Q(y) + 2aS(x,y)

4

5

3)

Rewriting 2), we find

(x,y) = l/2(Q(x + y) - Q(x) - Q(y))

which gives a clear indication of some of the difficulties

to arise in quadratic forms over fields of characteristic

2.

We associate a given quadratic form Q for an n-

dimensional space V with a matrix M = (aij) by choosing a

base for V, say {x1 , ... , xn}' and letting (xi, xj) = aij.

Since ( , ) is symmetric, M will also be symmetric and the

matrix derived in the same way by choosing a different base,

say {y1 , ... , yn} will be equivalent to Min the usual

way. So their determinants will be equal up to square

factors. To bypass this problem, we define the determinant

of V (sometimes called the determinant of Q) to be the

canonical image of the determinant of the matrices defined

above in {O} U (F/F2 ) where F is the multiplicative group

of non-zero elements of F. The determinant of V, written

det V or det Q, is then well-defined. It is also possible

to go in the other direction. If V is an n-dimensional

vector space over a field F of characteristic .J 2 and

M = (aij) any nxn symmetric matrix with entries in F, then

we can define a symmetric bilinear form ( , ) on v by

defining (xi, xj) = aij where {xl, . . . , xn} is a base for

6

V over F. In either case we say V has the matrix M in the

base {x1 , . , xn}. A vector space V with such a map

Q and its associated map ( , ) is called a quadrafic space.

If det V = 0 we say V is singular. Otherwise V is non-

singular.

A linear map from one quadratic space V into another

quadratic space W which preserves the quadratic form is

called a representation. In other words, a linear map

a:V+W is a representation if for all x E V, QV(x) = QW(crx)

where QV is the quadratic form on V and QW the quadratic

form on W. Whenever there is no confusion as to which

quadratic space and quadratic form Q we are considering,

we shall suppress all subscripts. If a : V+W is a one-to-

one representation into (respectively onto), we call a

an isometry into (respectively onto). If there is an

isometry a from V onto W we say V and W are isometric and

write v~w. If V=W and Qv=Qw, the set of isometries of V

onto V form a group under composition of functions, which

we call the orthogonal group of V for a particular form Q.

The structure of this group is of some interest and has

been investigated in detail [6]. A primary use for

isometries is in discussing the equivalence of forms over

a quadratic space V. If Q1 and Q2 are two different forms

over two quadratic spaces V and W respectively, then Q1 is equivalent to Q2 , written Q1~Q2 , if there is an isometry

7

of V onto W. If V=W, this is the same as requiring the

existence of a change of base transformation which

preserves the quadratic form.

In searching for equivalent quadratic forms many

invariants have been found. Among these are:

1) the dimension of the space

2) the range of the form

3) the determinant of the space

4) the spinor norm

5) the Hasse Algebra.

We shall be interested in seeing which of these

invariants carry over to the case where F has characteristic

2. For a more thorough treatment of quadratic forms over

a field of characteristic ~ 2, see chapters 4 and 5 of

O ' Meara [ 6 ] •

II. QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2

Hereafter, any field F will have characteristic 2

unless otherwise specified. Let V be an n-dimensional

vector space over F. Using the standard derivations of

( , ) and Q, the fact that F has characteristic 2 causes

difficulties and ambiguities almost from the beginning.

A quadratic space V is called non-trivial if there are at

least two elements x,y in V such that (x,y) f O. As a

first step to proving the existence of an orthogonal base,

a base {x1 , ... , xn} such that (xi, xj) = 0 for if j,

it is usually necessary to prove the existence of a vector

x in V such that Q(x) f 0. Once this is established,

orthogonalization easily follows. The usual proof goes

as follows:

Theorem: A non-trivial quadratic space V contains

a vector x such that Q(x) f O.

Proof: Suppose Q(x) = O for every x E V. Then

Q(x+y) = O for all x, y E V. So 2(x,y) = Q(x+y) - Q(x)

- Q(y) = o.

Now if V is over a field of characteristic f 2 we have the

implication (x,y) = 0 for all x, y E V and so the form on

8

9

Vis trivial, contrary to hypothesis. But if V is over a

field F of characteristic 2 we can draw no such conclusion,

for, given any o. e: F, 2o. = O. So since (x,y) is in F,

2(x,y) = 0 provides absolutely no information about (x,y)

and we cannot reach the contradiction necessary to prove

the theorem.

It therefore becomes apparent that if we hope to

have a non-ambiguous entity to work with we shall have to

make some adjustments in our definitions of ( , ) and Q.

However, we still wish to retain our relationships with

square symmetric matrices. We therefore pick a base

{x1 , .. . , xn} for V, a symmetric nxn matrix M = (aij)

and make the following definition. For x = o.1x1 + ...

+ o.nxn with o.ie:F, define Q:V+F by

Q( x) = Ei~j o.i o.j aij.

Then for o. e: F and x, y, yi e: V, these properties follow:

1) Q(o.x) = o.2Q(x)

2) Q(x+y) = Q(x) + Q(y) + (x,y) defines a symmetric

bilinear form on V.

Induction yields: n n

Q( 1: yi) = 1: Q(y.) + Ei<J.(yi,yJ.). i=l i=l 1

We also retain for our chosen base {x1 , ... , xn}, the

relationships (xi,xj) = aij for i ~ j and Q(xi) = aii"

10

Now for each x E V,

which implies

Q(x+x) = 4Q(x) = 0

0 = Q(x+x)

= Q(x) + Q(x) + (x,x)

= (x,x).

So (x,x) = 0 for each x E V. This makes the form we have

defined on V what is called an alternate form. Over a

field of characteristic ~ 2 this form would also be skew-

symmetric, i.e. (x,y) = -(y,x). But in a field of

characteristic 2, -1 = 1 and skew-symmetry degenerates

into symmetry.

If (x,x) = 0 for every x E V, it does not necessarily

follow that Q(x) = 0 for every x E V. Let V be a 2-dimen-

sional space with base {x1 ,x2 } and let

M = Then defining things as above (x1 ,x2 ) = Q(x1 ) = Q(x2 ) = 1.

But we have seen that (x1 ,x1 ) must be 0 which shows that

Q(x) and (x,x) are not always the same, as they were in the

case of fields of characteristic ~ 2. Consider the matrix

N of Vin the base {x1 ,x2}.

N = ~ ) . The two matrices we have associated with V are not only

11

different, but also their determinants differ modulo

squares. So in order to avoid confusion, if we speak of a

matrix associated with a form we will mean the matrix of

inner products. Instead of defining the quadratic form

from a matrix we will define the form from a homogeneous

quadratic polynomial. For example, we will consider the

form discussed above as defined by the polynomial x2 + xy

+ y2 rather than the matrix

0 i) We choose the polynomial x2 + xy + y2 since every element

z of V can be written in the form ax1 + Sx2 with a, S £ F

and so 2 2 . Q(z) = a Q(x1 ) + S Q(x2 ) + aS(x1 ,x2 )

= a 2 + aS + s2 .

Substituting x and y for a and S respectively to indicate

that a and S can take on any values in F, we arrive at the 2 2 polynomial x + xy + y . In this way we continue to have

the determinant of the space defined as the determinant of

the matrix of inner products modulo squares, and it is still

an invariant.

We now return to the question of orthogonalization.

If a non-trivial quadratic space with the alternate form

defined above has an orthogonal base, then the matrix of

inner products consists entirely of zeros which implies

12

the form is trivial, which is a contradiction. It is possible

to define non-alternate forms over fields of characteristic

2 so that orthogonalization is still possible. But these

forms closely parallel the characteristic ¥ 2 case and are

of little interest here. Thus, whenever we speak of a

quadratic space over a field of characteristic 2 we shall

mean a space with the alternate form defined above.

As before, a representation is a linear function

with certain preservative properties. There is, however,

some ambiguity in the literature concerning the exact

definition of a representation in the characteristic 2

case. A representation can be defined by:

1) Q(x) = Q(crx) for all x E V

2) (x,y) = (crx,cry) for all x, y E V.

As seen in O'Meara [6], these two definitions are equivalent

over fields of characteristic ¥ 2. But in the case of

characteristic 2, this is not necessarily the case. Consider

a 2-dimensional vector space V over F with a quadratic form

derived from the polynomial xy. That is, V has a base

{x1 ,x2} such that Q(x1 ) = Q(x2 ) = 0 and (x1 ,x2 ) = 1. Then

define a : v~v by

cr(x1 ) = x1 + x2 cr(x2 ) = x2 .

Extending a linearly we get a linear map from V to V. Let

13

axl + ax2 and yxl + ox2 be arbitrary members of v where

a, a, y, o £ F. Then (ax1 + ax2 , yx1 + ox2 ) = ao + Sy.

Also

(cr(axl + ax2),cr(yxl + ox2)) = (a(xl + x2) + ax2,

y(x1 + x2 ) + ox2 )

= ao(xl + x2' x2)

+ Sy(x2 , x1 + x2 )

= ao + Sy.

This implies that (x,y) = (crx,cry) for all x, y £ V. In

contrast, Q(x1 ) = 0 and

Q(crx1 ) = Q(x1 + x2 )

= Q(xl) + Q(x2) + (xl,x2)

= 1.

So the two definitions are not equivalent here.

From the general definition of Q,

Q(x + y) = Q(x) + Q(y) + (x,y)

and

Q(cr(x + y)) = Q(crx) + Q(cry) + (crx,cry).

If Q(z) = Q(crz) for all z £ V, then

Q(x + y) = Q(crx +cry).

Hence (x,y) = (crx,cry) for all x, y £ V. So preservation of

Q implies the preservation of ( , ) while the converse does

not always hold. We shall define a representation as a

linear map between quadratic spaces which preserves the

bilinear form. Properties which hold under this definition

14

will hold under the other definition. In addition, our

definition has the advantage of agreeing (at least in

spirit) with the ideas of representation and isometry in

analysis and topology in which ( , ) is considered as an

inner product.

Now we determine conditions which are necessary and

sufficient, in addition to the preservation of ( , ), for

the preservation of Q. Let V be an n-dimensional quadratic

space with base {x1 , ... , xn}. Then for each x £ V, n

x = E aixi with the ai £ F. Then Q(x) = Q(crx) for all i=l

x £ V if and only if n

Q( E aix.) i=l l

if and only if

n = Q( E ai crxi)

i=l

But a is a representation, so

Ei<jaiaj(xi,xj) = Ei<jaiaj(crxi,crxj).

This implies Q(x) = Q(crx) for all x £ V if and only if for

all ai £ F

n 2 E ai Q(x.) =

i=l l

if and only if for each xi in the base for V

15

Q(xi) = Q(crxi).

And so we have the following theorem:

Theorem 2.1: Let V be an n-dimensional quadratic

space over F with base {x1 , ... , xn}' Wan m-dimensional

quadratic space over F, and cr V+W a representation. Then

Q(x) = Q(crx) for each x E V if and only if Q(xi) = Q(crxi)

for each xi in the base for V.

Since a representation cr is a linear transformation,

if cr maps V to V, then cr can be represented as a transforma-

tion matrix between two matrices of V. We shall call the

determinant of the matrix associated with cr the determinant,

det cr, of cr.

A quadratic space V is said to have an orthogonal

splitting into subspaces vl' . , V m if V = V 1 EB . . .

EB Vm and (Vi,V.) = 0 for i "F j. m J

We write v = v1 .L. . . .l.

Vm or V = J_ v. and define ..L <P vi = {O}. A subspace u i=l J.

of V s:elits V if there is a subspace W of V such that

V = U ~ W. Let v1 , ... , Vm be quadratic spaces with

associated bilinear forms which we represent by ( , )i.

Then V = V1 (;9 •

by de fining ( ,

. )

• (:9 v m VxV+F

m ( E xi, i=l

can be made into a quadratic space

such that for xi' Yi E Vi

m m E yi) = E (xi,yi)i.

i=l i=l

16

Since each element of V can be written uniquely as a sum

of elements of the Vi, this map is well-defined. Also with

the bilinear form V = V1 .J. ••• ..l Vm since (Vi,vj·) = 0

for i F j and the bilinear form is unique. In addition,

if Vi has the matrix Mi, then V has the matrix

M =

M1 • • • 0 • M

2

0 • • • M

..L W , and m ai : Vi+Wi is a representation for each i=l, ... , m,

then a:V+W defined by

a(x) m

= a( E x.) = i=l l

m r a.x.

l l i=l

is a representation. We write a = a 1 .l. ... ..1.. am. If

Let U be a subspace of V. Then we define the orthogonal

complement U* of U to be {x E V:(x,U) = O}. We call V* the

radical of V and write V* = rad V. For a subspace U of V,

rad U = Un U*. Vis said to be regular if rad V = {O}.

For the following let V = v1 + . . . + Vm with

(Vi,Vj) = O for i ~ j.

Lemma 2.2: rad V =rad V1 + ... +rad Vm.

17

Proof: Let x be an element of rad V. Then m

x = I: xi with xi E Vi. For i = 1, ... , m: i=l

since (xj,Vi) = 0 for i ~ j. So

(x1 ,Vi) = (x,V1 ) = 0

since (x, V) = o. Thus each xi E rad vi and so x E rad v1

+ . . . + rad v m . Then

m Now let x = I: xi with each xi

i=l

m (x, V) = ( I: xii/)

i=l m

= ( . E xi, V l + . . . i=l

+ v ) m

E rad vi.

+(x,V). m m

All other terms drop out since (xi,Vj) = O for i ~ j.

But xi E rad Vi so (xi,Vi) = O and thus (x,V) = o. So

x E rad V. Therefore rad V = rad v1 + ... + rad Vm.

Corollary 2.2.1: V is regular if and only if all

the Vi are regular.

Proof: {O} = rad V = rad v1 + ... + rad Vm

implies rad Vi = {O} for each i.

18

Rad Vi = {O} implies

rad V = rad v1 + .•. + rad Vm

= {0} + ..• + {O}

= {0}.

So V is regular if and only if all the Vi are regular.

Lemma 2. 3: If V is regular then V = V1 ..1.. ••• ..L Vm.

Proof: V = V1 + ... + Vm and (Vi,Vj) = 0 for i ~ j,

so it suffices to show that V = V1 $ ..• W Vm. In other

words, we need only show that if

0 = x = x1 + • + x m with xi E Vi, then xi= 0 for each i = 1, .. , m. So

let 0 = x1 + •.. + xm. Then for each i,

0 = (O,Vi)

= (xl + . . + x m' Vi)

= (xi, Vi)

since (xj,Vi) = 0 for i ~ j. Thus xi E rad vi for each

But since Vis regular, Corollary 2.2.1 says xi = 0 for

each i.

i.

We now wish to relate the regularity of a quadratic

space to its matrix of inner products. A matrix Mis non-

singular if and only if the determinant of M is different

from O.

19

Theorem 2.4: Let V be an n-dimensional quadratic

space with base {x1 , ... , xn}. Let M be the matrix of

inner products in this base. Then V is regular if and only

if M is non-singular.

Proof: We shall prove the contrapositive, that V

is not regular if and only if M is singular. Suppose V

is not regular. Then there is an x E V, x ~ O, such that

(x,V) = O. Then

x = cx1x1 +

with cxi E F and not all cxi = 0.

O = (x,x1 )

. + ex x n n

Then for each i,

= cxl(xl,xi) + ... + cxn(xn,xi)

which implies that the rows of Mare linearly dependent.

So M is singular. The converse is proved by reversing the

steps above.

This theorem makes it reasonable to refer to a regular

quadratic space as non-singular and to a non-regular

quadratic space as singular.

Theorem 2.5: If U is a regular subspace of the

quadratic space V, then V = U .J.. U*. If V = U ..L W is

another splitting, then W = U*.

20

Proof: (U,U*) = 0 since (x,U) = 0 for each x E U*.

Un U* =. {O} since rad U = UnU* and rad U =. {O}. It remains

only to show that V = U + U*. Let {u1 , ... , un}.be a

base for U. Then we must find a1 E F and u* E U* such

that

x = a1u1 + ... + anun + u*.

Let M = (aij) be the matrix of inner products for U. Then

calculating (x,ui) for each i, we get the following system

of equations:

al(ul,ul) + . . . + an(un,ul) + (u*,u1 ) = (x~u1 ) . . al(ul,un) + . . . + an(un,un) + (u*,un) = ( X, Un).

But (ui,u*) = 0 for each i, so we reduce to the following

system of n equations inn unknowns a1 , ... , an:

alall + · · · + analn = (ul~x) .

where aij = (u1 ,uj). But U is regular so M = (a1j) is non-

singular and the system has a unique solution. Now if

V = U -1.. W, then (U,W) = 0. So W<U*. But V=U J.. U* implies

the dimension of U* equals the dimension of W. So W = U*.

In passing we also note that if U is a regular subspace of

V, then U** = U.

Suppose there is a subspace U of V such that

V = U@ rad V. Then V = U ~rad V since (U,rad V) = O. If

21

such a splitting exists, we call it a radical splitting.

Unless rad V = V or rad V =·{a}, it is not necessarily

true that U is unique. Also V = U .J.. rad V implies

rad V = rad U ..J.. rad (rad V).

But

rad (rad V) = rad V

since

(rad V, rad V) = (rad V, V) = O.

So

rad V = rad U .J... rad V

and thus rad U = {0}. Therefore U is regular. We now

proceed to use radical splittings to show we can limit our

examination of quadratic spaces to regular spaces.

Lemma 2.6: If V and W are quadratic spaces with V

regular and cr:v~w a representation, then cr is an isometry.

Proof: Let x E ker cr. Then

(x, V) = (crx,crV)

= ( 0 ,crV)

= o. So x E rad V. But V is regular so x = 0 and cr is one-to-

one.

22

Theorem 2.7 (Witt): Let the quadratic spaces V and

v1 have radical splittings U J_ rad V and u1 -L. rad v1 respectively. Then

1) There is a representation a:V+V1 if and only

if there is a representation y:U+U1 .

2) V ~ v1 if and only if U ~ u1 with rad V ~ rad v1 .

Proof: 1) Let y:U+U1 be a representation. Define

T:rad V+rad vl by

T(X) = 0

for all x £ rad V. Then T is a representation and cr ..L T:V+V1 is also a representation.

Let a:V+V1 be a representation. Define y:U+U1 by

cr(x) = y(x) + z

for x £ U, y(x) £ u1 , and z £rad V1 . Then for x, y £ U

(x,y) = (ax,cry)

= (yx + z1 ,yy + z2 )

= (yx,yy) + (yx, z2 ) + (z1 ,yy)

+ (zl,z2)

= (yx,yy).

So y is a representation.

2) Let y:U+U1 and T:rad+V rad v1 be isornetries onto.

Then y ..L T:V+V1 is also an isornetry onto.

23 .

Let cr:V+V1 be an isometry onto. Of necessity, cr carries

rad v onto rad v1 . So rad V ~ rad v1. Now, from part 1)

there is a representation y:U+U1 . But u is regular, so by

Lemma 2.6 y is an isometry. Since rad v ~ rad v1 , y must

be onto and so u ~ u1 .

This theorem does much for us. First, by letting

V = v1 we can see that although the "U" in a radical

splitting may not be unique, it is unique up to isometry.

Second, it allows us to limit our discussions to regular

spaces, which we shall do from this point on. Note that

this does not rule out the possibility of singular sub-

spaces, for a regular space can easily have singular

subspaces.

III. THE SYMPLECTIC GROUP

The set of all isometries of an n-dimensional

quadratic space V onto itself forms a group under

composition of functions. In the case of characteristic

t- 2, the structure on the quadratic space is an

orthogonal geometry and so we call this group the

orthogonal group, On(V), of V. In the characteristic 2

case our quadratic space has a symplectic geometry

structure and hence we call the group of isometries of

V onto itself the symplectic group, Spn(V), of y. As

Artin [2] does a thorough development of the symplectic

group, we shall state the major results here without

proof and then pass on.

Let V be a regular quadratic space. For each

y E v, define cry:v~v by

cry(x) = x + c(x,y)y

where c is a fixed element of F. Then cry is an isometry

onto. If c = o, cr = 1 y v for each y E V. If c t- o, then

cry leaves x fixed if and only if (x,y) = 0. So cry fixes

the elements of (Fy)*. An isometry of the form cry for

some y E V is called a symplectic transvection in the

direction of y.

We now state the following theorems, which

characterize Spn(V).

24

25

Theorem 3.1: Every element of Spn(V) has

determinant 1. The group Spn(V) is generated by the

symplectic transvections.

Theorem 3.2: The center of Spn(V) consists of lv·

These theorems characterize Spn(V) and show that

every isometry from an n-dimensional quadratic space V

onto itself is the composition of isometries of the form

cry(x) = x + c(x,y)y

for some y £ V. Also, if Spn(V) has more than one

element, then Spn(V) is non-commutative.

IV. EQUIVALENCE OF QUADRATIC FORMS

Prior to discussing equivalence of forms, w~ must

develop a little more information about the structure of

quadratic spaces over fields of characteristic 2. Let

V be an n-dimensional quadratic space. We define a

vector x E V to be isotropic if (x,x) = 0 and anisotropic

if (x,x) ~ 0. A space is totally isotropic if (x,y) = 0

for every x, y E V. At times, in the characteristic ~ 2

case, a vector x E V is called isotropic if Q(x) = O.

In this case, since Q(x) = (x,x) the two definitions are

equivalent. In characteristic 2, this is not the case.

Suppose V is a quadratic space in which every vector is

isotropic. If x being isotropic means Q(x) = 0, then for

any x, y E V,

0 = Q(x + y)

= Q(x) + Q(y) + (x,y)

= (x,y).

So V is totally isotropic. But if being isotropic means

(x,x) = O, then every vector in the 2-dimensional quadratic

space with the mat.rix

M = (~ 6) is isotropic. To see this, let {x1 ,x2} be the base for V

in which it has the matrix M; then for ax1 + ax2 an

26

27

arbitrary element of V;

= 0

and (x1 ,x2 ) = 1. So Vis not totally isotropic. So in

the characteristic 2 case, the two possible definitions

are not equivalent. We therefore follow our precedent

and adopt the weaker definition. That is, x E V is

isotropic if (x,x) = 0. Since our form is alternate

(x,x) = 0 for every x E V, and so every vector in a

quadratic space over a field of characteristic 2 is

isotropic.

A space having the matrix

M = is clearly isotropic. A quadratic space which has the

matrix M in some base is called a hyperbolic plane. Thus

all hyperbolic planes are regular, isotropic, and have

determinant 1. Since all hyperbolic planes have the

same matrix, they are all isometric.

Theorem 4.1: Every regular 2-dimensional quadratic

space V is a hyperbolic plane.

Proof: Vis regular so Vis not totally isotropic.

28

and (x1 ,x2 ) = a F O. Otherwise Vis totally isotropic.

Then {x1 ,x2/a} is also a base for V and

= l/a(a)

= 1.

So V has the matrix

M = G 6) in the base {x1 ,x2/a}. So Vis a hyperbolic plane.

A quadratic space V is universal if Q(V) = F. If

H is a hyperbolic plane over a field of characteristic ~

2, then H is universal. We exhibit an example of a non-

universal hyperbolic plane in the following lemma. We

shall make future use of this example.

Lemma 4.2: Let V be a 2-dimensional quadratic space

over z 2 (a) where z2 = {O,l} and a is transcendental over

Z2· Suppose the form Q on V is defined by the polynomial x2 + xy + y2 Then a is not an element of Q(V).

Proof: Suppose there is a z E V so that Q(z) = a.

This means there are x, y E z 2 (a) such that x2 + xy + y2

= a. Then

29

2 since if x = O, then y = a and la e z2 (a) which is impossible.

Similarly y ~ O. Also x I y since 2 x = (y + a)/(x + y).

Now since x, ye z2 (a), x = f (a) grar

h(a) y = JTciT

where f(a), g(a), h(a), j(a) E z2[a] and g(a) ~ 0 ~ j(a).

Then

(r(a)\ 2 + f(a)h(a) + (h(a)) 2 = lit'CiT) g(a)j(a) JTCiT a

Finding a common denominator and letting

f(a) = f(a)j(a)

h(o:) = h(a)g(a)

g(a) = g(a)j(a),

we reduce to the following equation:

f(a) 2 + f(a)h(a) + h(a) 2 = ag(a) 2 .

Also f(a) I h(a) so f(a) 2 ~ h(a) 2 and g(a) 2 ~ 0. So let

P(x) f(x) 2 + f(x)h(x) + h(x) 2 + - 2 = xg( x) .

Then P(x) is a polynomial in z2 [x] with a as a root. We

need only show that P(x) is not the zero polynomial to

arrive at the desired contradiction. We have

30

f(x) n + + f o = x . . . g(x) m + + = x . . . go h(x) k + + ho = x . . .

with f 0, go, ho E z2 . Since these polynomials are over a

field of characteristic 2, squaring them just squares

each term. So f(x) 2 , g(x) 2 , and h(x) 2 have first terms 2n 2m 2k x , x , and x respectively and all powers of x in each

polynomial are even. Then xg(x) 2 has only odd powers of

x, and so f(x)h(x) must cancel all the even powers of

f(x) 2 + h(x) 2 ~ 0. The first term of f(x)h(x) is xn+k.

Suppose n<k. Then

2n<n+k<2k.

So nothing cancels the term x2k and P(x) ~ O. A similar

argument holds if k<n.

This leaves only the case n = k. Then

2n = n+k = 2k

and so the highest even power of P(x) is x2n + xn+k + x2k = x2k

Again P(x) ~ 0. But if P(x) ~ O, then a is algebraic over

z2 , which is a contradiction. Therefore a is not an

element of Q(V).

We have seen that not all hyperbolic planes are

universal. However any forms equivalent to the zero form

and any forms over a perfect field are universal.

31

We now proceed to obtain the closest possible analogue

to an orthogonal base. To do this we first need a lemma.

Lemma 4.3: Every regular quadratic space Vis

split by a hyperbolic plane.

Proof: Let x be any vector of V, x ~ O. Then x is

isotropic. Since Vis regular, there is a y E V such that

(x,y) = a ~ 0. Then H = Fx + Fy

is a regular isotropic 2-dimensional subspace of V. So

H is a hyperbolic plane, and since H is regular, H splits

V by Theorem 2.5.

Theorem 4.4: Every regular n-dimensional quadratic

space V has a splitting V = H1 .1... ••• ...L.Hr where n = 2r and

each Hi is a hyperbolic plane. Furthermore, this splitting

is unique up to isometry.

Proof: Since V is regular, by Lemma 4.3, V is split

by a hyperbolic plane, say H1 . So

V = H1 J... H1 *. Then since Vis regular, Corollary 2.2.1 implies that H1* is regular and of dimension n-2. We then proceed by

induction to derive

V = H1 ..l.. ••• ..l.Hr.

2r = n since at each step, H1* is regular and so split by

a hyperbolic plane or H1* = {O}. Since all hyperbolic

32

planes are isometric, this splitting is unique up to

isometry.

Thus if V is a regular n-dimensional quadratic space,

then V has a base {x1 ,y1 , ... , xr"~} where 2r = n. Then

V has the nxn matrix

M =

0 1 • • • 1 0

0 1 1 0

0

·o 1 0 • . • 1 0

in this base. We call such a base a symplectic base. This

result can easily be extended to regular quadratic spaces

with alternate forms over fields of characteristic ~ 2

(see Kaplansky [5]).

M =

The matrix then assumes the form:

0 1 • • • 0 -1 0

·o 1 0 ... -1 0 .

Note that Theorem 4.4 implies that all regular quadratic

spaces with alternate forms over any field are even-

dimensional. Also every regular quadratic space over a

field of characteristic 2 has determinant 1.

We have now reached the point where we can begin to

consider the equivalence of quadratic forms. Let V and v1 be regular quadratic spaces with forms Q and Q1 respectively.

33

We shall say that Q is equivalent to Q1 , or what will mean

the same thing, V is equivalent to v1 , if there is an isometry

a from V onto v1 such that Q(x) = Q(crx) for each r E v. We

notice that the statement that v is equivalent to vl is the

same thing as the statement that V is isometric to v1 under the stronger definition of isometry discussed in

Section II. If Q is equivalent to Q1 , we write Q~Q1 or

v~v1 . If Q and Q1 are different forms on the same

quadratic space V, then the statement that Q~Q1 is the

same as the statement that there is a change of base for V

so that Q agrees with Q1 with respect to the new base .

That is, let {x 1, . , xn} be a base for V; then

and

If Q~Q1 , we must be able to find a new base {y1 , ... , yn}

for V such that

and if ( , ) and ( , )1 are the bilinear forms associated

with Q and Q1 respectively, then

(yi,yj) = (xi,xj)l.

Let V be a 2-dimensional quadratic space. Then Q on

V is defined by some binary quadratic polynomial, say

34

ax2 + bxy + cy2 with a, b, c E F, b ~ O. This means, if

{xl,x2} is a base for v and axl + ax2 an arbitrary element

of V, then 2 2 Q(axl + ax2) = aa + baa + ca .

Now (x1 ,x2 ) = b and {x1 ,x2/b} is also a base for V. In

fact, since

= l/b(b)

= 1,

{x~2/b} is a symplectic base for V and

Q(x2/b) = (l/b) 2Q(x2 )

= c/b2

= d.

Then if yx1 + ox2/b is an arbitrary element of V1 then

Therefore any binary quadratic form is equivalent to a form

defined by a polynomial with 1 as the coefficient of the

xy term. We shall therefore consider only forms generated 2 2 by polynomials of the form ax + xy + by . We shall

abbreviate this form by [a,b]. The form generated by xy,

we symbolize by [O,O] and call the zero form.

Now suppose V is regular and n-dimensional. Then

V is an orthogonal sum of hyperbolic planes and, since the

35

hyperbolic planes are orthogonal, the form on V is the

orthogonal sum of the forms on the hyperbolic planes.

That is,

where [ai,bi] is the form on Hi and V = H1 _L. . • ..L Hr'

2r = n. Then Q is generated by the polynomial 2 + + 2 + 2 + x2y2 + b2y2

2 + alxl xlyl blyl a2x2 . .

All other terms disappear due to orthogonality.

.

In discussing equivalence of forms, we wish to find

a system of invariants. Since we can have only a

symplectic base for a quadratic space V, the spinor norm

and Hasse Algebra are no longer invariants, as we lack the

orthogonal base prerequisite to the construction of the

necessary Clifford Algebras and Hasse Algebras [6]. From

our definition of equivalence, the dimension of the quadratic

space and the range of the quadratic form are useful

invariants. The determinant of V is also an invariant but

since every regular quadratic space has determinant 1, it

is of no use to us here. So we study an analogue to the

determinant, developed by Cahit Arf [l], called the Arf

Invariant. In doing this study, we follow the development

of Kaplansky [5]. Throughout this development, all forms

36

considered will be regular forms on quadratic spaces over

fields of characteristic 2.

Lemma 4.5: [a,b] J_ [c,d] ~ [c,d] ...L [a,b].

Proof: We have a symplectic base {x1 ,y1 ,x2 ,y2}

for V such that,

and

and

Q(x1 ) = a

Q(yl) = b

Q(x2 ) = c

Q(y2) = d

(xl,yl) = (x2,y2) = 1

Q(u1 ) = c

Q(v1 ) = d

Q(u2 ) = a

Q(v2 ) = b

(u1 ,v1 ) = (u2 ,v2 ) = 1

37

Let

ul = x2

vl = Y2

u2 = xl

v2 = Y1·

Then [a,b] 1... [c,d] "' [c,d] J_[a,b].

From the symmetry of the form, it is clear that

[a,b] "' [b,a]. If necessary, this fact can be proven in

the style of Lemma 4.5. In fact, proof of equivalence

using the definition will follow this method, and hence

will be completed in slightly less detail. Also, whenever

we prove two forms equivalent, we shall implicitly assume

they are forms on the same quadratic space.

Lemma 4.6: [a,b] ..L [c,d]"' [a+ c,d] J_ [a,b + d].

with

Q(ul) = a

Q(vl) = b

Q(u2) = c

Q(v2) = d.

38

Then {u1 + u2 ,v2 ,u1 ,v1 + v2} is also a symplectic base

for V and

= a + c + O

= a + c.

Similarly

Q(v1 + v2) = b + d.

So [a,b] _L [c,d] "' [a + c,d] J_ [a,b + d].

Lemma 4.7: If Q is a regular binary quadratic form

which represents zero non-trivially, then Q"' [O,O].

Proof: Since Q represents zero non-trivially, there

is a non-zero vector u E V such that Q(u) = O. Complete

u to a symplectic base {u,v}. Then (u,v) = 1 and Q(v) = a.

Then {u,w} where w = v + au is also a symplectic base

since

Also

(u,w) = (u,v + au)

= (u,v) + a(u,u)

= 1 + 0

= 1.

Q(w) = Q(v + au)

= Q(v) + a2Q(u) + a(v,u)

39

= o. So Q rv [O,O].

Corollary 4.7.1: [a,O] rv [O,O] rv [O,S] for all

a, a E F.

As a further example of this technique of proving

equivalence, we offer the following lemma:

Lemma 4.8: 2 [a,a] rv [a ,l] for any a E F.

Proof: {u,v} is a symplectic base for V such that

(u,v) = 1 and Q(u) = Q(v) = a. Then {s,t} where

s = au + (a + l)v

and

t = u + v

is also a base for v. (s,t) = (au + (a + l)v, u + v)

= a(u,u) + a(u,v) + (a +

+ (a + l)(v,v)

= 0 + a + (a + 1) + 0

= 1.

So {s,t} is a symplectic base for V.

Q(s) = Q(au + (a + l)v)

l)(v,u)

40

= a2Q(u) + (a + 1) 2Q(v) +

a(a + l)(u,v)

= a2 (a) + (a + 1) 2 (a) + a2 + .a

= a3 + a3 + a + a 2 + a

= 2 a .

Q(t) = Q(u + v)

= Q(u) + Q(v) + (u,v)

=a+a+l

= 1. 2 So [a,a] ~ [a ,l].

group.

Since F is a field, it is also an additive abelian

Let P = {x2 + x : x E F}. Then since F has

characteristic 2,

which implies P is a subgroup of F. What we shall be

interested in is the factor group

F0 = F/P.

Lemma 4.9: If [a,b] ~ [c,d] then ab+ P = cd + P.

Proof: V is a quadratic space with symplectic base

{u,v} such that Q(u) = a, Q(v) = b, and V has a second

symplectic base {s,t} such that Q(s) = c, Q(t) = d.

Suppose

Then the matrix

41

s = au + Sv

t = yu + ov.

represents the transformation from· {u,v} to.{s,t}. Now

1 = (s,t)

= (au + Sv,yu + ov)

= ay(u,u) + ao(u,v) + Sy(v,u) + So(v,v)

= ao + Sy

= det A.

So we know that A can be written as a product of elementary

matrices. So there are three cases we need consider:

Case 1. s = u, t = av. Then

1 = (s,t)

= (u,av)

= a(u,v)

= Cl. •

So t = v.

Case 2. s = v, t = u. Then Q(s)

surely ab = ba.

Case 3. s = u + av, t = v. Then

d = Q(t) = b

and

c = Q(s)

= b, Q(t) = a, and

Then

42

= Q(u + av)

= Q(u) + a2Q(v) + a(u,v) 2 = a + a b + a.

ab + cd = ab + (a + a2b + a)b

= ab + ab + Cl2b2 + ab

= (ab) 2 + ab e: p.

So ab + P = cd + P.

In the binary case, we will call the canonical

image of ab in Fa the Arf Invariant of the form [a,b].

The preceding lemma makes it reasonable to call this an

invariant. We would now like to extend the Arf Invariant

to an arbitrary regular n-dimensional quadratic space in

the following manner. If V is a 2r-dimensional quadratic

space, then a regular quadratic form on V has the form

[a1 , b1 ] J_ • • • .....L [ar, br]. We define the Arf Invariant

of this form to be the canonical image in Fa of a1b1 +

... + arbr. Before proving this to be an invariant, we

reduce our calculations to the 4-dimensional case.

Lemma 4.la: Let V be a regular n-dimensional

quadratic space with two symplectic bases {u1 ,v1 , ... ,

ur,vr} and {s1 ,t1 , ... , sr' tr}. Then it is possible

43

to pass from the first base to the second base by a

sequence of changes each affecting only two pairs of

ui, vi. Such changes we shall call dyadic changes·.

Proof: We shall prove the lemma by induction on r

where 2r = n is the dimension of V.

The case r = 2 is trivial.

Assume the lemma is true for r = k and let V have dimension

2k + 2. Then V = H1 ..L .

base {ui,vi}. Then

with ai,bi E Hi.

Case 1. Suppose for some i,

ai = bi = o.

-1.Hk+l where each Hi has a

Then without loss of generality,

Let W = H1 ...L ••• ..l.Hk. Then s 1 , t 1 E W and we can

complete s 1 , t 1 to a base for W and by induction move that

base to the base for W, {s 1 ,t1 ,u2 ,v2 , ... , uk,vk} by a

series of dyadic changes. Thus we can move from

{ul,vl, ... , uk+l'vk+l} to {sl,tl,u2,v2, .,uk+l'vk+l}

by a series of dyadic changes. Let Z = H2 ...L • • • _L Hk+l.

44

Then Z has bases {u2 ,v2 , ... , uk+l'vk+l} and.{s 2 ,t2 ,

... , sk+l'tk+l} and by induction we can move from one

to the other by a series of dyadic changes. And ·so

the lemma follows.

Case 2. Using the orthogonality of the Hi, we note that,

1 = (s1 ,t1 )

= (al,bl) + · · · + (ak+l'bk+l)

implies that at least one of the (ai,bi) ~ O. We may

assume (a1 ,b1 ) = a ~ O. Write D = H1 i. H2 and define

q = bl + b2.

Suppose (p,q) = o ~ o. Then (p,q/o) = 1 and.{p,q/o}

form a symplectic base for a hyperbolic plane which can

be completed to a symplectic base for D. A dyadic change

is possible from the base {u1 ,v1 ,u2 ,v2} to this new

base.

Then

sl = P + ~3 + · · · + ak+l

tl = q + b3 + . . . + bk+l

and we reduce to Case 1.

Suppose (p,q) = O. Then (a2 ,b2 ) = a and there is a dyadic

change from the base {u1 ,v1 ,u2 ,v2} to the symplectic base

{a1 + a 2 , b1/a, a2/a, b1 + b2 }. There is a third

symplectic base in which a 1 + a 2 and b1 + b2 are elements

45

of the same hyperbolic plane of D. After a dyadic change

to this base, we again reduce to Case 1. And so the lemma

is proved.

Theorem 4.11: The Arf Invariant is an invariant.

Proof: By Lemma 4.10, we need only consider the case

where Vis a 4-dimensional regular quadratic space. Let

V = H1 J_ H2 have a symplectic base· {u1 ,v1 ,u2 ,v2} and let

[a,b] ..L [c,d] be the form on V in this base. So the Arf

Invariant of this form is ab + ed. Let s 1 , t 1 start a

second symplectic base and, as in the preceding lemma,

write

We note that from Lemma 4.9, once we have a symplectic

base for a hyperbolic plane, we may replace it with any

other symplectic base for that hyperbolic plane without

affecting the Arf Invariant. Also if H is a 2-dimensional

regular quadratic space and x, y E H such that (x,y) = 1,

then {x,y} is a symplectic base for H.

Case 1. (a1 ,S1 ) = 1. Then {a1 ,s1} is a symplectic base

for Hi and so we can replace it with {u1 ,v1 }. Suppose

a2 = 0. Then if s2 = O,

sl = ul

tl = vl.

46

We then complete s 1 , t 1 to a base for V by choosing

s2 = u2

t2 = v2.

Then the Arf Invariant is ab + ed. If $2 ~ O, then we

can complete s2 to a base for H2 and then replace s2 by u2 . Then

sl = ul

tl = vl + u2

and we complete sl' tl to a symplectic base for V by

choosing

s2 = u2

t2 = ul + v2.

Then

Q(sl) = a

Q(tl) = b + c

Q(s2) = c

Q(t2) = a + d.

So the Arf Invariant is

a(b + c) + c( a + d) = ab + ac + ca + cd

= ab + ed.

47

and since cx2 I- 0 we can complete it to a symplectic base

for H2 and then replace cx 2 by u2 . Then

t 1 = v1 + yu 2 .

We then complete s 1 , t 1 to a symplectic base for V by

choosing

Then

e = Q(sl) = a + c

f = Q(tl) = b + y2c

g = Q(s2) = c

h Q(t2) 2 + y + b + d = = Y a

and ef + gh is our representative for the Arf Invariant

and

(ab + cd) + (ef + gh) = ab + cd + (a+ c)(b + y2c)

+ c(y2a + y + b + d)

= ab + cd + ab + ay 2c + cb

+ y2c2 + cy2a + cy + cb + cd

= (yc) 2 + ye E P.

48

So (ab + cd) + P = (ef + gh) + P. A symmetric argument

proves the case (a2 ,S 2) = 1.

Case 2. (a1 ,e1 ) = y ;i O, 1. Then (a2 ,e2 ) = 1 + y· ~ O, 1.

So {a1 ,e1/y,a2 ,e2/(l +y)} is a symplectic base for V.

Changing bases as before, we get the symplectic base

{s1 ,t1 ,s2 ,t2 } such that

Then

t 1 = yv1 + (1 + y)v2

s 2 = (1 + y)u1 + yu2

Q(s 1 ) = a + c

Q(tl) = y 2b + (1 + y) 2d

Q(s 2 ) = (1 + y) 2a + y 2c

Q(t2) = b + d.

Here our representative of the Arf Invariant is

(a+ c)(y2b + (1 + y) 2d) + ((1 + y)2a + y2c) (b

= ay2b + a(l + y) 2d + cy2b + c(l + y) 2d + (1 +

ab + (1 + 2 y) ad + y2cb + y2cd 2 + (1 + y) 2cd + (1 + y) 2ab + y2cd = y ab

= ab(y2 + (1 + y)2) + cd(y2 + (1 + y)2)

= (ab + cd)(y2 + (1 + y)2)

+ d) y)2

49

= (ab+ cd)(y2 + 1 + y2 )

= ab + ed.

So the Arf Invariant is invariant.

We now know that, if Vis any n-dimensional regular

quadratic space over a field of characteristic 2 with two

quadratic forms Q and Q1 , then Q ~ Q1 implies Q and Q1

have both the same determinant and the same Arf Invariant.

Over certain fields of characteristic 2, we get a stronger

result. A given form on a quadratic space V is equivalent

to itself no matter what base is chosen for V. So if V

has the form [a,b] in the symplectic base {u,v} and a is

a square in V, then by replacing {u,v} with {u/la,la v}

we get an equivalent .form [l,c] where c = ab. This result

leads to the following lemma and theorem and also gives

us a hint as to the location of non-equivalent forms with

the same Arf Invariant.

Lemma 4.12: If a+ b E P, then [l,a] ~ [l,b].

Proof: a + b E P implies 2 b = a + z + z

for some z E F. Let {u,v} be the symplectic base in which

the space has the form [l,a]. Then'{s,t} is also a base

where

50

s = u

t = zu + v.

Then

(s,t) = (u,zu + v)

= z(u,u) + (u,v)

= 1

and {s,t} is a symplectic base. Hence

Q(s) = 1

Q(t) = z 2Q(u) + Q(v) + z(u,v) 2 + a + z = z

= b.

So [l,a] ~ [l,b].

Theorem 4.13: .If V is a regular n-dimensional

quadratic space with two forms Q = [a1 ,b1 ] _L ••• _L

[ar,br] and Q1 = [c 1 ,d1 J J_ .•• J_ [cr,dr], 2r = n, and

.each ai and ci is a square in F, then Q~Q1 if and only if

their Arf Invariants are equal.

Proof: The necessity is a direct result of Theorem

4.11. For the sufficiency, we rely on the discussion

preceding Lemma 4.12 and repeated use of Lemma 4.6. Then r

[a1 ,b1 J J_ .•. J_[ar,br]~[l, I: a.bi] J_ [O,O] ..l.. i=l J.

•.. J_ [O,O]

and

Since

51

r [c1 ,d1 J J_ .•• J_ [cr,dr],\,[1, E cidi] _L[O,O] J_

i=l ••• _L [O,O].

r E aibi +

i=l

r .E cidi i=l

e: p' r r

then by Lemma 4.12, [l, E aib.J~[l, E c.di]. i=l l. i=l l.

Corollary 4.13.1: Two forms on a regular quadratic

space over a perfect field of characteristic 2 are

equivalent if and only if their Arf Invariants are equal.

Proof: This follows from the theorem since in a

perfect field of characteristic 2, every element is a square.

Every finite field and every algebraically closed

field is perfect, so we now have a set of necessary and

sufficient conditions for equivalence over every finite

or algebraically closed field of characteristic 2. Consider

the perfect field z2 = {O,l}. Then the four possible

regular binary quadratic forms on a 2-dimensional quadratic

space V over z2 are [O,O], [O,l], [l,O], and [1,1]. From

Corollary 4.13.1,

since each has Arf Invariant O. Also

52

[l,l]~[O,O]

since 1 ~ O. Therefore the only regular binary quadratic

forms over z2 up to equivalence are [1,1] and [O,OJ.

Expanding this we see that any regular form on a quadratic

space V over z2 is equivalent to an orthogonal sum of

copies of [1,1] and [O,O]. By Lemma 4.6,

[1,1] ~ [l,l]'V[O,l] _L [l,O]

'V[O,O] _J_ [O,O].

So a regular form over z2 is an orthogonal sum of copies

of [O,O] plus (possibly) one copy of [1,1]. For a

listing of all binary quadratic forms over certain other

finite fields of characteristic 2, see the Appendix.

We now show that the Arf Invariant is not always

sufficient for equivalence of forms. Consider the field

z2 (a) where a transcendental over z2 . We showed in

Lemma 4.2 that the form [1,1] over z2 (a) does not represent

a. That is, there is no vector u in V, a regular binary

quadratic space with base {s,t} over z2 (a), such that

Q(u) = a. The form [1,1] has Arf Invariant 1. The form

[a,l/a] also has Arf Invariant 1. If {u,v} is the

symplectic base for V in which V has the form [a,l/a],

then Q(u) = a. So [l,l]~[a,l/a] over z2(a) since there is

a transformation of {s,t} into another base· {w,z} such

that Q(w) = a. So the Arf Invariant is not always

53

sufficient. Notice that z2 (a) is not perfect since la is not an element of z2 (a).

Returning to the binary case, we notice that over

z2 no form [a,$] with a~ 0 ~ a is equivalent to [O,O].

Over z2 , the only form of this type is [1,1]. It is only

natural to ask when it is possible to extend this result

to other fields of characteristic 2. However, the

answer to our question is "never," as the following

lemmas show.

Lemma 4.14: Let V be a regular quadratic space over

a field F of characteristic 2; suppose there is an a E F

such that a ~ O, 1. Then [a, a + l]~[O,O].

So

Proof: Let {u,v} be a symplectic base for V. Then

Q(u) = a

Q(v) = a + 1

(u,v) = 1.

Q(u + v) = Q(u) + Q(v) + (u,v)

= a + (a + 1) + 1

= o. So by Lemma 4.7, [a,a + 1]~[0,0].

54

We note also that any forms [a,(z2 + z)/a] or

[Sz,(z + l)/aJ are equivalent to [O,O] where a, a E F,

a ~ 0 ~ a, and z E F. We observe that a field F of

characteristic 2 contains an a ~ O, 1 if and only if F

Corollary 4.14.1: If F is a field of characteristic

2, and a, a E F such that a ~ 0 ~ a, then [a,aJ~[O,O] for

all such a, a E F if and only if F = z2 .

Restating the corollary, we get the following.

Corollary 4.14.2: For any field F of characteristic

2 such that F ~ z2 , there are a, a E F, a ~ 0 ~ a such that

[a,aJ~[o,oJ.

We would now like to develop a set of necessary and

sufficient conditions for two quadratic forms over any

field of characteristic 2 to be equivalent. We will also

develop a technique to determine the equivalence of binary

quadratic forms.

Theorem 4.15 (Witt's Cancellation Theorem): If

V = U ..LU* and v2 = W J.. W* and U ~ W, then U* ~ W* 1 .

55

provided U and W are regular and v1 = v2 .

Proof: It is sufficient to prove the theorem in the

case that U and W are hyperbolic planes, since each regular

space is a sum of hyperbolic planes and all hyperbolic

planes are isometric. So let U < V1 and W < V2 be hyperbolic

planes with V1 = V2 . Then

V l = U .l H1 _L • • • .J_ Hn J_ I

and

V 2 = W .l. K1 .l.. • • • J_ Km .J.. J

where I and J are totally isotropic and Hi = Ki for each

i = 1, ... , min (m,n). We claim m = n and I= J. Now

and

V 2 = ( W ...L K1 _L • • • _J_ Km) _J_ J

are radical splittings of v1 and v2 respectively. Then

since V1 = V2 , Theorem 2.7 implies

and

I = J.

Hence 2n + 2 = 2m + 2 and m = n. We have now proved our

claim. But

56

and

with I= J and Hi= Ki for each i = 1, ... , n. So

U* = W*.

Theorem 4.16: Let q_, u2 , w1 , w2 be regular

quadratic spaces such that ul "' wl and for vl = ul _l_ u2

and V2 = W1 _L W2 , V1 = V2 . Let a 1 :u1~w1 be the isometry

which makes u1 "'w1 and let a2 :u2~w2 be the isometry proven

to exist in Theorem 4.15, and suppose v1 "' v2 by a = a 1 J_

a2 • Then u2 "' W2 by cr 2 .

Proof: Since u2 "' W2 it is sufficient to prove

that for all u2 £ u2

Q(u2) = Q(a2u2).

For each x £ V, x can be represented uniquely by

x = u1 + u2

where u1 £ lJ_, and u2 £ u2 . Notice that by picking

appropriate x £ V, we can involve every u2 £ u2 in an

expression of this type. By the definition of cr,

cru1 = cr1u1

Then

57

Q(x) = Q(ul + u2)

= Q(ul) + Q(u2) + (ul,u2)

and

Q(crx) = Q(cru1 + cru2 )

= Q(crlul + cr2u2)

= Q(crlul) + Q(cr2u2) + (crlul,cr2u2) ·

But since V1 'V V2 by cr,

Q(x) = Q(crx)

and since u1 'V w1

Q(ul) = Q(crlul).

So

Q(u2) + (ul,u2) = Q(cr2u2) + (crlul,cr2u2)

= Q(cr2u2 ) + (cru1 ,cru2 ).

But since a is an isometry

(ul,u2) = (cru1 ,cru2 )

and so

Q(u2) = Q(cr2u2)

Hence u2 'V w2.

Corollary 4.16.1: [a,b]'V[c,d] if and only if

[a,b] J.. [c,d]rv[O,O] J_ [O,O].

58

Proof: [a,b]~[c,d] if and only if

[a,b] l. [c,d]~[a,b] J_ [a,b]

~[a + a,b] J_ [a,b + b]

~[O,b] J_ [a,O]

~co,oJ J. co,oJ.

Corollary 4.16.2: Let Q1 = [a1 ,,b1 J _l ••• _L [ar,br]

and Q2 = [c1 ,d1 ] ..L .•• _L [cr,dr]. Then Q1 ~ Q2 if and

only if Q1 ..L Q2 is equivalent to the orthogonal sum of 2r

copies of [O,O].

Proof: This follows by repeated use of Lemma 4.6 and

Corollary 4.16.1.

Theorem 4.17: Let Q1 = [a,b], Q2 = [c,d], and

Q = Q1 ..LQ2 . Then Q is a regular form on a 4-dimensional

quadratic space V. If there are u1 , w1 E V,, u1 F 0 F w1 ,

u1 ~ w1 , such that

Q(u1 ) = Q(w1 ) = Q(u1 + w1 ) = O,

then Q ~ [ 0, 0] J_ [ 0, 0].

Proof: (u1 ,w1 ) = Q(u1 + w1 ) + Q(u1 ) + Q(w1 ) = O·

Now there is a v if (Fu1 + Fw1 ) such that (u1 , v) = a F 0.

Otherwise V is totally isotropic which is impossible. If

59

(v,w1 ) = O, then {u1 ,v1 } where v1 = v/a is a symplectic

base for a hyperbolic plane H1 < V such that (w1 ,H1 ) = 0.

If (v,w1 ) = S ~ O, then.{u1 ,v1 } is still a symplectic base

for H1 , but we must replace w1 with

w * 1 = w1 + Su1/a.

Then

Q(w1*) = Q(wl) + (S/a) 2Q(u1 ) + (S/a)(w1 ,u1 )

= 0

and

Q(w1 * + u1 ) = Q(w1 + (1 + (S/a))u1 )

= Q(w1 ) + (1 + (S/a)) 2Q(u1 )

+ (1 + (S/a))(u1 ,w1 )

= a. So w1* satisfies the hypothesis in conjunction with u1 .

We have shown (u1 ,w1 *) = 0 so

(v1,w1*) = (v/a,wl + Sul/a)

2 = l/a(v,w1 ) + Sia (v,u1 )

= l/a(S) + S/a2 (a)

= o. From this point on, we identify w1* with w1 . * So w1 e: H1 which is also a hyperbolic plane and we complete w1 to a

symplectic base.{w1 ,z} for H1* such that Q(z) = 0 using

60

Lemma 4.7. Similarly we get a base.{u1 ,v*} for H1 such

that Q(v*) = O. Then.{u1 ,v*,w1 ,z} is a symplectic base

for V such that

So [a,b] J_ [c,d]'V[O,O] J_ [O,O].

Corollary 4.17.1: Let Q = [a,b] _l_ [c,d] be a form

on a quadratic space V. Then [a,b]'V[c,d] if there are

u, w E V, u ~ w, u ~ 0 ~ w, such that

Q(u) = Q(w) = Q(u + w) = o.

We will now illustrate the use of Corollary 4.17.1

by verifying the results of Lemma 4.8. Let V be a 2 4-dimensional quadratic space with the form [a,a] J_ [a ,l].

Then V has a base.{u1 ,vpu2 ,v2} such that for a, B, y, o E F

Q(au1 + Sv1 + yu2 + ov2 ) = aa2 + aB + aB 2 + a 2y 2

+ yo + o2

If a = O, then

Q(au1 + Sv1 + yu2 + ov2 ) = aB + yo

and

If a ~ O, let

Then

and

Also

and

Hence

61

Q(x) = a(l) 2 + (1)(0) + a(0) 2 + a2 (1) + ·(l)(a) + a2

= a + a2 + a + a2

= 0

Q(y) = a(a2 ) + (a)(a) + a(a2 ) + a2 (0) + (O)(a) + a2

= a3 + a2 + a3 + a2

= 0.

Q(x + y) = a(a + 1) 2 + (a + l)(a) + a(a) 2

+ a2(1) + (1)(0) + 02

a3 + a + a2 + a + a3 + 2 = a

= o. 2 [a,a]"'[a ,l].

We would now like to characterize the non-equivalent

forms over certain fields F of characteristic 2. If F is

finite, then cr:F~P defined by 2 cr(x) = x + x

is a homomorphism of F onto P with kernel {O,l}. Hence

[F:P] = 2 by La Grange's Theorem and F0 has exactly 2

elements, and so over a finite field there are exactly

2 non-equivalent forms.

62

Lemma 4.18: If Fis a finite field, 1 E P if and k only if F has 2 elements where k is even.

Proof: Every finite field F of characteristic 2

has 2k elements. 1 E P if and only if for some x E F

x2 + x = 1

if and only if

x2 + x + 1 = 0 2 has a solution. x + x + 1 = 0 has a solution if and only

if x3 -1 = 0 has a solution other than 1 if and only if

F contains a primitive cube root of unity if and only if

F has order 2k where k is even.

Corollary 4.18.1: If V is a binary quadratic space k over a finite field F of order 2 where k is odd, then

[O,O] and [1,1] are the nonequivalent forms one V.

When k is even, we have shown that 1 E P and we

cannot get such a characterization. Attempts at further

characterizations in the cases where k is even or F is an

infinite field are recommended. We close with one final

set of necessary and sufficient conditions for

equivalence of binary quadratic forms.

63

Lemma 4.19: Let V be a regular binary quadratic space

over a finite field F with binary quadratic forms Q and Q1 .

Then Q ~ Q1 if and only if either both Q and Q1 represent

zero non-trivially or neither Q nor Q1 represent zero

non-trivially.

Proof: Clearly if a form is equivalent to [O,O],

it represents zero non-trivially. Suppose Q ~ Q1 and Q

represents zero non-trivially. Then by Lemma 4.7 Q ~ [O,O] which implies Q1 ~ [O,O] and Q1 represents zero

non-trivially. If Q does not represent zero non-trivially

and Q1 does, then Q1 ~ [O,O] which implies Q ~ [O,O]

which is a contradiction. Conversely if Q and Q1 both

represent zero non-trivially, then they are both

equivalent to [O,O], and hence equivalent. If neither

represents zero non-trivially, then neither is equivalent

to [O,O]. But V has only 2 non-equivalent forms, so

Q ~ Ql.

BIBLIOGRAPHY

1. Arf, Cahi t, "Unterschungen i.iber quadratische Formen in Korpern der Charakteristik 2 (Teil I)," J. filr reine und angew. Math., Vol. 183 (1940), pp.-148-167 .- --

2. Artin, Emil, Geometric Algebra, Interscience Publishers, Inc., New York, 1957.

3. Dieudonne, J., La geometrie des groupes classiques, Ergebnisse der Mathematik und ihrer Grenzgebiete, Neue Falge, Heft 5, Springer, Berlin, 1955.

4. Jones, Burton, The Arithmetic Theory of Quadratic Forms, Carus Monograph no. 10, The Mathematical Association of America, New York, 1950.

5. Kaplansky, Irving, Linear Algebra and Geometry; ~ Second Course, Allyn and Bacon, Inc., Boston, 1969.

6. O'Meara, O. T., Introduction to Quadratic Forms, Springer, Berlin, 1963.~

7. Witt, Ernst, "Theorie der quadratischen Formen in beliebigen Korpern," J. fur reine und angew. Math., Vol. 176(1937)~ pp:-31-44.

64

APPENDIX

65

Table I

~l

0 0 1

1 1 0

p = {0}

F - F 0 -

Possible Forms (up to symmetry)

[O,O]

[O,l]

[1,1]

66

f:-:1

0 0 0

1 0 0

ae Arf Invariant

0 0

0 0

1 1

67

Table II

F = {O,l,a,b}

+ 0 1 a b 0 1 a b

0 0 1 a b 0 0 0 0 0

1 1 0 b a 1 0 1 a b

a a b 0 1 a 0 a b 1

b b a 1 0 b 0 b 1 a

p = {O,l}

Fa = {P, a + P}

Possible Arf Invariants - O, a

Possible Forms al3 Arf Invariant (uE to s;ymmetr;y)

[O,O] 0 0

[0,1] 0 0

[O,a] 0 0

[O,b] 0 0

[1,1] 1 0

[l,a] a a

[l,b] b a

[a,a] b a

[a,b] 1 0

[b ,b] a a

68

Table III .. ,

F = {O,l,a,b,c,d,e,f}

+ 0 1 a b c d e f 0 1 a b c d e f

0 0 1 a b c d e f 0 0 0 0 0 0 0 0 0

1 1 0 c f a e d b 1 0 1 a b c d e f

a a c 0 d 1 b f e a 0 a b c d e f 1

b b f d 0 e a c 1 b 0 b c d e f 1 a

c c a 1 e 0 f b d c 0 c d e f 1 a b

d d e b a f 0 1 c d 0 d e f 1 a b c

e e d f c b 1 0 a e 0 e f 1 a b c d

f f b e 1 d c a 0 f 0 f 1 a b c d e

P =·{o,a,b,d}

F0 = {P, 1 + P}

Possible Arf Invariants - O, 1

Possible Forms (uE to s;:tmmetr;:t) a.a Arf Invariant

[O,O] 0 0

[O,l] 0 0

[O,a] 0 0

[O,b] 0 0

[O,c] 0 0

[O ,d] 0 0

[O,e] 0 0

69

Table III (cont)

Possible Forms (uE to s~mmetr~) as Arf Inva:riant

[O,f] 0 0

[1,1] 1 1

[l,a] a 0

[l,b] b 0

[l,c] c 1

[l,d] d 0

[l,e] e 1

[l,f] f 1

[a,a] b 0

[a,b] c 1

[a,c] d 0

[a,d] e 1

[a;e] f 1

[a,f] 1 1

[b,b] d 0

[b' c] e 1

[b ,d] f 1

[b ,e] 1 1

[b, f] a 0

[c,c] f 1

[c,d] 1 1

[c ,e] a 0

70

Table III 9cont)

Possible Forms (u;e to s~mmetr~::) a(3 Arf Invariant

[c,f] b 0

[d,d] a 0

[d,e] b 0

[d,f] c 1

[e ,e] c 1

[e, f] d 0

[f ,f] e 1

The vita has been removed from the scanned document

QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2

by

Lawrence Ervin Gosnell

(ABSTRACT)

This thesis is concerned with the study of

quadratic forms over fields of characteristic 2. First,

we consider the extension of quadratic forms to fields

of characteristic I 2. Then we make the adjustments

necessary to make the characteristic 2 case non-trivial,

and investigate the structure of quadratic spaces.

We define equivalence of quadratic forms and

investigate a set of invariants. We develop a set of

necessary and sufficient conditions for equivalence

and give a characterization of non-equivalent forms

over certain finite fields.