Quadratic Forms
Marco Schlichting
Notes by Florian Bouyer
26th October 2012
Copyright (C) Bouyer 2012.
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under the terms of the GNU Free Documentation License, Version 1.3
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Contents
1
2
Quadratic forms and homogeneous polynomial of degree 2
3
1.1
4
Free bilinear form modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Orthogonal sum
5
2.1
Witt Cancellation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2
Symmetric Inner Product space over
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
2.3
Witt chain equivalence theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
2.4
Witt Groups:
13
2.5
Second Residue Homomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
2.6
The Brauer Group and the Hasse Invariant
. . . . . . . . . . . . . . . . . . . . . . . . .
26
2.7
Tensor Product of Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
2.8
Quadratic Forms over
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
2.9
Integral quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
R
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
p-adic
numbers
1
10
In this course every ring is commutative with unit. Every module is a left module.
Denition 0.1. Let
M
R be a (commutative) ring and M a (left) R-module. Then a bilinear form on
β : M × M → R which is R-linear in both variables. i.e. β(ax + by, z) = aβ(x, z) + bβ(y, z)
β(x, by + cz) = bβ(x, y) + cβ(x, z) ∀x, y, z ∈ M, a, b, c ∈ R
is a map
and
Example. Standard Euclidean scalar product on
R
Rn . Rn ×Rn → R is a bilinear form on the R-module
n
Denition 0.2. A bilinear form
is called skewed symmetric if
β : M × M → R is called symmetric if β(x, y) = β(y, x) ∀x, y ∈ M . It
β(x, y) = −β(y, x) ∀x, y ∈ M . It is called symplectic if β(x, x) = 0 ∀x ∈ M
Rn is symmetric.
(x1 , y1 ) × (x2 , y2 ) 7→ x1 y2 − x2 y1
Example. Standard scalar product on
On
R
2
the bilinear form
is symplectic and skew-symmetric
Remark. Any symplectic bilinear form is also skew-symmetric because β symplectic ⇒ 0 = β(x+y, x+
y) = β(x, x) + β(x, y) + β(y, x) + β(y, y) = β(x, y) + β(y, x) hence it is skew-symmetric.
If 2 ∈ R is a non-zero divisor (i.e. 2a = 0 ⇒ a = 0 ∀a ∈ R) then any skew-symmetric form is also
symplectic because β skew-symmetric ⇒ β(x, x) = −β(x, x) ⇒ 2β(x, x) = 0 and 2 a non-zero divisors
we can divide by 2 hence β(x, x) = 0 ∀x ∈ M ⇒ β symplectic
Example. For
R = F2 , the form F2 × F2 → F2
β(1, 1) = 1 6= 0 ∈ F2
dened by
x, y 7→ xy
is skewed-symmetric but not
symplectic because
Denition 0.3. A bilinear form
β :M ×M →R
is called regular or non-degenerate or non-singular
if
1.
∀f : M → R
2.
β(x, y) = 0 ∀x ∈ M ⇒ y = 0,
a
R-linear
map
∃x0 , y0 ∈ M
similarly
such that
f (x) = β(x0 , x)
β(x, y) = 0 ∀y ∈ M ⇒ x = 0
r, l : M → M v = HomR (M, R) = {f : M 7→ R : f is R-linear}
β(x, −) : M → R : t 7→ β(x, t) and y 7→ l(y) = β(y, −) : M → R : t 7→ β(t, y).
Remark.
1. says
r, l
are surjective
2. says
r, l
are injective
f (x) = β(x, y0 )
and
In particular, if
M = V
a nite dimensional vector space over a eld
dim V = dim V v
and thus
1.
dened by
x 7→ r(x) =
Then
R = F
then
2. ⇒ 1.
(as
injectivitiy⇒ 2. surjectivity)
(M, β), (M 0 , β 0 ) be bilinear forms. An isometry from M to M 0 is an R-linear
0
0
isomorphism f : M → M such that β(x, y) = β (f (x), f (y)) ∀x, y ∈ M
0
0
Two bilinear forms (M, β), (M , β ) are isometric if there exists an isometry between them
Denition 0.4. Let
Exercise. Check that isometry is an equivalence relation
Check if
(M, β)
regular) if and only if
Denition 0.5. Let
1.
M
be an
isometric then
R-module.
(M, β)
is symmetric (skew-symmetric, symplectic,
A quadratic form on
M
is a function
q:M →R
such that
q(ax) = a2 q(x) ∀a ∈ R, x ∈ M
2. The form
βq
(M 0 , β 0 ) are
(M 0 , β 0 ) is.
and
βq : M × M → R
dened by
βq (x, y) = q(x + y) − q(x) − q(y)
is bilinear
is called the associated symmetric bilinear form.
q : M → R is called regular if βq is regular.
(M, q), (M 0 , q 0 ) be two quadratic forms modules. An isometry
0
0
isomorphism f : M → M such that q(x) = q (f (X)) ∀x ∈ M .
The quadratic form
Let
from
M
to
M0
is an
R-linear
β is a (symmetric) bilinear form then qβ (x) = β(x, x) is a quadratic form because qβ (ax) =
β(ax, ax) = a2 β(x, x) = a2 qβ (x) and βqβ (x, y) = qβ (x + y) − qβ (x) − qβ (y) = β(x + y, x + y) − β(x, x) −
β(y, y) = β(x, y) + β(y, x) is bilinear.
If β is symmetric then βqβ = 2β
Remark. If
2
Corollary 0.6. If 12
∈ R (i.e. 2 ∈ R is a unit) and M is an R-module then {quadratic forms on M } →
{symetric bilinear forms on M } by q 7→ βq is a bijection with inverse {symetric bilinear forms on M } →
{quadratic forms on M } dened by β 7→ 21 qβ
Proof. Exercise
1
Remark. If 2
∈R
then the theory of quadratic forms is the same as the theory of symmetric bilinear
forms.
1
2
But if
∈
/R
then the two theories may dier:
Z × Z → Z dened by x, y 7→ xy does not come from a quadratic
Z because if q : Z → Z is a quadratic form then q(a) = q(a · 1) = a2 q(1) and βq (x, y) =
q(x + y) − q(x) − q(y) = (x + y)2 q(1) − x2 q(1) − y 2 q(1) = 2xyq(1) 6= xy
Example. The symmetric bilinear form
forms on
Objective of this course: Understand classication of quadratic forms (or symmetric bilinear forms)
up to isometry:
How many quadratic forms exists (up to isometry)?
Given two quadratic forms how can I decide when they are isometric
A few applications of quadratic forms:
•
Algebra (quaternion algebras)
•
Manifold theory (as products pairing)
•
Number theory
•
Lattice theory (sphere packing)
1
Quadratic forms and homogeneous polynomial of degree 2
P
f =
ai1 ,...,in xi11 ...xinn ∈ R[x1 , ..., xn ] is called homogenous
i1
in
x1 ...xn (ai1 ,...in 6= 0) has degree i1 + ... + in = m
Denition 1.1. A polynomial
m
if all occurring monomials
x3 + x2 y + z 3 is homogeneous
x + x + zy 2 is not homogeneous.
Example.
3
of degree
of degree 3
2
Every homogeneous degree 2 polynomial
f ∈ R[x1 , ..., xn ]
(r1 , ..., rn ) 7→ f (r1 , ..., rn )
To every polynomial
R[x1 , ..., xn ] →
Remark. In general
one can associate a (polynomial)
Funtions
(Rn , R)
which maps to
Pn
P
ai x2i + i<j bij xi xj .
n
function f¯ : R
→ R by
f ∈ R[x1 , ..., xn ] has the form f =
f 7→ f¯
i=1
is not injective.
(nd
examples!)
But homogenous polynomials in
Claim. If
f ∈ R[x1 , ..., xn ]
Proof. Note: If
n-variables
m → Functions (Rn , R) is injective
n
then f¯ : R → R is a quadratic from
of degree
is homogeneous of degree 2
q1 , q2 are quadratic forms on M then q1 + q2
f = xi xj . Then f¯(r1 , ..., rn ) = ri rj so
and
aq1
are all quadratic forms
∀a ∈ R.
Thus can assume
1.
f¯(ar) = a2 f¯(r) ∀a ∈ R1 , r ∈ Rn
2.
f¯(r + s) − f¯(r) − f¯(s) = (ri + si )(rj + sj ) − ri rj − si sj = ri sj + si rj
Lemma 1.2. For any (commutative!) ring
R,
the map
(
homoegenous polynomials of degree
2 in n variables
dened by
f 7→ f¯ is
is bilinear in
bijective
Proof. Exercise
3
)
→ {quadratic
forms
Rn }
r, s ∈ Rn
RnP→
Rn
n
A = (aij ) ∈ Mn (R) we can dene a ring hoPn
momorphism A∗ : R[x1 , ..., xn ] → R[x1 , ..., xn ] by xj 7→ A∗ (xj ) =
j=1 aij xj which sends homogenous
polynomials of degree m to homogenous polynomials of the degree m.
If
ej 7→
i=1
aij ei
is an
R-linear
map given by a matrix
Denition 1.3. Two homogenous polynomials of degree 2
if
∃A ∈ Mn (R)
invertible with
f, g ∈ R[x1 , ..., xn ] are (linearly) equivalent
A∗ (f ) = g
Lemma 1.4. The map
(
homoegenous polynomials of degree
2 in n variables
)
/linear
equivalence
→ {quadratic
forms on
Rn }/isometry
is bijective
Proof. Exercise
1.1
Free bilinear form modules
R-module (M, β) is called free of rank n (n ∈ N) if M ∼
= Rn
If (M, β) is free of rank n then M has a basis e1 , ..., en and we can dened an associated bilinear form matrix B = (β(ei , ej )). Note that B = (β(ei , ej )) determines β since if x, y ∈ M have
P
P
coordinates (with respect to e1 , ..., en ) x1 , ..., xn , y1 , ..., yn ∈ R i.e. x =
xi ei , y =
yi ei then
P
P
P
β(x, y) = β( xi ei , yj ej ) = xi β(ei , ej )yj = (x1 , ..., xn )B(y1 , ..., yn )T
Denition 1.5. A bilinear
Example.
Standard scalar product on


1
..

B=
Rn
has bilinear form matrix with respect to standard basis
0


.
0
1
Lemma 1.6. Let
bilinear form module of rank
non-degenerate
bilinear form matrix
(M, β) be a free
⇐⇒ the associated
Proof. Recall that
β
is non degenerate if and only if
x
n with basis e1 , . . . , en , then β
B = (β(ei , ej )) ∈ Mn (R) is invertible
r, l : M → HomR (M, R)
is
dened by
7→ r(x) = β(x, −), r(x)(y) = β(x, y)
7→ l(x) = β(−, x), l(x)(y) = β(y, x)
(
are bijective.
M
has basis
e1 , . . . , e n .
Then HomR (M, R)
#
#
has basis e1 , ..., en where
e#
i ej
=
1 i=j
0 i=
6 j
P
e#
xj ej ) = xi . Let (rij ) receptively (lij ) be the n × n of r, l with respect to basis e1 , ..., en of M
i (
Pn
Pn
#
#
and e1 , ..., en of HomR (M, R) i.e.
rkj e#
= r(ej ) and k=1 lkj e#
k=1
k
k = l(ej ) so β(ej , ei )r(ej )(ei ) =
Pn
#
T
= transpose of B . Similarly for (eij ) = B . So, β non
k=1 rkj ek (ei ) = rij ∀i, j ⇒ (rij ) = B
T
degenerated ⇐⇒ r, l are R-linear isomorphism ⇐⇒ (rij ) and (lij ) are invertible ⇐⇒ B , B are
invertible ⇐⇒ B is invertible
i.e.
(M, β), (M 0 , β 0 ) be two free bilinear form modules over R of rank n with basis
M and e01 , ..., e0n for M 0 then (M, β) and (M 0 , β 0 ) are isometric ⇐⇒ associated bilinear
0
0 0 0
form matrices B = (β(ei , ej )) and B = (β (ei , ej )) are congruent. i.e. ∃A ∈ Mn (R) invertible such
T 0
that B = A B A
Lemma 1.7. Let
e1 , ..., en
for
” ⇒ ”: Let f : M → M 0 be an isometry.P Let (fij ) be the associated matrix with respect
n
0
0
to the basis e1 , ..., en and e1 , .., en i.e. f (ej ) =
fkj e0k . Then
⇒ f isomorphism
k=1P
Pnf isometry P
n
0
0
⇒ (fij ) invertible and β(ei , ej ) = β (f (ei ), f (ej )) = β ( k=1 fki e0k , l=1 flj e0l ) =
fki β 0 (e0k , e0l )flj =
T 0
T 0
(A B A)ij . Hence B = A B A
P
” ⇐ ”: A = (fij ) ∈ Mn (R) denes an isomorphism f : M → M 0 by f (ej ) =
fkj e0k such that
0
0
β(ei , ej ) = β (f (ei ), f (ej )) (Calculation as above). Hence f : M → M is an isometry
Proof.
4
Denition 1.8. Let
R
B ∈ Mn (R) we let hBi stand for the bilinear form module (Rn , β), β : Rn ×Rn →
dened by

x , y 7→ β(x, y) = xT By
x1  y1 



xn
 
 
 
yn



e1 , . . . , en with ei
β(ei , ej ) = eTi Bej = Bij
This is a free bilinear form module with basis
associated bilinear form matrix
Remark.
hBi ∼
= hB 0 i
for
B.
Note
B, B 0 ∈ Mn (R) ⇐⇒ ∃A ∈ Mn (R)
has an
1
in the
invertible such that
i-th
position and
B 0 = AT BA
Denition 1.9. The determinant of a non-degenerate free bilinear form module
M = (M, β) with
det M = det(β(ei , ej ))i,j=1,...,n ∈ R∗ /R2∗ . Here R2∗ ⊂ R∗ is the set
∗
2∗
of units which are squares. Note that det M ∈ R /R
does not depend on the choice of basis because if
0
0
0 0
T
e1 , . . . , en is another basis, then (β(ei , ej )) = A (β(ei , ej )A ⇒ det(β(e01 , e0j ) = (det(A))2 det(β(ei , ej ))
basis
e1 , . . . , e n
is the determinant
Example. Recall
•
If
hai
is
R×R→R
∗
h1i = h2i ⇒ deth1i = deth2i ∈ R /R
=1
x, y 7→ axy .
dened by
2∗
⇒2
is a square
⇒ h1i h2i
over
Q
=2
0 1
1
1 0
∗
2∗
∼
⇒ 1 = −1 ∈ R /R ⇒ (−1) is a square ⇒
• If
=
1
0
0
0
1
0
1
0 1
1 0
∼
over C (see below)
over Z, Q, R but
=
1 0
1 0
0 1
Remark. if
(M, β)
is a free of rank
• (M, β)
is non-degenerate
• (M, β)
is symmetric
• (M, β)
is skew-symmetric
• (M, β)
is symplectic
n
with basis
e1 , . . . , e n
and bilinear form matrix
B
0
1
then
⇐⇒ det B ∈ R∗ .
⇐⇒ B = B T
⇐⇒ B T = −B
⇐⇒ B T = −B
and all diagonal entries of
B
are 0
Proof. Exercise.
Lemma 1.10. Let
phism. Then
n≥
R
be (commutative!) ring and f
l. In particular, Rn ∼
= Rl ⇒ n = l
: Rn → Rl
is a surjective
R-module
homomor-
m ⊂ R be a maximal ideal. Then R/m = k is a eld. Reducing f mod m yields a surjective
f : (R/m)n → (R/m)l of nite dimensional k -vector space ⇒ n = dimk (R/m)n ≥ dimk (R/m)l =
Proof. Let
map
l
Remark. Lemma does not hold for non-commutative ring in general. For example let
space of
2
∞
dimension and
A = Endk (V )
then
A⊕A∼
=A
as
A
V
be a
k -vector
module
Orthogonal sum
Denition 2.1. Let (M, β) and (M 0 , β 0 ) be two bilinear form modules. Their orthogonal sum
(M 0 , β 0 ) has underlying module M ⊕ M 0
(x, u), (y, v) 7→ β(x, y) + β 0 (u, v).
Remark.
•
• B ∈ Mn (R), B 0 ∈ Mm (R)
(M, β)
(M 0 , β 0 )
If
and
(M 0 , β 0 )
Denition 2.2. Let
and bilinear form
then
hBi ⊥ hB 0 i =
(M ⊕ M 0 ) × (M ⊕ M 0 ) → R
B
0
0
B0
are regular (symmetric,skew-symmetric,symplectic) then so is
(M, β) be a symmetric
β(y, x) = 0). Let N ⊂ M be a sub-module.
⊥
module N
= {x ∈ M |β(x, y) = 0 ∀y ∈ N }
or skew-symmetric bilinear form (so
The orthogonal complement of
5
(M, β) ⊥
dened by
N
(in
(M, β) ⊥
β(x, y) = 0 ⇒
M ) is the sub-
Lemma 2.3. Let
(M, β) be symmetric or skew-symmetric and N ⊂ M
M = N ⊥ N⊥
a sub-module such that
(N, βN )
is non degenerate. Then
x ∈ N ∩ N ⊥ then β(x, y) = 0 ∀y ∈ N (as x ∈ N ⊥ ).
⊥
is non degenerate we have x = 0. So M ⊃ N + N
= N ⊕ N ⊥.
⊥
check N + N
= M . Let x ∈ M then β(x, −)|N ∈ HomR (N, R)
⇒
Proof. Have to check that
since
x∈N
and
βN
We next need to
N ∩ N ⊥ = 0.
If
But
βN non−degenrate
∃x0 ∈ N : β(x, y) = β(x0 , y) ∀y ∈ N
then
β(x−x0 , y) = 0 ∀y ∈ N ⇒ x−x0 ∈ N ⊥ ⇒ x = x0 +x − x0 .
|{z} | {z }
∈N
N + N⊥ = M ⇒ N ⊕ N⊥ = M
⊥
⊥
⊥
Lastly we need to check that (N, βN ) ⊥ (N , βN ) = (M, β). If x, y ∈ N, u, v ∈ N
⊥
u, y + v) = β(x, y) + β(x, v) + β(v, y) + β(u, v) = βN (x, y) + βN (u, v)
| {z } | {z }
∈N ⊥
Thus
=0
then
β(x +
=0
Corollary 2.4. If (M, β) is a nitely generated symmetric bilinear
hu2 i ⊥ · · · ⊥ huk i ⊥ N where ui ∈ R∗ and β(x, x) ∈ R \ R∗ ∀x ∈ N
form module. Then
M = hu1 i ⊥
M0 = M and if β(x, x) =∈ R \ R∗ ∀x ∈ M0 = M then take N = M0 = M and we are
∗
done. So assume ∃x ∈ M0 : β(x, x) ∈ R then ax 6= 0 ∈ M ∀a ∈ R \ {0} (if ax = 0 ⇒ β(ax, x) =
aβ(x, x) ⇒ a = 0). So Rx ⊂ M is a free module of rank 1 with basis x. Rx has bilinear form matrix
(β(x, x)) ∈ R∗ invertible. So, β|Rx is non degenerate. ⇒ M = |{z}
Rx ⊥ (Rx)⊥ with u1 = β(x, x) ∈ R∗
| {z }
Proof. Set
hu1 i
=:M1
contradicting Lemma 1.10
Repeat with M1 in place of M0 to obtain M = hu1 i ⊥ · · · ⊥ huk i ⊥ Mk . We can repeats as long as
∃x ∈ MK : β(x, x) ∈ R∗ . But the procedure stops because K > n impossible otherwise there exists a
n
k
surjective map R M = hu1 i ⊥ · · · ⊥ hun i ⊥ Mk R ⇒ n ≥ k
|
{z
}
Rk
Remark. If
β(x, x) ∈ R \ R∗ ∀x ∈ N 6= 0 and β
is non-degenerate then
(N, β) cannot has an orthogonal
basis
Proof. If
N = hu1 i ⊥ · · · ⊥ hun i
are non-degenerate
⇒ ui ∈ R∗ .
N is( non-degenerate with respect to base e1 , . . . , en
ui i = j
∗
β(ei , ej ) =
in particular β(e1 , e1 ) = u1 ∈ R
0 i 6= j
and if
then
hui i
Theorem 2.5 (Existence of orthogonal basis over elds of char
6= 2). Let k be a eld of char 6= 2
(M, β) a nite dimensional symmetric bilinear form. Then M = hu1 i ⊥ · · · ⊥ hul i ⊥ N such that
β|N = 0 (β(x, y) = 0
∀x, y ∈ N ). In particular (M, β) has an orthogonal basis, e1 , . . . , el , el+1 , . . . en

ui j = i = 1, . . . l
such that β(ei , ej ) =
0 j = i = l + 1, . . . n


0 j 6= i
and
Proof. From corollary we have
(R = k) M = hu1 i ⊥ · · · ⊥ hul i ⊥ N
with
ui ∈ R∗ , β(x, x) ∈ R \ R∗ ∀x ∈
k\k∗ =0
N.
Need to show
β|N = 0. β(x, x) = 0 ∀x ∈ N ⇒ associated quadratic form q(x) = β(x, x) = 0
β(x, y) = 21 (q(x + y) − q(x) − q(y)) = 0
Want to generalize theorem on existence of orthogonal basis to rings.
Denition 2.6. A ring
Note that
Notation.
R/m
(R, m, k)
m⊂R
is a local ring if
Proof. Need to show
We see that
is called local if it has a unique maximal ideal
is a eld as
Remark. In a local ring
•
R
(R, m, k)
is a maximal ideal.
k = R/m, m ⊂ R
we have
is the maximal ideal
∗
R =R\m
m = R \ R∗ .
m ∩ R∗ = ∅
because
m.
m ( R ⇒ m ⊂ R \ R∗ .
6
⇒
char6=2
a ∈ R \ R∗ , then (a) = Ra ( R (proper ideal because Ra = R ⇒ ∃b : ba = 1 contradicting
a∈
/ R∗ ) Every proper ideal is contained in a maximal ideal ⇒ Ra ⊂ m ⇒ a ∈ m ⇒ R \ R∗ ⊂ m
•
If
(R, m, k) local then A ∈ Mn (R) is invertible if and only if A mod m ∈ Mn (k) is invertible because
A ∈ Mn (R) invertible ⇐⇒ det A ∈ R∗ = R \ m ⇐⇒ det(A mod m) 6= 0 ∈ k = R/m ⇐⇒ A
mod m ∈ Mn (k) is invertible.
Example.
•
Fields are local rings with
m=0
• Z(p) = { ab ∈ Q : a, b ∈ Z, p - b} where p ∈ Z
m = { ab ∈ Q : p - b, p | a}. Then Z(p) /m = Fp
• k
eld
k[T ]/T n
is prime. This is a local ring with maximal ideal
is a local ring with maximal ideal
Denition 2.7. A nitely generated
Rn , n ∈ N
i.e.
R-module P
∃R-module N : M ⊕ N ∼
= Rn
Theorem 2.8. Let
M∼
= Rl
for some
(R, m, k)
l∈N
f
∃N : M ⊕N ∼
= Rn .
M
and let
p be the composition of all these maps.
projective so
is called projective if it is a direct factor of some
M
be a local ring and
Proof.
←
Let
(T )
a nitely generated projective
R-module,
f
p : Rn → Rn be the linear map Rn → M ⊕ N
Note that
then
f −1
q
→
x,y7→(x,0)
M ⊕ N → Rn
p2 = (f −1 qf )2 = f −1 qf f −1 qf = f −1 q 2 f =
|{z}
=q
f
→
p
and
im p ∼
=M
←
g
∃ basis of Rn
1l 0
p=U
U −1 .
0 0
Claim.
with respect to which
Claim is true over a eld (i.e.
kl ∼
= M/mM , k r ∼
= N/mN
→
mod m).
has the form
1l
0
0
0
i.e.
∃U ∈ Mn (R)
such that
1l 0
= Rl
0 0
U
M/mM and N/mN are nite dimensional k -vector spaces
Note that claim implies theorem because
so
p
∼
=
im p ← im
→
g1
g2


1
 l
kl ⊕ kr
0
0
0
g1−1 ⊕g2−1
g1 ⊕g2
M/mM ⊕ N/mN
f
Then we get
1l
0
q
mod m ∼
=
(R/m)n
/ kl ⊕ kr
O
/ M/m ⊕ N/mN
O
f −1
p
/ (R/m)n
0
1
= A−1 pf (g1 ⊕ g2 ), p mod m = A l
0
0
| {z }
0
A−1 ,
0
now lift all entries of
A
to
A∈Mn (k)
entries of
R
(under the surjective
R/m → k )
to obtain a matrix
S ∈ Mn (R)
A = S
0
mod m so
0
such that
1l
0
mod m. A ∈ Mn (k) invertible ⇒ S ∈ Mn (R) invertible. S −1 pS mod m =
T B
−1
S pS =
with T = 1l mod m and B, C, D = 0 mod m. ⇒ after base change (given
C D
T B
by S ) p becomes
as above. Idea: Want to perform row and column operation to make
C D
7
1l 0
2
after base changes. Now, T = 1 mod m ⇒ T ∈ Mn (R) invertible. p =
0
0
2
T B
T B
T B
T + BC ∗
2
−1
p ⇒
=
=
. T = T + BC ⇒ 1 = T + T
BC .
C D
C D
C D
∗
∗
−1
−1
T B
1 ∗
1 T B
1 −T B
1 B
⇒after bases change p becomes
, B, C, D =
=
C D
∗ ∗
0
1
0
1
C D
{z
}
{z
}
|
|
X
X −1
1
B
1 B
1 B
2
0 mod m. p = p ⇒
=
⇒ BC = 0, BD = 0, DC = 0, D = CB + D2 .
C D
C D
C D
3
2
3
2
2
⇒ D2 = DCB
| {z } + D ⇒ D = D ⇒ D (1 − D) = 0 ⇒ D = 0 ⇒ D = CB . The second to last implicaT
C
B
D
into
0
1
0
1
tion is because 1 − D is invertible as D ≡ 0 mod m.
−C 1
C
| {z }
Y −1
1 0
1 B
1 B
1 −B
=
an other base change gives
0 0
0 1
0 0
0
1
| {z }
| {z }
R
Example.
R=Z
1 0
1
=
C 1
0
| {z }
B
0
Then
Y
Z −1
Z
Remark. If
B
CB
is a Euclidean domain (e.g.
R = Z)
every nitely generated
is nitely generated projective over
then
Z
then every nitely projective
R-module
is free.
Z-module is isomorphic to Zn ⊕nite abelian group. If P
P ⊆ Zm ⇒ P has no element of nite order. ⇒ P = Z n ⊕ 6
nite
Remark. For a general commutative ring, projective
R-modules
may not be free
√
R = Z[ −5] = Z[T ]/(T 2 + 5) . Fact: R is a Dedekind
domain,
√
Dedekind domain is a projective R-module. Let I = (2, 1 +
−5) ⊂ R. From
If I was free then I ∼
= Rn , n 6= 0 because I 6= 0. Let's compute R/I :
Example.
R/I
Assume
=
Z[T ]/(T 2 + 5, 2, 1 + T )
=
F2 [T ]/(T 2 + 5, T + 1)
=
F2 [T ]/(T 2 + 1, T + 1)
=
F2 [T ]/((T + 1)2 , T + 1)
=
∼
=
F2 [T ]/(T + 1)
every ideal
the fact
I
I ⊂R
of a
is projective.
F2
n = 1 then I = Rt for some t ∈ R. Now t
R/I = F2 . If t ∈
/ R∗ : I = Rt ⇒ 2 = at
contradicting
0 = R/Rt
a ∈ R∗ ⇒ I = Rt = R a1 2 = R · 2
is not a unit because otherwise
⇒
2 irreducible
and
R/I
=
Z[T ]
/ I
T 2 + 5 |{z}
=
Z[T ]/(T 2 + 5, 2)
=
F2 [T ]/(T 2 + 1)
=
F2 [T ]/(T + 1)2
6=
F2
2R
⇒ I ∼
= Rn ⇒ n ≥ 2. So Rn ∼
= I ⊂ R, let F = eld of fraction of R. Then I ,→ I ⊗R F and
R ,→ F = R ⊗R F so we get F n = I ⊗R F ⊂ F = R ⊗R F (since F is a localization of R) ⇒ F n ⊂ F
contradiction so n 6≥ 2.
Denition 2.9. An inner product space is a non-degenerate bilinear form module
nitely generated and projective.
Remark. Over a local ring any inner product space is free
8
(M, β)
where
M
is
Denition
2.10. Let R be
0 1
2
i.e. H = (R , β).
1 0
The hyperbolic plane
a ring.
His
the symmetric inner product space
(
0 i=j
H has basis e1 , e2 with respect to which we have β(ei , ej ) =
.
H is a symmetric
1 i 6= j
T
0 1
0 1
0 1
space because
=
, it is non-degenerate because det
= −1 ∈ R∗ . Does
1 0
1 0
1 0
∼ h1i ⊥ h−1i because −1 0 =
H have an orthogonal basis? If 12 ∈ R (i.e. 2 ∈ R∗ ) then H =
0 1
1
1
0 1
1 − 12
1
/ R (2 ∈
/ R∗ ) then H has no
and det A = 1 and thus A is invertible. If
1
2 ∈
1 0
− 12 12
1
2
| {z }
| {z }
=A AT
0 1
x
x
y
orthogonal basis. In H 3
then x
= 2xy ∈
/ R∗ ∀x, y ∈ H since 2 ∈
/ R∗ . If (H, β)
y
1 0
y
∗
had an orthogonal basis e1 , e2 then β(ei , ei ) ∈ R ⇒ H has not orthogonal basis.
a 1
Example. If (R, m, k) is a local ring with chark = 2 then for all a, b ∈ m
has no orthog1 b
a 1
onal basis (otherwise, any orthogonal basis would yield an orthogonal basis mod m but
1 b
0 1
mod m =
= H has no orthogonal basis.
1 0
Theorem 2.11. Let
•
If char(k)
6= 2
(R, m, k)
then
M
be a local ring and
M = (M, β)
a symmetric inner product space.
has an orthogonal basis
• If char(k) = 2 then M = hu1 i ⊥ · · · ⊥ hul i ⊥ N1 ⊥ · · · ⊥ Nr
ai 1
, ai , bi ∈ m
1 bi
where
ui ∈ R∗
and
Ni =
M nitely generated, β is symmetric ⇒ M = hu1 i ⊥ · · · ⊥ hul i ⊥ N such that ui ∈ R∗
∗
and β(x, x) ∈ R \ R = m ∀x ∈ N .
Recall R local and M nitely generated projective ⇒ M ∼
= Rn+l , same for N so N ∼
= Rn
n
If n = 0 done. So assume n ≥ 1.Then β non-degenerate ⇒ for ϕ : R = N → R linear, ∃x0 ∈ N :
Proof. Recall
(x1 ,...,xn )7→x1
β(x0 , y) = ϕ(y) ∀y ∈ N ⇒ ∃x, y ∈ N : β(x, y) = 1 (y = e1 , x = x0 ).
If char(k) 6= 2 (2 ∈
/ m = R \ R∗ ⇒ 2 ∈ R∗ ). If N 6= 0 ⇒ ∃x, y ∈ N, β(x, y) = 1. Then x + y ∈ N
so β(x + y, x + y) = β(x, x) + 2β(x, y) + β(y, y) ⇒ 2 ∈ m ⇒ N = 0 (due to the contradiction of
{z
}
| {z } | {z } | {z }
|
∈m
char(k)
∈m
∈m
=2
6= 2)
= 2. We are going to prove that N = N1 ⊥ · · · ⊥ Nr with Ni as in
n (N ∼
= Rn ). n = 0 ⇒ N = 0 and we are done. n = 1 ⇒ N ∼
= R
then β|N is a non-degenerate symmetric form on R but any rank 1 inner product space is ∼
= hui
∗
∗
for u ∈ R because β : R × R → R, β(x, y) = xy · β(1, 1) and β non-degenerated ⇒ β(1, 1) ∈ R .
This contradict our assumption that β(x, x) ∈ m ∀x ∈ N . So assume n ≥ 2: Since β|N is non
degenerate and N free (of rank n) ⇒ ∃x, y ∈ N : β(x, y) = 1 (because N ∼
= Rn , ϕ : Rn → R by
n
x1 , . . . , xn 7→ x1
⇒
∃x : ϕ(y) = β(x, y) ∀y ∈ R so in particular ∃x ∈ N 1 = ϕ(e1 ) = β(x, e1 ))
Now assume that char(k)
the theorem by induction on
β non−deg
The subspace
Rx + Ry ⊂ N
has bilinear form matrix with respect to
and by assumption
a, b ∈ m
because
β(x, x) β(x, y)
a
=
β(y, x) β(y, y)
1
β(z, z) ∈ ∀z ∈ N .
{x, y}
1
b
This has determinant
ab − |{z}
1 ∈ R∗ ⇒ x, y
|{z}
∈m
|
9
∈m
/
{z
∈m
/
}
linearly independent and
N1 := Rx + Ry ⊂ N
is isometric to
a 1
1 b
⇒ N = N1 ⊥ N1⊥
and apply
N1⊥ and we are done
2 1
2 1
p
Example. (M, β) =
= 3 and 3 ∈ (Z(2) )∗ , hence
over Z(2) = { ∈ Q|2 - q}. det
q
1 2
1 2
M is non-degenerate. But M has no orthogonal basis because
/2 = F2 and any orthogonal basis
Z(2)
0 1
over Z(2) induces a orthogonal basis over Z(2) /2 but M =
over F2 which we have seen has
1 0
induction hypothesis to
no orthogonal basis.
p ∈ Z is prime, p 6= 3 (otherwise M is degenerate as 3 ∈
/ (Z(3) )∗ ). Then M is
∗
non-degenerate since det M = 3 ∈ (Z(p) ) . If furthermore p 6= 2 then by theorem M has an orthogonal
1
∗
basis: For instance x =
then β(x, x) = 2 ∈ (Z(p) ) (p 6= 2) ⇒ Rx ⊂ M non-degenerate subspace so
0
2 1
a
a
⊥
⊥
2
M = Rx ⊥ (Rx) . Now (Rx) = {y ∈ M : β(x, y) = 0} =
∈R | 1 0
=0 =
b
1 2
b
1
1
1
R
⇒
,
is an orthogonal basis of M if R = Z(p) p 6= 2, 3 (Also works for Q, R, C)
−2
0
−2
Over
R = Z(p)
where
Denition 2.12. Let (M, β) be a symplectic inner product space (β symplectic if β(x, x)
= 0 ∀x ∈ M ).
M is a basis x1 , y2 , x2 , yx , . . . , xn , yn such that M = (Rx1 + Ry1 ) ⊥ · · · ⊥
(Rxn + Ryn ) and β(xi , yi ) = 1 ∀i (⇒ β(yi , xi ) = −1) i.e. the bilinear form matrix of β with respect to
the basis x1 , y1 , ..., xn , yn is


0 1
0
−1 0





0
1




−1
0




..


.



0 1
0
−1 0
A symplectic basis of
Theorem 2.13. Let
(M, β)
(R, m, k)
be a local ring and
(M, β)
a symplectic inner product space.
Then
has a symplectic basis
⇒ free
(since R local)⇒ ∃x, y ∈
β(x, x) β(x, y)
M : β(x, y) = 1. So the inner product matrix of β with respect to {x, y} is
=
β(x, y) β(y, y) β sympletic
0 1
. Thus set N1 = Rx + Ry ⊂ M is a non-degenerate free submodule of rank 2 with symplectic
−1 0
basis {x, y}. (N1 , β|N1 ) non-degenerate ⇒ M =
N1
+ N1⊥ repeating the same argument with
|{z}


0 1

−1 0
0 1
N1⊥ instead of M we obtain M = N1 ⊥ N2 ⊥ · · · ⊥ Nn where Ni =
−1 0
Proof.
(M, β) inner product space ⇒ β
non-degenerate,
M
projective
Corollary 2.14. Over a local ring any symplectic inner product space has even dimension. Furthermore any two symplectic inner product spaces are isometric if and only if they have the same rank.
2.1
Witt Cancellation
Motivation: Let
V1 , V2
dim W = dim V2 .
k , If V1 ⊕ W ∼
= V2 ⊕ W for some nite
dim V1 = dim V1 ⊕ W − dim W = dim V2 ⊕ W −
be nite dimensional vector spaces over
dimensional vector space
W
then
V1 ∼
= V2
because
The same is true for free modules of nite rank (over a commutative ring), and also
over nitely generated projective modules over local rings.
Question: If
V1 , V2 , W
are symmetric inner product spaces does
10
V1 ⊥ W ∼
= V2 ⊥ W ⇒ V1 ∼
= V2
?
Example.
*−1
over
1
R = F2 (or
+R local
 with 2
*
0
0
−1 0

−1 0  ∼
0 0
=
0 −1
0 1
0
0
F2 . For
∼
∈
/R
)+then h−1i ⊥ h−1i ⊥ h−1i = h−1i ⊥ H, that is,
0
1 but h−1i ⊥ h−1i H because H has no orthogonal basis
0
the isometry see the example below.
Denition 2.15. We say Witt cancellation holds for a ring
spaces over
R
RM ⊥P ∼
=N ⊥P ⇒M ∼
=N
∀M, N, P
if
symmetric inner product
Example. Witt cancellation does not hold over elds of char2 (or for local rings
*−1
Note.
+
−1


1
*−1
∼
= 
+
0
1
*−1
1 (∗),
0
β =
because
−1

R where 2 ∈
/ R∗ ).
+

has orthogonal
1
(
e1 , e2 , e3
basis
with
e3 , e1 + e3 , e2 + e3 ,
β(ei , ej ) = 0
the inner product
cancellation holds then
2∈
/R
∗
i 6= j
β
has inner product
h−1i ⊥ h−1i ∼
=H
−1 i = 1, 2
.
1
i=3
*
−1

0
matrix
1
β(ei , ei ) =
and
but this is not the case for eld
In the basis
e1 + e2 +
+
1 ⇒ (∗). So If Witt
0
char2 (or local rings R with
)
Denition 2.16. Let
then
for
M = N ⊥ N⊥
M
be an symmetric inner product space and
and the reection of
M
at
N
N ⊆M
a non-degenerate subspace
is the isometry
rN : M = N ⊥ N ⊥ → N ⊥ N ⊥
x ∈ N, y ∈ N ⊥
(x, y) 7→ (x, −y),
Remark.
rN
is
R-linear,
an isomorphism
(rN ◦ rN = id)
and preserves inner product hence
rN
is an
isometry
Lemma 2.17. Let
β(y, y) ∈ R∗ .
If
Proof. Consider
R
(M, β) be a symmetric inner product space and x, y ∈ M such that
1
is local with 2 ∈ R then there is a reection r of M such that r(x) = y
u = x + y, v = x − y ∈ M
then
β(x, x) =
x = 12 (u + v), y = 12 (u − v)
• u ⊥ v : β(u, v) = β(x + y, x − y) = β(x, x) − β(y, y) = 0
(Since by assumption
β(x, x) = β(y, y))
• β(u, u) or β(v, v) ∈ R∗ : 4β(x, x) = β(2x, 2x) = β(u + v, u + v) ∈ R∗ (since β(x, x) ∈ R∗ and
2 ∈ R∗ ). By the rst point β(u + v, u + v) = β(u, u) + β(v, v). If β(u, u), β(v, v) ∈ m = maximal
∗
ideal of R ⇒ β(u, u) + β(v, v) ∈ m. Contradiction hence β(u, u) or β(v, v) ∈ R
•
If
β(u, u) ∈ R∗ then Ru ⊆ M
1
2 (u − v) =
If β(v, v) ∈
y
R∗
then
Rv ⊆ M
non-degenerate subspace
1
rRu (x) = rRu ( u+v
2 ) = 2 rRu (u+v)
non-degenerate subspace
r(Rv)⊥ (x) = 21 r(Rv)⊥ (u + v)
=
v∈(Ru)⊥
=
u∈(Rv)⊥
v∈((Rv)⊥ )⊥ =Rv
1
2 (u
− v) = y
Theorem 2.18. Let
is
∀M, N, P
(R, m, k)
be a local ring with
symmetric inner product space over
R
2 ∈ R∗ .
we have
Then Witt cancellation holds for
M ⊥P ∼
=N ⊥P ⇒M ∼
= N.
R.
That
M, N, P be symmetric inner product spaces over R such that M ⊥ P ∼
= N ⊥ P . By
∗
∗
our assumption R local and 2 ∈ R ⇒ P = hu1 i ⊥ · · · ⊥ hur i for ui ∈ R . Thus it suces to
∼
=
∼ N ⊥ hui ⇒ M ∼
show M ⊥ hui =
= N . Let f : M ⊥ hui → N ⊥ hui be an isometry and let
x ∈ M ⊥ hui and y ∈ N ⊥ hui be a generator for hui i.e. M ⊥ hui = M ⊥ Rx and N ⊥ hui =
N ⊥ Ry . β(x, x)M ⊥hui = u = β(y, y)N ⊥hui . f isometry: β(f (x), f (x))N ⊥hui = β(x, x)M ⊥hui = u =
β(y, y)N ⊥hui ⇒ f (x), y ∈ N ⊥ hui satisfy hypotheses of lemma 2.17 ⇒ ∃ reection r : N ⊥ hui → N ⊥
Proof. Let
11
hui
r(f (x)) = y⇒ r ◦ f : M ⊥ hui → N ⊥ hui
such that
∼
=
r ◦ f : Rx → Ry .
is an isometry such that
x7→y
∼
=
∼
=
⇒ r ◦ f : (Rx)⊥ → (Ry)⊥ ⇒ M → N
r◦f
| {z }
| {z }
M
2.2
N
Symmetric Inner Product space over
R
M over R has an orthogonal basis M = hu1 i ⊥ . . . hun i , n =
dimR (M ), ui ∈ R∗ = R \ {0}. If u > 0 then u = a2 for some a ∈ R, hui ∼
= h1i. If u < 0 then
←a
2
∼
u = −a and hui = h−1i. So M = r h1i ⊥ s h−1i (and r + s = dimR (M ))
Any symmetric inner product space
←a
Proposition 2.19 (Inertia Theorem). Over
R
we have
r h1i ⊥ s h−1i ∼
= m h1i ⊥ n h−1i ⇒ r = m
and
s=n
r + s = dimR (M ) = n + m. Assume without loss of generality
r ≤ m then n ≤ s. Witt cancellation tells us that (s − n) h−1i ∼
= (mP− r) h1i. Note that
if m − r = s − n 6= 0 then ∀x 6= 0 ∈ (s − n) h−1i we have β(x, x) = −
x2i < 0. However
P 2
∀x 6= 0 ∈ (m − r) h1i we have β(x, x) = xi > 0. Contradiction. Hence s − n = m − r = 0
Proof. The equation implies that
that
Corollary 2.20. The numbers
r, s
in
M∼
= r h1i + s h−1i
M ∼
= r h1i ⊥ s h−1i
the negative index of M
Denition 2.21. If
−
do not depend on the choice of an orthogonal
M
basis for
s=i M
of M
is called
We have showed that if (over
over
and
R) M ∼
=N
R then r = i+ M is called the positive index of M .
i+ M − i− M = r − s = sgn(M ) is called the signature
i+ N = i+ M, i− N = i− M
then
Corollary 2.22. Two symmetric inner product-spaces
+
+
−
−
i M = i N, i M = i N ⇐⇒
2.3
M = rank
rank
M, N over R
N, sgn M = sgn N
sgn(N ) = sgn(M )
are isometric
M ∼
= N
⇐⇒
Witt chain equivalence theorem
Notation. Let
hu1 , . . . , ul i
u1 , . . . , ul ∈ R
∗
write
hu1 , . . . , ul i
for

* u1

hu1 i ⊥ · · · ⊥ hul i = 
0
0
..

+


.
. We say
ul
is a diagonal form
Denition 2.23. We say
(M, β)
represent
a∈R
if
∃x ∈ M
Example. A diagonal form
··· +
and
ul x2l has a solution
Lemma 2.24. Let
such that
β(x, x) = a
hu1 , . . . , ul i represents u1 , . . . , ul , u1 + u2 , . . . . The
x1 , . . . , xl ∈ R ⇐⇒ a is represented by hu1 , . . . , ul i
equation
a = u1 x21 +
R
be a local ring (or a ring in which every direct summand of a nitely generated
ha, bi and hc, di be non-degenerate diagonal forms (a, b, c, d ∈ R∗ ). Then
ab = cd ∈ R∗ /(R∗ )2 and ∃e ∈ R∗ which represent ha, bi and hc, di
free module is free) Let
ha, bi ∼
= hc, di ⇐⇒
Proof. ⇒: We've already done. (They obviously need the same determinant modulo squares, and
need to represent the same numbers)
e ∈ R∗ represents ha, bi and hc, di ⇒ ∃x, y ∈ R2 such that β(x, x)ha,bi = e = β(y, y)hc,di ⇒
ha, bi =
Rx
⊥ (Rx)⊥ and hc, di =
Ry
⊥ (Ry)⊥ with u1 , u2 ∈ R∗ . Now e · u1 =
|{z}
|{z}
| {z }
| {z }
non−degenerat
non−degenerat
rank1
rank1
|
{z
}
|
{z
}
⇐:
hei⊥hu1 i
hei⊥hu2 i
det(hei ⊥ hu1 i) = det ha, bi = det hc, di = det he, u2 i = eu2 ∈ R∗ /(R∗ )2 ⇒ eu1 = eu2 g 2
g ∈ R∗ ⇒ u1 = u2 g 2 for some g ∈ R∗ , hu1 i ∼
= hu2 i ⇒ ha, bi ∼
= he, u1 i ∼
= he, u2 i ∼
= hc, di
Denition 2.25. Two non-degenerate diagonal forms
n)
are called simply (chain) equivalent (with notation
12
ha1 , . . . , an i and hb1 , . . . , bn i
≈s ) if either
for some
(of the same rank
• n≥2
•
or
and
∃1 ≤ i < j ≤ n
such that
hai , aj i ∼
= hbi , bj i
and
al = bl ∀l 6= i, j
n = 1 ha1 i ∼
= hb1 i
Two non-degenerate diagonal forms ha1 , . . . , an i and hb1 , . . . , bn i are
≈) if ∃M1 , . . . , Mr non degenerated diagonal forms of rank n such
hb1 , . . . , bn i and M1 ≈s M2 ≈s · · · ≈s Mr
chain equivalent (with notation
that
M1 = ha1 , . . . , an i , Mr =
ha1 , . . . , an i ≈ hb1 , ..., bn i ⇒ ha1 , . . . , an i ∼
= hb1 , . . . , bn i
P
Example. σ ∈
group on n letters. ha1 , . . . , an i ≈ aσ(1) , . . . aσ(n) because true
n = permutation
P
P
a 0
b 0
∼
⇒ true for all σ ∈ n because n is generated
for transpositions because
=
0 b
0 a
{z
}
|
Remark.
hai⊥hbi∼
=hbi⊥hai
by transpositions.
Witt's Chain Equivalence Theorem. Let
forms over a local ring
R
with
2 ∈ R∗
ha1 , . . . , an i and hb1 , . . . , bn i be non-degenerate diagonal
ha1 , . . . an i ≈ hb1 , . . . , bn i ⇐⇒ ha1 , . . . , an i ∼
= hb1 , . . . , bn i
then
Proof. ⇒: We seen this in the remark
⇐: We use induction on
n.
For
n=0
there is nothing to say.
n = 1, n = 2
is true by denition
of chain equivalence.
n ≥ 3 and ha1 , . . . , an i ∼
= hb1 , . . . , bn i
Claim: ∃ non-degenerate diagonal form hc1 , . . . , cn i ≈ ha1 , . . . , an i with c1 = b1 .
Note that the claim implies the theorem because ha1 , . . . , an i ≈ hc1 , . . . , cn i , hc1 , . . . , cn i = hb1 , c2 , . . . , cn i ∼
=
hb1 , . . . , bn i
⇒
hc2 , . . . , cn i ∼
= hb2 , . . . bn i. So by hypothesis ⇒ hc2 , . . . , cn i ≈ hb2 , . . . , bn i
Witt cancellation
*
+
Assume
hence
ha1 , . . . , an i ≈
c1 , c2 , . . . , cn
|{z}
≈ hb1 , . . . , bn i.
b1
Proof of claim: Let P = {(c, p)|c = hc1 , . . . , cn i , 1 ≤ p ≤ n such that hc1 , . . . , cp i represents
b1 }. Note that (ha1 , . . . , an i , n) ∈ P so P 6= ∅. Let p = min{l|∃(c, l) ∈ P } well dened and
1 ≤ p ≤ n because P 6= ∅. Choose (c, p) with p minimal as above. c = hc1 , . . . , cn i has prop∗
2
2
erty that hc1 , . . . , cp i represent b1 ⇒ ∃x1 , . . . xp ∈ R such that b1 = c1 x1 + · · · + cp xp ∈ R . As2
x
2
2
2
2
∗
sume that p ≥ 2. If ∀i 6= j ci xi + cj xj ∈ m = R \ R then (c1 x1 + c2 x2 ) + (c2 x2 + c3 x3 ) +
|
{z
} |
{z
}
∈m
· · · + (cp x2p + c1 x21 ) = 2b1
{z
}
|
which is a contradiction as
2 ∈ R∗
and
∈m
b1 ∈ R∗ ⇒ ∃i < j
such that
∈m
P
ci x2i + cj x2j ∈ R∗ . Since hc1 , . . . , cp i ≈ cσ(1) , . . . , cσ(p) ∀σ ∈ p we can assume d = c1 x21 + c2 x22 ∈ R∗ .
Then hc1 , c2 i ∼
= hd, dc1 c2 i because both represent d ∈ R∗ and both have the same determinant
∗
2∗
(in R /R ), ⇒ hc1 , . . . , cp i ≈ hd, dc1 c2 , c3 , . . . , cp i ≈ hd, c3 , . . . , cp , dc1 c2 i but hd, c3 , . . . , cp irepresent
b1 because b1 = c1 x21 + c2 x22 + · · · + cp x2p ⇒ (hd, c1 , . . . , cp , dc1 c2 , cp+1 , . . . , cn i , p − 1) ∈ P which
|
{z
}
d·12
p⇒ p = 1 ⇒ ∃ hc1 , . . . , cp i ≈ ha1 , . . . , an iand hc1 i
b1 = c1 x2 ⇒ hb1 i ∼
hc
i⇒
∃
hb1 , c2 , . . . , cp i ≈ ha1 , . . . , an i
= 1
contradicts minimality of
2.4
represents
b1 ,
i.e.,
Witt Groups:
Goal: Dene
R}/metabolic
W (R) = abelian group to be {isometry classes of symmetric inner
1
forms (=hyperbolic if
2 ∈ R) with group operation given by ⊥
Denition 2.26. A symmetric inner product space
summand
N ⊆M
Remark. N ⊆ M
N ∩ P = 0)
such that
N = N ⊥.
(M, β )
is called metabolic (or split) if
Such a direct summand
is a direct summand if
∃P ⊆ M
product space over
such that
N
N ⊕P = M
(i.e.
direct
0 1
1
Example.
• H=
is metabolic with Lagrangian
: R → R2
1 0
0
2
i.e. L = {(x, 0) ∈ R |x ∈ R} ⊂ H is a Lagrangian. Because
N +P = M
13
∃
is called Lagrangian.
dened by
and
x
x 7→
.
0
∗ L
P = {(0, y)|y ∈ R} ⊆ H,
0 1
x
= 0 ∀z ∈ R} = {(x, y) ∈ R2 |y = 0} = L
∗ L⊥ = {(x, y) ∈ R2 |(z, 0)
y
|{z} 1 0
is a direct summand with complement
∈L
|
{z
}
zy=0 ∀z ⇐⇒ y=0
• In is the identity matrix in Mn (R). A ∈ Mn (R) then

0
In
In
An
is metabolic with Lagrangian

I
 n
the image of the map
0
Rn −→ Rn ⊕ Rn
(proof is the same as above)
•
Lemma 2.27. Let
(M, β) be a symmetric inner product space then (M, β) ⊥ (M, −β) is metabolic.
Proof. The submodule
(M, −β)
∗ L
L = {(x, x) ∈ M ⊕ M |x ∈ M } ⊆ M ⊥ M
is a Lagrangian for
(M.β) ⊥
because
P = {(y, 0) ∈ M 2 |y ∈ M } as L ∩ P = 0
(a, b) = (b, b) + (a − b, 0) so L ⊕ P = L + P = M 2
| {z } | {z }
is a direct summand with complement
element
(a, b) ∈ M
2
is
∈L
and every
∈P
a
∗ L = L⊥ because let
∈ L⊥ ⊆ M 2 ⇐⇒ β(a, x) − β(b, x) = 0 ∀x ∈ M ⇐⇒ β(a − b, x) =
b
a
0 ∀x ∈ M
⇐⇒
a−b=0⇒a=b⇒
∈ L. Hence L⊥ = L
b
β non degenerate
Denition 2.28. A free symmetric inner product space is called hyperbolic if it is isometric to
Hn
Note. M, N are metabolic (or hyperbolic) then so is M ⊥ N . If M, N are metabolic with Lagrangian
L1 ⊆
M, L2 ⊆
N then M ⊥ N has Lagrangian L1 ⊥ L2 ⊆ M ⊕ N .
0 In
∼
= Hn (by change of basis). Let the basis of the rst one to be e1 , . . . , en , en+1 , . . . , e2n
In 0
then the basis of the second one is (e1 , en+1 ), (e2 , en+2 ), . . . (en , e2n ) where each pairs gives a copy of
H
0 In
0 In
∗
∼
∼
Lemma 2.29. If 2 ∈ R then for all A ∈ Mn (R),
=
= Hn
In A
In 0
0 In
Proof. We need to nd a base change, that is ∃X ∈ Mn (R) which is invertible such that X
XT =
In A
I
0
In
0
0 In
−1
. Take X =
which is invertible with inverse X
= 1n
In 0
A
I
− 21 A In
n
2
Lemma 2.30. Let
local or
n ∈ N,
R = Z)
and
R
R-module
are free
0 In
isometric to
In A
be a ring for which all nitely generated projective
then any metabolic inner product space
A ∈ Mn (R).
If moreover,
2∈R
∗
(M, β)
is
(e.g..,
R
for some
then every metabolic space is hyperbolic.
(M, β) be metabolic with Lagrangian L ⊆ M . L ⊆ M being a direct summand ⇒ P ⊆ M
L∩P = 0 and L+P = M . By assumption M projective ⇒ P, L projective
⇒
P, L
assumption on R
0 B
are free. In a basis for L, and P , the inner product space β has inner product matrix
, with
BT C
C = C T . The upper left corner is 0 because L = L⊥ we have β(x, x) = 0 ∀x ∈ L.
Claim: The matrix B is invertible.
∗
Proof of claim: B is the matrix of the linear map P → L = HomR (L, R) with respect to the basis
Proof. Let
such that
x7→β(x,−)
of
P
and the dual basis of
L.
Need to show
P → L∗
14
dened by
x 7→ β(x, −)
is an isomorphism.
x ∈ P : β(x, y) = 0 ∀y ∈ L ⇒ x ∈ L⊥ = L ⇒ x ∈ L ∩ P = 0 ⇒ x = 0
∗
Surjectivity: Let φ ∈ L , φ : L → R. Dene φ̄ : M = L ⊕ P → P by (x, y) 7→ φ(x). Now β is
non-degenerate ⇒ ∃ a , b ∈ M = L ⊕ P such that φ̄(x, y) = β(a + b, x + y) ∀x ∈ L, y ∈ P ⇒ φ(x) =
Injectivitiy:
∈L ∈P
φ̄(x, 0) = β(a + b, x) =
+ β(b, x) = β(b, x) ∀x ∈ L.
β(a, x)
| {z }
So
b∈P
is sent to
φ
under the
=0 since a,x∈L=L⊥
P → L∗ . This
shows surjectivity.
0
In
1
Notice that
=
T −1
−1
I
(B
)
CB
0
n
0 In
T −1
. For A = (B )
CB −1
In A
0 In
∗
∼
Note. If 2 ∈ R then
= Hn
In A
map
Corollary 2.31. Over a local ring
R,
0
(B T )−1
0
BT
B
C
1
0
0
B −1
⇒M ∼
=
0
BT
every metabolic space has even dimension and if
B
C
∼
=
2 ∈ R∗ , R
local, then every metabolic space is hyperbolic
Denition 2.32. Let
M, N be symmetric inner product spaces over R then M are N are called Witt
∃ metabolic spaces P, Q such that M ⊥ P ∼
= N ⊥ Q. Denote by W (R) be the
equivalence classes [M ] of symmetric inner product spaces M over R
∼ N)
Equivalent (M
set of Witt
if
Lemma 2.33 (Denition). Orthogonal sum
[N ] = [M ⊥ N ]
and
⊥ makes W (R) into an abelian group
−[M, β] = [M, −β]. W (R) is called the Witt group of R.
with
0 = [0], [M ]+
• + is well dened because if M ∼ M 0 , N ∼ N 0 then ∃P, P 0 , Q, Q0 metabolic such that
M ⊥ P ∼
= (M 0 ⊥ N 0 ) ⊥
= M 0 ⊥ P 0, N ⊥ Q ∼
= N 0 ⊥ Q0 . Then (M ⊥ N ) ⊥ (P ⊥ Q) ∼
| {z }
Proof.
metabolic
(P 0 ⊥ Q0 ) ⇒ M ⊥ N ∼ M 0 ⊥ N 0
| {z }
metabolic
•
We have
[M ] + [N ] = [M ⊥ N ] = [N ⊥ M ] = [N ] + [M ] (since M ⊥ N ∼
= N ⊥ M ) and the group
law is commutative
• [0] + [M ] = [0 + M ] = [M ]
because
0⊥M ∼
=M
• [M, β] + [M, −β] = [(M, β) ⊥ (M, −β)] = 0 because (M, β) ⊥ (M, −β) is metabolic for any inner
product space (M, β).
W : (commutative) rings → abelian groups, dened by R 7→ W (R) is a functor. For f : R → S
W (f ) : W (R) → W (S) by [M, β] 7→ [MS , βS ]
where MS = S ⊗R M and βS : MS × MS → S is dened s1 ⊗ x1 , s2 ⊗ x2 7→ s1 s2 β(x1 , x2 ). Note that if
(M, β) is non-degenerate then so is (MS , βS ) : if M is free choose an R-basis of M , say x1 , . . . , xn ∈ M
then MS is free with S -basis 1⊗x1 , . . . , 1⊗xn . Then (M, β) non-degenerate ⇐⇒ (β(xi , xj )) ∈ Mn (R)
is invertible
⇒
(f (β(xi , xj )) = (βS (1 ⊗ xi , 1 ⊗ xj )) ∈ Mn (S) is invertible ⇐⇒ (MS , βS ) is non∗
∗
Remark.
a ring homomorphism, we dene a map of abelian groups
f :R →S
degenerate. In the case
g
(M, β)
is projective do it as an exercise
f
R → S → T ring homomorphism, note that W (f ) ◦ W (g) = W (f ◦ g) because T ⊗S (S ⊗R M ) ∼
=
(T ⊗S S) ⊗R M ∼
= T ⊗R M
| {z }
For
∼
=T
x⊗y7→xf (y)
Proposition 2.34. Let
R
2 ∈ R∗ . Then two symmetric
N and [M ] = [N ] ∈ W (R)
be a local ring with
are isometric if and only if rank
M = rank
Proof. ⇒: Obvious
inner product spaces
M, N
⇐: [M ] = [N ] ∈ W (R) ⇒ M ∼ N ⇒ ∃ metabolic P, Q such that M ⊥ P ∼
= N ⊥ Q. R local,
2 ∈ R∗ ⇒ metabolic = hyperbolic so P ∼
= Hp , Q ∼
= Hq . Now rank M = rank N and rank M ⊥ P =
p
rank N ⊥ Q ⇒ p = q ⇒ M ⊥ H ∼
= N ⊥ Hp so by Witt cancellation ⇒ M ∼
= N.
15
Denition 2.35. Let
R be a local ring. The rank homomorphism rk : W (R) → Z/2Z is dened by
M . Note this map is well dened as M ∼ 0 ⇐⇒ ∃P metabolic such that M ⊥ P is
metabolic⇒ M ⊥ P and P have even rank ⇒ rk M = 0 ∈ Z/2Z. The rank map is surjective because
rk(h1i) = 1.
Let I(R) = ker(rk) =set of equivalence classes of even rank inner product spaces
rk M
∗
2∗
2
det M
The discriminant is the homomorphism disc: I(R) → R /R
dened
by [M ] 7→ (−1)
0 In
which is well dened because disc P = 1 for P metabolic as det
= (−1)n
In A
[M ] 7→
rank
Note that disc map is surjective because disc(hu, −1i)
Proposition 2.36. Let
char
F =2
be a eld in which every element is a square (e.g. F algebraic closed, or
∼
=
perfect, e.g., nite and char F = 2) then rk : W (F ) → Z/2Z is an isomorphism.
F
and
= u ∀u ∈ R∗
F
h1i 7→ 1. Recall that every
hu1 i ⊥ · · · ⊥ hul i ⊥ N1 ⊥ · · · ⊥ Nr
∗
2∗
with Ni = H (in the case of a eld). So W (F ) is generated by hui , u ∈ F /F
as an abelian group.
2∗
Consider the map Z → W (F ) dened by 1 7→ h1i. Since h1i + h1i = h1i + h−1i (as −1 ∈ F ), we
have h1i + h1i = 0 ⇒ This map factors as Z/2Z → W (F ) with 1 7→ h1i. As W (F ) is generated by
hui , u ∈ F ∗ /F 2∗ = {1} this means that the map Z/2Z → W (F ) is surjective and it is injective as
Proof.
rk : W (F ) → Z/2Z
is surjective (for any commutative ring) since
symmetric inner product space over
rk
Z/2Z W (F ) Z/2Z
sends
F
a eld is isometric to
∼
=
rk
Z/2Z
1 7→ h1i 7→ 1 ⇒ Z/Z2 → W (F ) ⇒ W (F ) →
∼
=
{z
}
|
id
Corollary 2.37.
W (Fq ) = Z/2Z
for
q
even,
∼
=
rk : W (C) → Z/2Z
Example.
W (R) → Z dened by [M ] 7→ sgn M is
∼
=
Claim: sgn : W (R) → Z is an isomorphism
Surjective: sgn(n h1i) = n sgn(h1i) = n · 1 = n
well dened because
sgn(H) = 0
R is M ∼
= n h1i + m h−1i.
n = m ⇒ M = n(h1i + h−1i) = 0 ∈ W (R)
Injectivitiy: Every symmetric inner product space over
sgn(n h1i + m h−1i) = n − m
then
If
sgn M =
Recall:
rk
• I(F ) = ker(W (F ) → Z/2Z) =
disc
• I(F ) → F ∗ /F 2∗
fundamental ideal map of abelian groups dened by
Note. The disc map extends to all of
where
r = rk M ,
W (F )
by
M 7→ (−1)
W (F ) → F ∗ /F 2∗
rk M
2
det M
M 7→ (−1)
dened by
r(r−1)
2
det M ,
but, in general, this is not a map of abelian groups so we don't often use this.
Proposition 2.38. Let
F
be a nite eld. Then the discriminant map is an isomorphism:
∼
=
I(F ) →
F ∗ /F 2∗
Proof. If char
∗
2∗
F /F
F =2
this is true because both sides are equal to
0.
(Since
F
nite and char
F =2⇒
= 0)
F is odd.
ha, bi ∼
= hab, 1i
So assume char
Claim:
We have to prove the following special case
The claim implies the proposition: dene the map
ρ : F ∗ /F 2∗ → I(F )
a 7→ ha, −1i this is easily
ρ(ab) = hab, −1i =
ha, bi + 2 h−1i = ha, −1i +
by
seen to be a well dened map of sets. This is a map of abelian groups because
habi + h−1i = habi + h1i + h−1i + h−1i = hab, 1i + h−1i + h−1i
=
claim
hb, −1i = ρ(a) + ρ(b). The maps is surjective because every ha1 , . . . , a2n i ∈ I(F ) is ha1 , . . . , a2n i =
ha1 . . . a2n , 1, 1, . . . , 1i = ha1 . . . a2n , −1i+2n h1i = ρ(a1 . . . a2n )+nρ(−1) because ρ(−1) = h−1, −1i =
claim
h1, 1i.
The map
claim
∗
ρ is injective because F /F
disc
2∗ ρ
∗
→ I(F ) → F /F
2∗
dened by
a 7→ ha, −1i 7→ a, hence
|
{z
}
id
we are done.
ha, bi ∼
= hc, di ⇐⇒ ab = cd ∈ F ∗ /F 2∗ and ∃e ∈ F such that e is represented
by both forms. Obviously ha, bi ∼
= hab, 1i has the same determinant, so the claim is equivalence to
2
2
the fact, since hab, 1irepresent 1, that ha, bi represent 1, i.e., ∃x, y ∈ F such that 1 = ax + by . This
Proof of claim: Recall:
follows from the following lemma
16
Lemma 2.39. Let
has a solution
F be
x, y ∈ F
a nite eld of order
q =odd.
Then
∀a, b ∈ F ∗
the equation
1 = ax2 + by 2
1 − by 2 = ax2 . We use the pigeon hole principle.
2∗
Let ϕ : F → F : x 7→ x hence F
= im ϕ = F ∗ / ker(ϕ) = F ∗ /{±1} (The last equality holds
∗
|F
|
q−1
q+1
2∗
2∗
since char F 6= 2)⇒ |(F )| =
2 = 2 ⇒ number of square in F = |F |+1 (for zero) = 2 . Hence
q+1
2
2
n1 = |{ax2 |x ∈ F }| = |{x2 |x ∈ F } = q+1
2 . Similarly n2 = |{1 − by |y ∈ F }| = |{y |y ∈ F } = 2 .
2
2
2
2
Hence n1 + n2 = q + 1 > |F | ⇒ {ax |x ∈ F } ∩ {1 − by |y ∈ F } =
6 ∅ ⇒ 1 − by = ax has a solution.
Proof. Need to nd x, y such that
∗
∗
2
Theorem 2.40. Let
Proof. char
Assume
F
q elements then


char F = 2
Z/2Z
W (F ) = Z/2Z ⊕ Z/2Z q ≡ 1 mod 4 ( ⇐⇒ −1 ∈ F 2∗ )


Z/4Z
q ≡ 3 mod 4 ( ⇐⇒ −1 ∈
/ F 2∗ )
be a nite eld with
∼
=
F = 2 we have already done (in this case rk : W (F ) → Z/2Z)
char F odd, so q odd. We have an exact sequence
rk
0 → I(F ) → W (F ) → Z/2Z → 0
∼
=F ∗ /F 2∗
Since
q
is odd we have
|F ∗ /F 2∗ | = 2
because
.2
F 2∗ = im(F ∗ → 2F ∗ )
x7→x
and
.2
ker(F ∗ → F ∗ ) = {±1}.
⇒ |W (F )| = 4 ⇒(by the structure theorem of nite groups)
2W(F ) = Z/2Z ⊕ Z/2Z or W (F ) = Z/4Z∗ .
2∗
2
If −1 ∈ F , (−1 = a ), ⇒ 2 hui = hui + hui = hui + a u = hui + h−ui = 0 ∈ W (F ) ∀u ∈ F .
W (F ) generated as an abelian group by hui , u ∈ F ∗ ⇒ every element in W (F ) has order ≤ 2.
⇒ W (F ) 6= Z/4Z ⇒ W (F ) = Z/2Z ⊕ Z/2Z.
If −1 ∈
/ F 2∗ ⇒ if 2 h1i = 0 ∈ W (F ) then h1i + h1i = h1i + h−1i ∈ W (F )
⇒
h1i + h1i ∼
=
char F odd
h1i + h−1i
⇒
h1i ∼
= h−1i ⇒ 1 = det h1i = det h−1i = −1 ∈ F ∗ /F 2∗ ⇒ −1 ∈ F 2∗
Witt Cancellation
a contradiction to the assumption
−1 ∈
/F
∗2
which is
⇒ 2 h1i 6= 0 ⇒ W (F ) 6= Z/2Z ⊕ Z/2Z ⇒ W (F ) = Z/4Z.
(Then the theorem follows from the following lemma)
Lemma 2.41. Let
F
be a nite eld of odd characteristic, with
q = |F | elements.
Then
−1 ∈ F 2∗ ⇐⇒
q ≡ 1 mod 4.
−1 ∈ F ∗ ∼
= Z/(q − 1)Z is the only element of order 2. ⇒ −1 ∈ F 2∗ ⇐⇒ F ∗ = Z/(q − 1)Z
element of order 4 ⇐⇒ 4|(q − 1) ⇐⇒ q ≡ 1 mod 4.
Proof.
an
p ∈ Z is an odd prime, then p
in Fp ( ⇐⇒ p ≡ 1 mod 4)
Remark.
square
To see this:
can be written as
a2 + b2 = (a + ib)(a − ib) ∈ Z[i].
Recall that
p = a2 + b2
Z[i]
with
a, b ∈ Z ⇐⇒ −1
has
is a
is a Euclidean domain, hence a UFD
(unique factorization domain) and thus, irreducible elements and prime elements are the same.
If
a2 + b2 = (a + ib)(a − ib) ∈ Z[i], we have p = a2 + b2 ⇒ p not prime in Z[i]. The converse also holds:
if p is not a prime in Z[i] then p = xy ∈ Z[i] for non-units x, y ∈ Z[i]. But then (if x = a + ib)
N (x) = a2 + b2 has the properties N (xy) = N (x)N (y), N (x) = 1 ⇐⇒ x unit. So p = xy ⇒ N (p) =
| {z }
=p2
N (x)N (y) ⇒ N (x) = p = N (y) ⇒ a2 +b2 = N (x) = p. So p can be written as p = a2 +b2 with a, b ∈ Z
⇐⇒ p is not a prime in Z[i]. But p is a prime in Z[i] ⇐⇒ Z[i]/p is a domain. But Z[i] = Z[t]/(t2 + 1)
Z
2
2
2
2
2
so Z[i]/p = Z[t]/(t + 1, p) =
p [t]/(t + 1) = Fp [t]/(t + 1). Now Fp [t]/(t + 1) is a eld ⇐⇒ t + 1
2
2
irreducible ⇐⇒ t + 1 has no solution in Fp ⇐⇒ −1 ∈
/ F2∗
p . On the other hand t + 1 reducible
2 2
2
⇐⇒ −1 ∈ F2∗
,
−1
=
a
,
t
+
1
=
(t
+
a)(t
−
a)
.
Then
F[t]/(t
+
1)
=
F
[t]/((t
−
a)(t + a)) =
p
p
CRT
Fp [t]/(t + a) × Fp [t]/(t − a) = Fp × Fp not a domain.
2
2
Hence p = a + b
⇐⇒ p not a prime in Z[i] ⇐⇒
2∗
⇐⇒ −1 ∈ Fp ⇐⇒ p ≡ 1 mod 4
17
Z[i]/p = Fq [t]/(t2 + 1)
not a domain
Corollary 2.42 (of Theorem 2.40 ). Two symmetric inner product spaces over a nite eld of odd
characteristic are isometric if and only if they have the same rank and the same determinant
(∈
F ∗ /F 2∗ ).
Theorem 2.43. Let
F
be a eld then
W (F )
is generated as an abelian group by
hui , u ∈ F ∗
subject
to the relations:
1.
hui = a2 u ∀a, u ∈ F ∗
2.
hui + h−ui = 0 ∀u ∈ F ∗
3.
hui + hvi = hu + vi + huv(u + v)i ∀u, v ∈ F ∗ , u + v ∈ F ∗
Remark. The theorem asserts that
rank 1 free abelian group with basis {a}
dened by
basis
z }| {
∼
Z{a}
⊕a∈F ∗
=
→ W (F )
2
{u} − {a u}, {u} + {−u}, {u} + {v} − {u + v} − {uv(u + v)}
{a} 7→ hai is an isomorphism. (Here Z{a} ∼
= Z denotes the free Z-module
of rank
1
with
{a}.)
Proof. We already know that
W (F )
is generated by
hui , u ∈ F ∗
and that
1, 2, 3
holds in
W (F ) ⇒
rank 1 f.a.g.w/ basis {a}
z }| {
⊕a∈F ∗
Z{a}
ρ:
→ W (F )
{u} − {a2 u}, {u} + {−u}, {u} + {v} − {u + v} − {uv(u + v)}
ρ is injective. Will give a
F 6= 2 (the char F = 2 case needs a dierent, longer proof ) P
n
PnUsing relation 2.P(m{u} = −{−u}) we can write every element in LHS as i=1 {ui }. Given U =
i=1 {ui } and V =
j=1 {vj } in the LHS such that ρ(U ) = ρ(V ) ∈ W (F ) ⇒ n = rk ρ(u) = rk ρ(V ) =
m ∈ Z/2Z. So m ≡ n mod 2 and without loss of generality say m ≥ n so m − n = 2k, k ≥ 0. Then
U = U +k({1} + {−1}) ∈ LHS. Replacing U ∈ LHS with U +k({1}+{−1}) we can assume that m = n.
{z
}
|
is a well dened surjective map of abelian groups. So we need to check that
proof when char
=0 by 2.
ρ(U ) = ρ(V ) ∈ W (F )
Then
and
rk ρ(U ) = rk ρ(V ).
⇒ hu1 , . . . , un i ∼
= hv1 , . . . , vn i
1
∗
2 ∈F
⇒
chain equivalence thm
hu1 , . . . , un i ≈ hv1 , . . . , vn i ⇒ ∃ diagonal forms c1 , . . . , cl such that hu1 , . . . un i ≈S c1 ≈S · · · ≈S
cl ≈S hv1 , . . . , vn i ⇒ it suces to show that {u1 } + · · · + {un } = {v1 } + · · · + {vn } in the case
hu1 , . . . , un i , hv1 , . . . , vn i are simply chain equivalence (i.e., they dier in two places). So, we can
assume n = 2, we need to show that hu1 , u2 i ∼
= hv1 , v2 i then {u1 } + {u2 } = {v1 } + {v2 } in LHS.
Assume hu1 , u2 i ∼
= hv1 , v2 i ⇒ u1 u2 = v1 v2 a2 for some a ∈ F ∗ and u1 = v1 x2 + v2 y 2 for some x, y ∈ F .
2
2
2
2
2 2
2
2
If x, y 6= 0 then {v1 } + {v2 } = {v1 x } + {v2 y } = {v1 x + v2 y } + {v1 v2 x y (v1 x + v2 y )} = {u1 } +
1.
{ a12 u1 u2 x2 y 2 u1 } = {u1 } + {u2 } ∈
1
u 1 = v2 y
2
and
2
2
v1 v2 a = v2 y u2 ⇒
3.
LHS. If
v1 ( ay )2
x
or
y = 0,
= u2 .
Then
say
x=0
then
y 6= 0
{u1 }+{u2 } = {v2 y
2
since
u1 ∈ F ∗ ,
}+{v1 ( ay )2 } =
1.
then we get
{v2 }+{v1 } ∈
LHS.
So
2.5
ρ(U ) = ρ(V ) ∈ W (F ) ⇒ U = V ∈
LHS⇒
ρ
injective
Second Residue Homomorphism
For a DVR (Discrete valuation ring )
π ∈ R,
W (Q), W (Z), W (Qp ) . . .
formizing element
R
F , residue
∂π : W (F ) → W (k)
with eld of fraction
we will construct maps
Denition 2.44. A discrete valuation ring (DVR) is a local ring
•
Noetherian
•
A domain (ab
=0∈R⇒a=0
or
b = 0)
18
(R, m, k)
eld
k = R/m
and uni-
which will help compute
which is:
• m 6= 0
(m = πR
is a principal ideal
for some
π ∈ R)
There are other (equivalent) characterizations of DVR (which we won't need):
• R
is a DVR
⇐⇒
local 1-dimensional integrally closed noetherian domain
• ⇐⇒
Local
• ⇐⇒
Local PID
• ⇐⇒
Local domain with principal
• ⇐⇒
valuation ring of a discrete valuation on a eld
1-dimensional
The
p-adic
m
such that
∩n≥0 mn = 0
• Z(p) = { ab ∈ Q|p - a}, p ∈ Z
Example (Of DVR).
•
noetherian regular domain
prime.
m = pZ(p) , k = Fp .
Fraction eld
Q.
Zp = lim Z/pn Z = {(xn )n∈N≥1 , xn ∈ Z/pn Z : xn+1 = xn mod pn }, m =
integers
←
n→∞
pZp , k = Zp /pZp = Fp
• D
and eld of fractions
= Dedekind domain,
p⊂D
• K is a eld, f ∈ K[T ]
f K[T ](f ) , k = K[T ]/f .
• R
is a UFD,
f ∈R
m,
i.e.
R
/ p}, m = pDp , k = D/p
Dp = { ab ∈ FracD|b ∈
K[T ](f ) = { ab ∈
an irreducible element (=prime element)
(R, m, k)
πR = m
Lemma 2.46. Let
a prime ideal then
is irreducible.
Denition 2.45. Let
generating
Zp = Qp
Frac(K[T ])
R(f ) = { ab ∈ Frac R|f - b}
is a choice
π ∈ m ⊂ R
then every element
a ∈ R, a 6= 0
be a DVR, a uniformizing element of
be a DVR with uniformizing element
a = π n u for some n ∈ N and u ∈
can be written uniquely as
= K(T )|f - b}, m =
π ∈ R,
R∗
R
π n u = π m v with u, v ∈ R∗ . Without loss of generality assume m ≥ n. R
domain ⇒ π
= vu ∈ R. If m 6= n ⇒ π n−m ∈ πR = m but vu−1 ∈ R∗ = R \ m which is a
−1
contradiction ⇒ n = m ⇒ 1 = vu
⇒u=v
n
Existence :Let a ∈ R, a 6= 0. If a ∈ ∩n≥0 m
= ∩n≥0 π m R then a = π n bn ∀n.
⇒
bn =
Proof. Uniqueness :Assume
m−n
−1
R domain
πbn+1 ⇒ (bn ) ⊂ (bn+1 ) ⊂ (bn+2 ) ⊂ . . . R, is an ascending chain of ideals which has to stop because R is
noetherian. ∃n such that (bn ) = (bn+1 ) in particular bn+1 ∈ (bn ) ⇒ bn+1 = cbn but bn = πbn+1
⇒
R domain
1 = cπ ⇒ π ∈ R∗ = R \ m which contradicts the fact that π ∈ m. Hence a = 0 ⇒ ∩n≥0 π n R = 0.
n
Hence ∃n such that a ∈ π R but a ∈
/ π n+1 R ⇒ a = π n u, u ∈
/ πR = m hence u ∈ R∗ .
Remark.
∩n≥0 π n R = ∩n≥0 mn .
Corollary 2.47. Let
a ∈ F, a 6= 0
R
For all Noetherian
R : ∩n≥0 mn = 0
be a DVR with uniformizing element π and
a = π n u where u ∈ R∗ .
F
its eld of fractions, then every
can be written uniquely as
So we can dene a function
ν : F∗ → Z
dened by
a = π n u 7→ n = ν(a)
(with
u ∈ R∗ )
with the
properties
1.
ν(ab) = ν(a) + ν(b)
2.
ν(a + b) ≥ min(ν(a), ν(a))
3. Setting
ν(0) = ∞
we have
R = {a ∈ F : ν(a) ≥ 0}, R∗ = {a ∈ F |ν(a) = 0}, m = {a ∈ F |ν(a) >
0}
F is a function ν : F ∗ → Z satisfying
R = {a ∈ F |ν(a) ≥ 0} where ν(0) = ∞.
Denition 2.48. A discrete valuation on a eld
The valuation ring of
Denition 2.49. Let
ν
is the ring
(F, ν) be a discrete valuation on a eld F with associated DVR R and choice
π ∈ R. The second residue homomorphism is the map ∂π : W (F ) → W (R/π)
of uniformizing element
dened by
(
hai 7→
where
1., 2. above.
hui n = ν(a) odd
0
n = ν(a) even
a ∈ F ∗ , a = π n u, n = ν(a), u ∈ R∗ , u = u mod m = πR, u ∈ R/π
19
∂
Note.
depends on the choice of the uniformizing element
π.
Lemma 2.50. The second residue homomorphism is well dened
Proof. Recall that
W (F )
is generated by
hai, a ∈ F ∗
subject to:
1.
hui = x2 u , u, x ∈ F ∗
2.
hui + h−ui = 0
3.
hui + hvi = hu + vi + huv(u + v)i, u, v, u + v ∈ F ∗
(
We need to check that
∂ hui = ν(u) φ
∂π
where
preserves these relations. First dene
u = π ν(u) φ, φ ∈ R∗ , φ = φ mod πR.
i =
0 i even
,
1 i odd
Then:
u = π n φ, x = π m ψ where φ, ψ ∈ R∗ . Then x2 u = π 2m+n φψ 2
n φ ∈ W (R/π) as required ∂ hui = n φ .
1. Let
u = π n φ, −u = π n (−φ)
n ( φ + −φ ) = 0 ∈ W (R/π)
{z
}
|
2. Let
with
φ ∈ R∗ .
Then
so that we can write
so
∂ x2 u = 2m+n φψ 2 =
∂ hui + ∂ h−ui = n φ + n −φ =
=0
u = π n φ, v = π m ψ ,
π (π n−m φ + ψ)
3. Let
without loss of generality assume
n ≥ m. u + v = π n φ + π m ψ =
m
Case 1.
n > m :
Then
n − m > 0 ⇒ t = π n−m φ + ψ ∈ R∗ , u + v = π m t,
| {z } ∈m
/
note that
∈m
uv(u + v) = π n+2m φψt . So ∂ hu + vi + ∂ huv(u + v)i = m hti +
|{z}
∈R∗
2
n+2m φψt = m ψ + n φψ = ∂ hvi + ∂ hui
t ≡ ψ mod πR.
Case 2.
n = m:
Now
Now
u + v = π n (φ + ψ)
and
l = 0: φ + ψ = t ∈ R∗ .
φ + ψ = πl t
where
Case i.
uv(u + v) = π 3n φψt . Then
|{z}
∈R∗
∂ hu + vi+∂ huv(u + v)i = n hti+3n φψt = n φ + ψ +n φψ(φ + ψ) =
n ( ψ + φ ) ∈ W (R/π) which is what we wanted.
Case ii.
l > 0: u + v = π l+n t ∈ πR. In particular ψ + φ = 0 so ψ = −φ ∈ R/πR.
3n+l
So uv(u + v) = π
φψt. Then
∂hu + vi + ∂ huv(u + v)i = n+l
hti +
3n+l φψt = n+t hti + n+l ∂ −ψ 2 t = n+l (hti + h−ti) = 0 = n ( −ψ +
ψ ) = n φ + n ψ = ∂ hui + ∂ hvi
Theorem 2.51. Let
D
Now
u + v = πn t
t ∈ R∗ .
be a Dedekind domain with eld of fractions
and
F.
Then the sequence of abelian
group
L
∂℘
M
0 −→ W (D) −→ W (F ) −→
W (D/℘)
℘⊂D
max. ideal
is exact.
We will prove the above theorem in the special cases:
Lemma 2.52. Let
(R, m, k)
D = Z,
be a DVR with eld of fraction
k
DVR,
and uniformizing element
Then the composition
W (R)
/ W (F )
is zero
20
∂π
k[T ].
/ W (R/π)
π ∈ m ⊂ R.
a 1
⇒ W (R) is generated by hui , u ∈ R∗ and
, a, b ∈ m. We have ∂ hui = 0 by
1 b
(
0
ν(u) even
ν(u)
where u = π
v, v ∈ R∗ .
denition of ∂ : ∂ hui =
hvi ν(u) odd
0 1
a 1
a 1
If a = 0 then
is metabolic so, W (F ) 3
=0⇒∂
=0
1 b
1 b
1 b
a 1
∗
If a 6= 0 then a ∈ F
so
= hai + ha(ab − 1)i , a = π ν(a) v, v ∈ F ∗ , a(ab − 1) =
1 b
*
+
a
1
=
= ∂ hai + ∂ ha(ab − 1)i = ν(a) hvi + ν(a) v (ab − 1)
π ν(a) v(ab − 1), ab ∈ m ⇒ ∂
1 b
| {z }
| {z }
Proof.
R
local
∈R∗
−1 mod m
ν(a) (hvi + h−vi) = 0
∂p
Corollary 2.53. The composition
W (Z/pZ)
is the
uation ring
∂p
W (Z) → W (Q) → ⊕p∈Z W (Z/pZ) (p prime) is zero, where W (Q) →
2nd residue homomorphism associated with the p-adic valuation on Q which has val= { ab ∈ Q|p - b}
Z(p)
Proof. We have dened
Q
W (Q)
∂p
/
Q
W (Z/pZ)
O
' ?
⊕W (Z/pZ)
?
∂p has image in ⊕p W (Z/pZ). This is the case because if hui ∈ W (Q), u ∈ Q∗ then
a
u = b and {p ∈ Z prime |ν(u) 6= 0} ⊂ {primes in the factorization of a, b} nite ⇒ ∀ξ ∈ W (Q), ∂p ξ = 0
for all but nitely many p ∈ Z prime. For the composition to be zero, it suces to check that the
∂p
composition W (Z) → W (Q) → W (Z/pZ) is zero for all p. The composition is zero because it factors
∂π
as W (Z) → W (Z(p) ) → W (Q) → W (Z/pZ)
Need to see that
Q
{z
|
}
0
Denition 2.54. A principal ideal domain (PID) is a commutative ring
is a domain (ab
0⇒a
some
or
b = 0)
and for which every ideal is a principal ideal (I
R which
⊂ R ⇒ I = Rx for
=
x ∈ I)
Example.
Z, k[T ] (k a eld) are PIDs (Euclidean domain⇒PID)
R is a PID: Let π be a uniformizing element, so m = πR and let I ⊂ R be any ideal. I = 0
n
n
is principal, so assume I 6= 0. Let n = min{ν(a)|a ∈ I, a 6= 0} ∈ N≥0 . Then I = π R because I ⊂ π R
≥0
z }| {
ν(a) − n u) ∈ π n R, and π n R ⊂ I since ∃a ∈ I, a 6= 0, ν(a) = n
ν(a)
since if a ∈ I, a 6= 0, a = π
u = π n (π
|
{z
}
A DVR
∈R
so
a = π n u, u ∈ R∗ , a ∈ I ⇒ π n R = π n uR = aR ⊂ I .
Hence
I = πn R
is principal.
Remark. A PID is noetherian
R be a PID, I1 ⊂ I2 ⊂ · · · ⊂ I2 ⊂ · · · ⊂ R be an ascending chain of ideals. Then
I = ∪In ⊂ R is an ideal ⇒ I = xR for some x ∈ I ⇒ x ∈ In for some n ⇒ I = Rx ⊂ In ⊂ I ⇒ In =
Im = I ∀m ≥ n
Proof. Let
Denition 2.55. An
R-module M
is called cyclic if
Fact. Every nitely generated module
is,
M∼
= ⊕ni=1 R/ai
for some non-units
Corollary 2.56. Let
R
1. Every submodule
M∼
= R/I
M over a PID R
a1 , . . . , an ∈ R
be a PID with eld of fractions
M
of a nitely generated free
2. Every nitely generated
R-submodule M ⊂ F n
21
for some ideal
I ⊂ R.
is a nite direct sum of cyclic
F
R-module
is also free
is free
R-modules,
that
Proof.
M
R
is a PID then
is nitely generated
R is noetherian, M ⊂ Rn ⇒ M is nitely generated so in 1 and 2 the module
⇒ M = ⊕ni=1 R/ai for some non-unit ai . But R/a ⊂ Rn or R/a ⊂ F n ⇒ a = 0
because:
If
a 6= 0
and
a ∈
/ R∗
then the composition
a
R/a ⊂ R ,→ R
zero at the same time (a contradiction). Similarly the composition
a 6= 0, a ∈
/R
∗
) and equals
a=0
R/a → R/a ,→ F
n
a=0
R/a → R/a ,→ R
→ F n is injective (if
is injective and
R/a ⊂ F n
a
zero, again a contradiction
Corollary 2.57. Every inner product space over a PID is free
Lemma 2.58. Let
R
[M, β] 7→ (MF , βF )
F, x, y ∈ M .
is injective, where
F
be a PID with eld of fractions
MF = M ⊗R F
W (R) → W (F ) dened by
βF (x ⊗ a, y ⊗ b) = abβ(x, y) for a, b ∈
then the map
and
0 1
Proof. Assume [MF , βF ] = 0 ∈ W (F ) then (MF , βF ) ∼ 0 ⇒ ∃ metabolic V =
with
1 A
T
A ∈ Mn (F ), A = A such that (MF , βF ) ⊥ V is metabolic. There exist d ∈ R such that dA ∈ Mn (R).
Then
d−1
0
0
d
0
1
1
A
d
0
0
=
0 d−1
1
1
d2 A
⇒
0
1
1
A
∼
=
0
1
1
d2 A
⇒ (M, β) ⊥
0
1
1 d2 A
{z
}
|
metabolic over R
F . Hence can assume (M, β) to be metabolic over F .
(M, β) be a symmetric inner product space over R such that (M, β)F = (MF , βF ) is metabolic
over F . MF = M ⊗R F , βF (x ⊗ a, y ⊗ b) = abβ(x, y), x, y ∈ M, a, b ∈ F . Note M ⊂ MF (as
R ⊂ F, M ∼
= Rn , MF ∼
= F n ). Now (MF , βF ) metabolic⇒ ∃N ⊂ MF Lagrangian
Claim: M ∩ N ⊂ M is a Lagrangian for (M, β)
M ∩ N is a direct summand of M because M/M ∩ N ⊂ MF /N ∼
= F m is a nitely generated Rm
submodule of F
⇒ M/M ∩ N is nitely generated free R-module, so M/M ∩ N ∼
= Rl . Any section
PID
s : M/M ∩N → M of g : M → M/M ∩N ∼
= Rl (that is gs = 1) yields a direct sum decomposition (M ∩
N ) ⊕ im(s) = M ⇒ M ∩ N ⊂ M is a direct summand. We now need to check that (M ∩ N )⊥ = M ∩ N .
⊥
Let x ∈ M , then x ∈ (M ∩ N )
⇐⇒ β(x, y) = 0 ∀y ∈ M ∩ N ⇐⇒ β(x, y) = 0 ∀y ∈ N (because
∀t ∈ MF = M ⊗R F ∃a ∈ R, a 6= 0 such that at ∈ M , in particular ∀y ∈ N, ∃a ∈ R, a 6= 0, ay ∈ M ∩ N ,
β(x, y) = 0 ⇐⇒ β(x, ay) = 0). But β(x, y) = 0 ∀y ∈ N ⇐⇒ x ∈ N ⊥ = N ⇐⇒ x ∈ M ∩ N .
is metabolic over
Let
N =N ⊥
a6=0
M ∩N ⊂M
x∈M
⇒ (M, β) is metabolic ⇒ [M, β] = 0 ∈ W (R)
To nish the proof, take [M, β] ∈ W (R) such that (M, β)F = 0 ∈ W (F ) ⇒ ∃V metabolic symmetric
inner product space over R such that (M ⊥ V )F is metabolic over F ⇒ M ⊥ V metabolic over
R ⇒ [M ] = [M ] + [V ] = [M ⊥ V ] = 0 ∈ W (R).
Hence
is a Lagrangian
Theorem 2.59. The sequence of abelian group
⊕p ∂ p
M
0 → W (Z) → W (Q) −→
W (Fp ) → 0
p∈Z≥2 prime
is exact and the map
W (Z) → W (R)
dened by
M 7→ M ⊗Z R
is an isomorphism
W (Z) → W (Q) is injective since Z is a PID and the composition
W (Z) → W (Q) → ⊕p W (Fp ) is zero.
For n ∈ Z≥1 , let Pn be the set Pn = {a ∈ Z \ {0}|∀ prime p : p|a ⇒ p ≤ n} (e.g. P1 =
{+1, −1}, P2 = {+2n , −2n }, Pn ⊂ Pn+1 ). Note that Pn−1 = Pn unless n is prime. Let Ln ⊂ W (Q)
be the subgroup generated by hai with a ∈ Pn . So Ln−1 ⊂ Ln and Ln−1 = Ln unless n is prime. The
Proof. We have already proved that
composition
∂p
Lp−1 ⊂ Lp → W (Fp )
a map of abelian groups
Claim:
is zero because, for
a ∈ Pp−1 , νp (a) = 0
so
∂p (a) = 0.
Hence we get
∂p
Lp /Lp−1 → W (Fp ).
∂p
Lp /Lp−1 → W (Fp )
is an isomorphism for all primes
p ∈ Z≥2 .
The claim will follow from:
Lemma 2.60. If
0 < |n|, |n1 |, . . . , |nk | < p,
and
n ≡ n1 . . . nk mod p,
Lp /Lp−1
22
then
hpni = hpn1 . . . nk i ∈
k . The case k = 1 follows from k = 2 with n2 = 1.
k ≥ 2: Write n1 n2 = pl + r, 0 6= |r| < p. |pl| = |n1 n2 − r| ≤ |n1 ||n2 | + |r| ≤ (p − 1)(p −
1) + p − 1 = p2 − p ≤ p2 ⇒ |l| < p. We will show that hpn1 . . . nk i = hprn3 . . . nk i ∈ Lp /Lp−1 . This is
clear for l = 0 as then both sides are the same. Assume l 6= 0 and write m = n3 . . . nk .
Proof. We use induction on
Assume
hpn1 . . . nk i − hprn3 . . . nk i = hpn1 n2 mi − hprmi
= hpn1 n2 mi − hlmi − hprmi
+ *
*
= hpn1 n2 mi −
p2 lm
| {z }
−
mod Lp−1
+
prm
|{z}
=:v
mod Lp−1
=:u
=
hpn1 n2 mi − hu + vi − huv(u + v)i
hpn1 n2 mi − hpm(pl + r)i − p4 m3 lr(pl + r)
hpn1 n2 mi − hpmn1 n2 i − p4 m3 lrn1 n2
=
− hmlrn1 n2 i = 0
=
=
mod Lp−1
hpn1 n2 . . . nk i = hprn3 . . . nk i where n1 n2 = pl+r, |r| < p. This proves the case k = 2 (and hence
k = 1). Now, the product rn3 , ..., nk has k − 1 factors, and we can apply the induction hypothesis.
Hence
We now construct an inverse of the map in the claim. We dene the map
by
{u} 7→ hpni
where
that is our choice of
1.
n ∈ Z \ {0}, |n| < p, u ≡ n mod p.
n
Note that
does not matter. Need to check that
hui = a2 u , a, u ∈ F∗p .
Fp , |a0 |, |u0 |, |n0 | < p.
Choose
Then
φ
a0 , u0 , no ∈ Z \ {0}
φ
φ : ⊕u∈F∗p Z{u} → Lp /Lp−1
is well dened by the lemma,
preserves the three relations for
W (Fp )
a0 = a, u0 = u, n0 = a2 u ∈
hpu0 i − pa20 u0 = 0 ∈ Lp /Lp−1
such that
φ
{u} − {a2 u} 7→ hpu0 i − hpn0 i
=
lemma
φ
2.
hui + h−ui = 0 ∈ W (Fp ). Choose u0 ∈ Z \ {0}, |u0 | < p, u0 = u ∈ Fp .
hpu0 i + h−pu0 i = 0 ∈ Lp /Lp−1
3.
hui + hvi = hu + vi + huv(u + v)i ∈ W (Fp ), u, v, u + v ∈ F∗p . Choose −p < u0 < 0 < v0 < p, (then
|u0 +v0 | < p, |n0 | < p), |n0 | < p such that u0 , v0 , n0 ∈ Z\{0}, u0 = u, v0 = v, n0 = uv(u+v) ∈ Fp .
Then
Then
{u} + {−u} 7→
φ
{u} + {v} − {u + v} − {uv(u + v)} 7→ hpu0 i + hpv0 i − hp(u0 + v0 )i − hpn0 i
=
lemma
hpu0 i +
hpv0 i − hpu0 + pv0 i − hpu0 pv0 (pu0 + pv0 )i = 0 ∈ Lp /Lp−1
From this it follows that
map
φ̄
lemma
φ
induces a well dened map of abelian groups
is surjective because
hpmi ∈ im(φ̄).
Lp /Lp−1
generated by
It is injective because
hpmi,
φ̄ : W (Fp ) → Lp /Lp−1 . The
q of m are q < p. By the
all prime divisors
∂p
W (Fp ) → Lp /Lp−1 → W (Fp )
is the identity
(∗).
Hence
hui7→hpni7→hni=hui
φ̄
is an isomorphism⇒
(∗)
∼
=
∂p : Lp /Lp−1 → W (Fp )
is an isomorphism. This nishes the claim.
n ≥ 1 that Ln /L1 → ⊕p≤n W (Fp ) is an isomorphism.
n = 1 is clear as both sides are 0
The case n = 2 is true by the claim
n − 1 to n: If n is not a prime then LHSn =LHSn−1 =RHSn−1 =RHSn
If n is a prime, we have a map of short exact sequences:
We prove by induction on
The case
∼
= by induction
/ ⊕p≤n−1 W (Fp )
0
/ Ln /Ln−1
/ ⊕p≤n W (Fp )
/ W (Fn )
sequence
0
/ L1
+
W (Z)
/0
∼
= by claim
/0
∂p
Ln /L1 → ⊕p≤n W (Fp ) is also an isomorphism.
∼
=
W (Q)/L1 = ∪n≥1 Ln /L1 → ∪n≥1 ⊕p≤n W (Fp ) = ⊕p≥1,prime W (Fp ).
By the ve lemma we have that
Hence
/ Ln /L1
/ Ln−1 /L1
0
/ W (Q)
8
/ ⊕p∈Z≥2 prime W (Fp )
3
0
23
So we get the exact
/0
W (Z) ⊂ W (Q) and (⊕∂p )W (Z) = 0 ⇒ W (Z) ⊂ L1 . But L1 ⊂ W (Z) because L1 is generated
h1i , h−1i ⇒ W (Z) = L1 , and we have exactness of 0 → W (Z) → W (Q) → ⊕p W (Fp ) → 0. Finally
the map W (Z) → W (R) ∼
= Z is an isomorphism. This is due to the fact it is surjective since if
Since
by
sgn
U ∈ W (R)
U = n h1i + m h−1i but n h1i + m h−1i ∈ W (Z). It is also injective since W (Z) = L1
is generated by h1i , h−1i, so every element V of W (Z) has the form V = n h1i + m h−1i which is zero
in W (R) ⇐⇒ n − m = sgn(n h1i + m h−1i) = 0 ∈ Z ⇐⇒ n = m ⇐⇒ V = n h1i + n h−1i =
n(h1i + h−1i) = 0 ∈ W (Z). Hence W (Z) → W (R) is an isomorphism
then
Corollary 2.61. The map
M
W (Q) → W (R) ⊕
W (Fp )
p∈Z≥2 prime
dened by
M 7→ (M ⊗Q R,
P
p
∂p M )
is an isomorphism.
0 → W (Z) → W (Q) → ⊕p W (Fp ) → 0,
Proof. This follows from the exact sequence
exact via
W (Z) 
which is split
/ W (Q)
∼
=
&
W (R)
Corollary 2.62. Two symmetric inner product spaces M, N over Q are isometric ⇐⇒ sgn M =
sgn N, rk M = rk N , ∂p M = ∂p N ∈ W (Fp ) ∀p ∈ Z prime. (In terms of quadratic forms, any two
regular quadratic forms are equivalent over Q if and only if the previous condition are fullled)
Proof.
M∼
= N ⇐⇒ rk M = rk N
and
[M ] = [N ] ∈ W (Q) ⇐⇒ rk M = rk N
and
[M ] = [N ] ∈ W (R)
|
{z
}
sgn M =sgn N
and
∂p M = ∂p N
Corollary 2.63 (Weak Hasse Principle). The map1
Y
W (Q) ,→ W (R) ⊕
W (Qp )
Z3p prime
M, N over Q
p ∈ Z prime.
is injective. In particular two inner product spaces
M
and
N
are isometric over
R
and
Qp
for all
are isometric over
Proof. We have the following commutative diagram by denition of
∼
=
W (Q)
(a)
/ W (R) ⊕
L
W (R) ⊕
Q / W (R) ⊕
Q p prime
if and only if
∂p
p prime
(b)
Q
W (Fp )
(c)
W (Qp )
idW (R) ⊕
Q
∂p
p prime
W (Fp )
(a) isomorphism and (c) injective implies (b) injective.
M∼
=Q N ⇒ M ∼
=R N and M ∼
=Qp N for all p ∈ Z prime.
Assume M ∼
N
=R and M ∼
=Qp N for all p ∈ Z prime. Then rk M = rk N and [M ] = [N ] ∈ W (R) and
[M ] = [N ] ∈ W (Qp ) for all p. But (b) injective ⇒ rk M = rk N, [M ] = [N ] ∈ W (Q) ⇐⇒ M ∼
=N
Now
char Q6=2
1 In the lectures I carelessly wrote
p instead of
p but the image of W (Q) does not lie in
the image of h1i which is 6= 0 ∈ W (F ) for any eld F ?
L
Q
24
p,
L
otherwise, what is
associated form matrix
B = (βij )
Pn
2
i=1 ai xi
(with respect to e1 , e2 , e3 , . . . , en ) where βij =
Example. We start with two side remarks: The quadratic form
q=
P
+ i<j aij xi xj has
q(ei + ej ) − q(ei ) −


aij
q(ej ) = aij


2ai
That is,
i<j
j<i
i=j

2a1
 a12

B= .
 ..
a1n
a12
2a2
...
..
a1n



.

.
2an
a1 a12 ... a1n
E
D
a12 a2
 is hBi = d1 , d2 , . . . , dn
The diagonalisation of a symmetric matrix B =  ..
..
d1
dn−1
.
.
a1n
an
where di = determinant of the upper left corner of size i × i of B , provided d1 , . . . , dn−1 6= 0.

1. Does
15 = x2 + 2xy + 3y 2 − 4yz
2
x, y, z ∈ Q?
2
q = x + 2xy + 3y − 4yz . Does q represent 15? The
β(u, v) = q(u + v) − q(u) − q(v), u, v ∈ Q3 has form matrix


2 2
0
B = 2 6 −4
0 −4 0
Solution: Let
form
have a solution

which has determinant −32 hence it is non-degenerate.
+
*
2, 1, −1
| {z }
⇒
exercise
q
associated symmetric bilinear
It has diagonalisation
∼
hBi ∼
= 2, 28 , −32
=
8
isotropic and represent any rational number. In particular there exists
x, y, z
H
such that
Note
2. Does
q(x, y, z) = 15
that hBi ∼
= h2, 4, −4i ⇒ q ∼
= x2 + 2y 2 − 2z 2 ∼
= x2 + yz
15 = x2 + 4xy − 2xz + 7y 2 − 4yz + z 2 =: q
Solution: The associated bilinear form
β
of
q
has a solution
x, y, z ∈ Q.
has matrix form

2
B= 4
−2

−2
−4
2
4
14
−4
= 0 (since Be3 = −Be1 ). So q is degenerate, and we can eliminate a variable as
hBi has diagonalisation hBi ∼
(Qe1 + Qe2 )
+
=
{z
}
|
non−degenrate as det( 2 4 )=126=0
4 14
2 4
⊥
12
∼
∼
∼
(Qe1 + Qe2 )
⊥ h0i = 2, 2 ⊥ h0i = h2, 6i ⊥ h0i. This means that
=
4 14
{z
}
|
with determinant
follows: The inner product space
dim=1,degenerate as det B=0
q ∼
= x2 + 3y 2 , so does this represent 15? This is equivalent to asking
a ∈ Q∗ . Then det LHS = det RHS modulo square units ⇐⇒ h1, 3i
h15, 5i ∈ W (Q) (because h1, 3i and h15, 5i have the
∂p h1, 3i ∼
= ∂p h15, 5i ∈ W (Fp ) for all p prime.
same rank)
h2, 6i ∼
= h30, ai for some
∼
= h15, 5i ⇐⇒ h1, 3i =
⇐⇒ h1, 3i ∼
= h15, 5i ∈ W (R)
and
•
If
•
if
p 6= 3
or
5
then
∂p h1, 3i = 0 = ∂p h15, 5i
p = 3 then ∂3 h1, 3i = ∂3 h1i + ∂3 h3i = 0 + h1i, and ∂3 h15, 5i = ∂3 h15i + ∂3 h5i = h5i + 0 =
h−1i. Do they agree in W (F3 )? No because h1i =
6 h−1i ∈ W (F3 ) ∼
= Z/4Z generated by
h1i⇒ h1i − h−1i = 2 h1i =
6 0 ∈ W (F3 )
h1, 3i 6= h15, 5i ∈ W (Q) ⇒ q
x, y, z ∈ Q
We have showed that
no solution in
25
does not represent 15 and the equation has
2.6
The Brauer Group and the Hasse Invariant
Recall from MA377 (Rings and Modules):
Denition 2.64. Let
k
be a eld, a
∼
=
•
central: if
k → Z(A),
•
simple: if
A 6= 0
•
nite dimensional : if
Fact. Let
A, B
where
k -algebra A
Z(A)
is called:
denotes the center of
and the only ideals of
A
are
0
A = Mn (D)
2.
A ⊗k B
3.
A ⊗k Aop ∼
= Mn (k)
where
D
A
dimk A < ∞
be nite dimensional central simple
1.
and
A
k -algebras.
k -algebra
is a nite dimensional division
is also a nite dimensional central simple
where
Then:
k -algebra
n = dimk A
Denition 2.65. Let
F be a eld. The Brauer group, Br(F ), is the set of Brauer equivalence classes
[A] of nite dimensional central simple F -algebras A, where A ∼ B (A is Brauer equivalent to B ) if
Mm (A) ∼
= Mn (B) as F -algebras for some m, n ∈ N≥1 .
Br(F ) is a group with group law: [A][B] := [A ⊗F B], with 1 = [F ] and [A]−1 = [Aop ]. Indeed
Br(F ) is an abelian group: [A][B] = [A ⊗F B] = [B ⊗F A] = [B][A], 1[A] = [F ][A] = [F ⊗F A] = [A],
[A][Aop ] = [A ⊗F Aop ] = [Mn (F )] = [F ] = 1
Fact
Example. (From MA377)
• Br(C) = Br(F ) = {F } = 0
• Br(F ) = 0
if
F
where
F = F̄
is algebraically closed
is a nite eld
• Br(R) = {R, H} = Z/2
F be a eld with char F 6= 2 and a, b ∈ F ∗ .
1, i, j, k such that i2 = a, j 2 = b, k = ij = −ji
Denition 2.66. Let
F -algebra
Note.
with basis,
( a,b
F )
be the
4-dimensional
k 2 = ij(−ji) = −ab
Fact. For
a, b ∈ F ∗ , ( a,b
F )
Denition 2.67. An
is a
4-dimensional
F -algebra
which is
(generalized) quaternion algebra (over
central simple
F -algebra
F -algebra.
isomorphic to
( a,b
F )
for some
a, b ∈ F ∗
is called
F)
Structure theorem for Quaternion algebras. Let
( c,d
F )
Let
F
be a eld with char
F 6= 2.
Then
∼
( a,b
F ) =
⇐⇒ ha, b, −abi = hc, d, −cdi ∈ W (F )
A, B be nite dimensional central simple F -algebras. Then A ∼
= B ⇐⇒ dimF A =
dimF B and [A] = [B] ∈ Br(F ).
a,b
c,d
In particular [(
F )] = [( F )] ∈ Br(F ) ⇐⇒ ha, b, −abi = hc, d, −cdi ∈ W (F )
0 1
1 0
1,1 ∼
Example. ( F ) = M2 (F ) by i 7→
, j 7→
1 0
0 −1
−1,−1
( R ) = Real quaternion algebra.
−1,−1
M2 (R) = ( 1,1
1, −1i 6= h−1, −1, −1i ∈ W (R)
R ) ( R ) because h1,
| {z } |
{z
}
Remark. Let
sgn=1
Remark.
( a,b
F )
sgn=−3
⇐⇒ ( a,b
F ) M2 (F ) ⇐⇒ ha, b, −abi h1, 1, −1i ⇐⇒ ha, b, −abi
represent 0)
is a division algebra
is an isotropic (i.e., does not
∼ a,b op by 1 7→ 1, i 7→ −i, j 7→ −j, k 7→ −k . This means that [( a,b )] has order 2 in
( a,b
F ) = ( F )
2
a,b ∼ a,b
a,b op ∼
a,b
a,b
Br(F ) because ( a,b
)⊗
(
)
(
)⊗
(
)
M
(F
)
⇒
[(
)][(
)]
=
[M
(F
)]
=
[F ] = 1 ∈ Br(F ).
=
=
F
F
4
4
F
F
F
F
F
F
a,b
2
Hence [(
)]
∈
Br(F
)
,
where
for
an
abelian
group
G
we
denote
G
=
{x
∈
G|x
=
1}
2
2
F
Remark.
26
Lemma 2.68. Let
F
be a eld with char
1.
∼ a,−ab ∼ b,−ab
( a,b
F )=( F )=( F )
2.
a,c ∼ a,bc
( a,b
F ) ⊗F ( F ) = ( F ) ⊗F M2 (F )
Proof.
1.
∼ a,−ab
( a,b
F )=( F )
because
F 6= 2.
Then:
ha, b, −abi ∼
= a, −ab, a2 b
a,c
A = ( a,b
F ), B = ( F ) have basis BA = {1, iA , jA , kA } and BB = {1, iB , jB , kB } respectively.
Then A⊗F B has basis {u⊗v|u ∈ BA , v ∈ BB }. Let ΣA = {1⊗1, iA ⊗1, jA ⊗jB , kA ⊗jB } ⊂ A⊗F B
0
0
and ΣB = {1 ⊗ 1, 1 ⊗ jB , iA ⊗ kB , −ciA ⊗ iB }. Then ΣA , ΣB are the basis of A , B ⊂ A ⊗F B 2
0 ∼ c,−a c
0
) because
subalgebras with A ∼
= ( a,bc
F ) and B = ( F
2. Let
• (iA ⊗ 1)2 = i2A ⊗ 1 = a(1 ⊗ 1) = a
2
2
• (jA ⊗ jB )2 = jA
⊗ jB
= b ⊗ c = bc(1 ⊗ 1) = bc
• (1A ⊗ 1)(jA ⊗ jB ) = kA ⊗ jB = −(jA ⊗ jB )(iA ⊗ 1)
and
2
• (1 ⊗ jB )2 = 1 ⊗ jB
=c
2
= a · (−ac) = −a2 c
• (iA ⊗ kB )2 = i2A ⊗ kB
• (1 ⊗ jB )(iA ⊗ kB ) = −(iA ⊗ kB )(1 ⊗ jB ) = iA ⊗ −ciB = −ciA ⊗ iB
c,−a2 c ∼ 1,1
But (
) = ( F ) = M2 (F ) because c, −a2 c, a2 c2 ∼
= hc, −c, 1i ∼
= h1, 1, −1i. Every element
F
0
0
of A commutes with every element of B (one checks that ΣA commutes with ΣB )⇒ The map
φ :A0 ⊗F B 0 → A ⊗F B dened by x ⊗ y 7→ xy is a well dened map of F -algebras. The elements
{xy|x ∈ ΣA , y ∈ ΣB } are linearly independent in A ⊗F B (check!) ⇒ φ is injective. Since
∼
dimF A0 ⊗F B 0 = 8 = dimF A ⊗F B , this means that φ is an isomorphism ⇒ ( a,bc
F ) ⊗F M2 (F ) =
a,b
a,c
0
0 ∼
∼
A ⊗F B = A ⊗F B = ( F ) ⊗F ( F )
Denition 2.69. Let
F
with diagonalisation
F be a eld with char F 6= 2, and V be a symmetric inner
V ∼
= ha1 , . . . , an i. The Hasse invariant of V is the algebra
Y ai , aj Hasse(V ) =
∈ 2 Br(F )
F
product space over
1≤i<j≤n
Lemma 2.70.
Hasse(V ) ∈2 Br(F )
does not depend on diagonalisation
ha1 , . . . , an i
of
V
used to dene
Hasse(V )
Proof.
1.
Hasse(ha1 , . . . ai , ai+1 , . . . , an i) = Hasse(ha1 , . . . , ai−1 , ai+1 , ai , ai+2 , . . . , an i).
This is be-
cause

LHS
=
RHS
=

a , a a , a i i+1
r s 

F
F
r<s,{r,s}6={i,i+1}


a , a a , a Y
r s 
i+1 i

F
F
Y
r<s,{r,s}6={i,i+1}
Since
∼
( a,b
F ) =
b,a
F
⇒
LHS = RHS. Hence for all
σ ∈ Σn =
permutation group, we have
Hasse(ha1 , . . . , an i) = Hasse( aσ(1) , . . . , aσ(n) )
2. char
6= 2,
if
ha1 , . . . , an i
and
hb1 , . . . , bn i
are diagonalisation of
V
then
ha1 , . . . , an i ≈ hb1 , . . . , bn i
Hasse(ha1 , . . . , an i) =
(Chain equivalence Theorem on page 13). Hence it suces to show that
Hasse(hb1 , . . . , bn i)
for
ha1 , . . . , an i ≈s hb1 , . . . , bn i
27
simply chain equivalent.
By 1.
it suces
Hasse(ha, b, e1 , . . . , en i) = Hasse(hc, d, e1 , . . . , en i) where ha, bi ∼
= hc, di.
∼
ha, bi = hc, di ⇐⇒ ab = cd · x2 , a = cy 2 + dz 2 for some x, y, z ∈ F . Now
n
a, b Y a, ei b, ei
LHS
=
Hasse(he1 , . . . , en i)
F i=1 F
F
n
c, d Y c, ei d, ei
RHS
=
Hasse(he1 , . . . , en i)
F i=1 F
F
to show
Recall that
Note that
a, e b, e F
F
ab, e
∈ Br(F )
F
cd, e
2
since ab = cdx
F
c, e d, e F
F
=
=
=
and
∼ c,d
( a,b
F )=( F )
Lemma 2.71.
because
ha, b, −abi = ha, bi + h−abi = hc, di + h−cdi = hc, d, −cdi
W
Hasse(V ⊥ W ) = Hasse(V ) Hasse(W ) · ( det V,det
)
F
Proof. Exercise
So Hasse(−) does not dene a group homomorphism W (F ) → 2 Br(F ). Hasse(H) = Hasse(h1, −1i) =
2
∼ 1,1
∼
( 1,−1
F ) = ( F ) = M2 (F ) because h1, −1, 1i = h1, 1, −1i. But Hasse(H ) = Hasse(H) Hasse(H) ·
−1,−1
1,1
det H,det H
) = ( F ) 6= ( F ) = M2 (F ) = F ∈ Br in general.
(
F
2.7
Tensor Product of Inner Product Spaces
Denition 2.72. Let
(M, β), (B, γ) be symmetric bilinear forms over R. We dene (M ⊗R N, β ⊗R γ)
β ⊗ γ : M ⊗R N × M ⊗R N → R dened by (x ⊗ u, y ⊗ v) 7→ β(x, y) · γ(u, v),
β ⊗ γ(x ⊗ u, y ⊗ v) = β(x, y)γ(u, v) = β(y, x)γ(v, u) = β ⊗ γ(y ⊗ v, x ⊗ u)
to be the bilinear form
which is symmetric:
Lemma 2.73. Let
P, Q be nitely
HomR (Q, R) → HomR (P ⊗R Q, R)
generated
dened by
R-module then
f ⊗ g 7→ f · g
φ : HomR (P, R) ⊗R
(f · g)(x ⊗ u) = f (x)g(u), is an
the following map
where
isomorphism
φ is an isomorphism for (P, Q) = (R, R).
φ is an isomorphism for (P1 , Q) and (P2 , Q) then φ is an isomorphism for (P1 ⊕ P2 , Q) because
(P1 ⊕ P2 ) ⊗ Q = P1 ⊗ Q ⊕ P2 ⊗ Q, Hom(P1 ⊕ P2 , R) = Hom(P1 , R) ⊕ Hom(P2 , R) and φ1 ⊕ φ2 is an
m
n
isomorphism if and only if φ1 and φ2 are isomorphisms. ⇒ φ is isomorphism for (P, Q) = (R , R )
m, n ∈ Z≥0 .
n
A nitely generated projective module is a direct factor of R for some n. If φ1 is a direct summand
of a map φ which is an isomorphism then φ1 is an isomorphism ⇒ φ is an isomorphism for P, Q nitely
Proof.
If
generated projective modules.
Lemma 2.74. Let
(M, β), (N, γ)
R
be symmetric inner product spaces over
R.
Then
(M ⊗R N, β ⊗ γ)
is an inner product space over
M, N is nitely generated projective ⇒ M ⊗R N is nitely generated projective. We need
β ⊗ γ is non-degenerate. Now β, γ non-degenerated ⇐⇒ M → HomR (M, R) dened by
x 7→ β(x, −) and N → HomR (N, R) dened by y 7→ γ(y, −) are isomorphisms. β ⊗ γ is non-degenerate
⇐⇒ M ⊗R N → HomR (M ⊗R N, R) dened by x ⊗ y 7→ β(x, −)γ(y, −) is an isomorphism, but this
Proof.
to show
map is the composition of the following two maps
∼
=
∼
=
M ⊗R N −→ HomR (M, R) ⊗ HomR (N, R) −→ HomR (M ⊗R N, R)
Lemma
x ⊗ y 7→ β(x, −) ⊗ γ(y, −) 7→ β(x, −) · γ(y, −)
28
Lemma 2.75 (Denition). The Wilt group
a a commutative ring
with multiplication
and unit
W (R) of
[M, β] · [N, γ] = [(M, β) ⊗R (N, γ)]
h1i. W (R)
R
is a commutative ring
is called the Witt ring of
R.
(M, β) is metabolic and (N, γ) arbitrary then (M, β) ⊗ (N, γ) is
L ⊂ M of (M, β) denes a Lagrangian L ⊗ N ⊂ M ⊗ N of (M ⊗ N, β ⊗ γ)
Proof. We need to show that if
metabolic. But a Lagrangian
(exercise)
Remark.
hui · hvi = huvi ∈ W (R)
Denition 2.76. Let
R
be a local ring then the rank map
ring homomorphism. The kernel
ker(rk)
is an ideal
I(R)
W (R) → Z/2
dened by
M 7→ rk M
is a
which is called the fundamental ideal.
I(F ) is generated by even dimensional forms, hence additively generated by 2 dimensional
ha, bi = ha, 1i − h−b, 1i ⇒ I(F ) is additively generated by ha, 1i , a ∈ F ∗ ⇒ I 2 (F ) is additively
∗
2∗
generated by ha, 1i ⊗ hb, 1i = hab, a, b, 1i, the discriminant map, disc : I(F ) → F /F
dened by
dim V
2
2
∗
2∗
V 7→ (−1) 2 det V , in our case we have hab, a, b, 1i 7→ a b = 1 ∈ F /F hence disc(I 2 ) = 0 and
I(F )/I 2 (F ) → F ∗ /F 2∗ well dened surjective map of abelian groups
Remark.
forms
Theorem 2.77 (Pster). The map
I(F )/I 2 (F ) → F ∗ /F 2∗
Proof. The map is surjective because
2
In
If
W (F )/I
disc
I(F ) → F ∗ /F 2∗
is an isomorphism for all elds
sends
ha, −1i
to
a
for
F.
a ∈ F ∗.
we have:
1.
hai + hbi = h−abi + h−1i
2.
3 h−1i = h1i
because
ξ = hu1 , . . . , un i ∈ I/I 2
because
hab, a, b, 1i ∈ I 2
h1, 1, 1, 1i = h1, 1i ⊗ h1, 1i ∈ I 2 ,
then
n = 2m.
For
ξ = hu1 , . . . , u2m i =
1.
u = dicsξ
hence
4 h−1i = 0.
we have
h−(−1)m u, −1, . . . , −1i
| {z }
in
I/I 2
2m−1
=
2.
=
hu, −1, −1, −1i m
hu, −1i
m
even
odd
h−u, 1i
m is even, and when m is odd we have hu, −1i = h−u, 1i
hu, u, −1, −1i = h−u2 , −1, −1, −1i = h−1, −1, −1, −1i = 0 ∈ I/I 2 , by 1 and 2. Thus, if ξ is
∗
2∗
in the kernel of the discriminant map then 1 = disc(ξ) = disc(−u, 1) = u ⇒ u = 1 ∈ F /F
⇒ξ=
2
2
2
∗
2∗
h−u, 1i = h−1, 1i = 0 ∈ I/I ⇒ ξ = 0 ∈ I/I and the map I/I → F /F is injective.
where the last equation follows from 2 when
because
• I 2 (Fq ) = 0
Example.
• I 2 (F ) = I(F ) = 0
• I(R) 
/ W (R)
because disc
∼
=
: I(Fq ) → F∗q /F2∗
q
for any algebraically closed eld
rk
F
is an isomorphism.
because
∼
=
rk : W (F ) → Z/2Z
/ Z/2Z
>
sgn
reduction mod 2
2Z ⊂
•
Z
We'll see later:
I 2 (Qp ) = Z/2Z
Denition 2.78. Let
(=
det V ) = 1.
V
The Signed
4k -dimensional symmetric inner product space over F with
(−1)k −1
−1,−1 k
Hasse Invariant is s(V ) = ( F ) Hasse(V ) = (
) Hasse(V )
F
be a
29
discV
Note. If
V, W
have dimension divisible by 4 and discV
s(V ⊥ W )
= discW = 1
then
=
−1, −1 dim V +dim W
4
Hasse(V ⊥ W )
(
)
F
dim4 V dim4 W
−1, −1
−1, −1
det V, det W
Hasse(V ) Hasse(W )
F
F
F
{z
}
|
1,1
(F )
=
s(V )s(W )
=
s(H2 ) = ( −1,−1
) Hasse(H2 ) = ( −1,−1
)( −1,−1
) = [F ] ∈ Br(F ) ⇒ s(H2k ) = [F ] ∈ Br(F ). Hence
2
F
F
2
s : I (F ) → 2 Br(F ) is a well dened map of abelian groups. (as I 2 is generated by 4-dimesional
and
spaces)
Lemma 2.79.
s(I 3 F ) = 0
I is generated
ha1 , . . . , a8 i. So,
Proof.
s(ha1 , . . . , a8 i =
=
=
h1, ai ⇒ I 3
by
=
=
=
=
=
=
=
=
of char
is generated by
6= 2
h1, ai ⊗ h1, bi ⊗ h1, ci = h1, a, b, c, ab, ac, bc, abci =
8
by Lemma
F
4
2.68
a, a3 b4 c4
b, a3 b3 c4
c, a3 b3 c3
ab, a2 b2 c3
ac, ab2 c2
bc, abc
1, a4 b4 c
F
F
F
F
F
F
F
1, 1
a, a
b, ab
c, abc
ab, c
ac, a
bc, abc
removing powers of 2
F
F
F
F
F
F
F
a, a b, −a c, −ab ab, c a, −c bc, −a a, b
a, −ab
by using the relation
=
F
F
F
F
F
F
F
F
a, a a, −c b, −a bc, −a c, −ab c, ab by rearranging
F
F
F
F
F
F
c, −a
c, −1
a, −ac
pairing o and Lemma 2.68
F
F
F
a, −ac c, a Lemma 2.68 on the last two pairs
F
F
a, −ac2
Lemma 2.68
F
a, −a
removing powers of 2
F
1, 1
2
because a, −a, a
= h1, 1, −1i
F
0
=
F
(−1) 4 Hasse(ha1 , . . . , a8 i)
Y ai , aj F
i≤i<j≤8
Q
Y ai i<j≤8 aj 1≤i≤7
=
for every eld
Corollary 2.80. The signed Hasse invariant gives a well dened map of abelian groups I 2 (F )/I 3 (F )
2
→
Br(F )
Theorem 2.81 (Merkurev, 1981). The map
Remark.
∼
=
I 2 F/I 3 F →
2
Br(F )
I 0 /I 2 = W (F )/I = Z/2Z, I/I 2 = F ∗ /F 2∗ , I 2 /I 3 =
For any eld
F
Br(F ),
what about
6= 2)
I k /I k+1 =?
H n (F, Z/2Z), sometimes called Galois coH (F, Z/2Z) = Z/2Z, H 1 (F, Z/2Z) = F ∗ /F 2∗ and H 2 (F, Z/2Z) =
there are dened cohomology groups
homology groups, which satisfy
2
is an isomorphism (char(F )
0
30
2
Br(F )
for any eld
F
of characteristic
6= 2.
This makes the statement of the following theorem
plausible. For its proof and the development of the tools needed in the proof (motivic cohomology and
motivic homotopy theory), Voevodsky was awarded the elds medal in 2002.
Theorem 2.82 (Voevodsky, conjectured by Milnor). Let
F
be a eld of char
6= 2
then
I n (F )/I n+1 (F ) ∼
= H n (F, Z/2Z)
Lemma 2.83. Let
F
≤ 3.
Hasse W ∈ Br(F )
Then
of dimension
6= 2 and V, W symmetric inner product spaces over F
V ∼
= W ⇐⇒ dim V = dim W, det V = det W ∈ F ∗ /F 2∗ and Hasse V =
be a eld with charF
Proof. ⇒ is clear
dim V = dim W = 1: V ∼
= hdet V i = hdet W i ∼
= W , hence we are done.
dim V = dim W = 2: Then V ∼
= ha, bi , W ∼
= hc, di (charF 6= 2). ( a,b
F ) = Hasse(V ) = Hasse(W ) =
∼
∼
ha,
bi
hc,
di
(Witt
cancellation)
)
⇒
ha,
b,
−abi
hc,
d,
−cdi
⇒
( c,d
=
=
F
∗
2∗
⇐:
ab=cd∈F /F
dim V = dim W = 3: V ∼
) Hasse(h−ab, −ac, −bci)
= ha, b, ci , W ∼
= hx, y, zi , a, b, c, x, y, z ∈ F ∗ . Hasse(ha, b, ci) = ( −abc,−1
F
(Exercise).
Hasse(V ) = Hasse(W )
⇒
but
,
abc = det V = det W = xyz
⇒ Hasse(h−ab, −ac, −bci) = Hasse(h−xy, −xz, −yzi)
(∗)
−ab, −bc
−ac, −bc
−xy, −xz
−xz, −yz
−xz, −yz
−ab, −ac
=
F
F
F
F
F
F
−ab, ab
−ac, −bc
−xy, xy
−xz, −yz
⇒
=
F
F
F
F
h−ab, ab, 1i ∼
= h1, 1, −1i
−ac, −bc
−xz, −yz
∼
⇒
=
F
F
2
∼
⇒ −ac, −bc, −abc
= −xz, −yz, −xyz 2
⇒ h−abci ⊗ h−ac, −bc, −abi ∼
= h−xyzi ⊗ h−xz, −yz, −xyi
( −ab,ab
) = ( 1,1
F
F )
because
⇒ hb, a, ci ∼
=
Proposition 2.84. Let
as
h− det V i = h− det W i
hy, x, zi
6= 2. Assume that every 5-dimensional symmetric
0 non-trivially. Then for symmetric inner product spaces
V, W over F , V ∼
= W ⇐⇒ dim V = dim W, det V = det W ∈ F ∗ /F 2∗ , Hasse(V ) = Hasse(W ) ∈ Br(F )
F
be a eld with charF
inner product space is isotropic, i.e., represent
Remark. Proposition applies when
F = Qp
(See below).
(Also if
F =
any local eld, or non-real
number eld)
Proof. Induction on
n ≤ 3:
n = dim V = dim W
This case is the previous lemma
Assume n ≥ 4. V ⊥ h−1i has dimension ≥ 5
V ∼
= V0 ⊥ h1i. Similarly W ∼
= W0 ⊥ h1i. Now
hence it is isotropic.
⇒ V ⊥ h−1i ∼
= V0 ⊥ h1, −1i ⇒
• dim V0 = dim W0 = n − 1.
• det V0 = det V0 · det h1i = det V = det W = det W0
• Hasse(V0 ⊥ h1i) = Hasse(V ) = Hasse(W ) = Hasse(W0 ⊥ h1i) ⇒ Hasse(V0 ) · Hasse(h1i) ·
( detFV0 ,1 ) = Hasse(W0 ) · Hasse(h1i) · ( det FW0 ,1 )
⇒
Hasse(V0 ) = Hasse(W0 )
det V0 =det W0
So by induction hypothesis
V0 ∼
= W0 ⊥ h1i = W
= W0 ⇒ V = V0 ⊥ h1i ∼
31
Corollary 2.85. Let
F
be a eld with charF
6= 2
for which every 5-dimensional form is isotropic.
I 3F = 0
Then
V be a symmetric inner product space over F , [V ] ∈ I 3 F ⊂ I(F ) ⇒ dimF V = 2k .
If 4 - dim V replace V with V ⊥ H, this doesn't change [V ] = [V ⊥ H]. Hence we can assume
dim V = 4 · l for some l ∈ N. Now
Proof. Let
• dim V = 4l = dim H2l
• det V = (−1)
dim V
2
discV
=1
[V ] ∈ I 2
because
l
and
l
• Hasse(V ) = ( (−1)F ,−1 ) s(V ) = ( (−1)F ,−1 ),
| {z }
since
(−1)
dim V
2
= (−1)2l = 1.
But
[V ] ∈ I 3 ⊂ ker(s : I 2 → Br).
det H2l = 1
But
Hasse H2l =
[F ]∈Br
l
( (−1)F ,−1 )
So by the proposition we have
2.8
V ∼
= H2l ⇒ [V ] = 0 ∈ W (F )
Quadratic Forms over
Denition 2.86. The
p-adic
Zp
=
=
=
p-adic
integers
numbers
are (p
Zp
∈Z
prime)
n
lim Z/p Z
n→∞
{(xn )n∈N≥1 |xn ∈ Z/pn Z, xn+1 ≡ xn
∞
X
{
ai pi |ai ∈ {0, . . . , p − 1}}
mod pn }
i=0
=
Zp
completion of
Z
with repsect to
is a Discrete Valuation Ring with maximal ideal
Denition 2.87. The
p-adic
Qp
rational numbers
Qp
=
eld of fractions of
=
completion of
= {
∞
X
Q
pZp
||a||p = p−νp (a)
and residue eld
Zp /pZp = Fp
are
Zp
with respect to
||a||p = p−νp (a)
ai pi |ai ∈ {0, . . . , p − 1}, N ∈ Z}
i=N
P∞
Pk
i
i
n
Zp Z/pn Z by
i=1 ai p P7→
i=1 ai p mod p
∞
i
⇐⇒ x ∈ Zp /pZp = Fp is a unit (Zp local). So i=0 ai p ∈ Zp is a unit
We have the surjective ring homomorphism
(k ≥ n − 1). x ∈ Zp
⇐⇒ a0 6= 0 in Fp .
is a unit
∗
2∗
For p = 2 we rst look at Z2 /Z2 .
Q∗p /Q2∗
p . If p is odd this is an exercise.
P∞
i
We have a ring homomorphism Z2 → Z/8Z dened by
i=0 ai 2 7→ a0 + a1 2 + a2 4. Therefore
P∞
(Z2 )∗ (Z/8Z)∗ is surjective by the map 1 + i=1 ai 2i 7→ 1 + a1 2 + a2 4. Now (Z2 )2∗ → (Z/8Z)2∗ =
{12 , 32 , 52 , 72 } = {1}, so we have a well dened group homomorphism (Z2 )∗ /(Z2 )2∗ (Z/8Z)∗ .
P∞
∗
i
Proposition 2.88. The map Z∗2 /Z2∗
2 → (Z/8Z) dened by
i=0 ai z 7→ a0 + a1 2 + a2 4 is an isomorWe want to understand
phism.
a1 , a2 = 0 ⇐⇒ x = 1 + 8y
for some
y∈
z
P∞
i
i=1 ai 2 ∈ kernel of the map
Z2 . We need to sow that x is a square in Z2 .
Proof. We already know that the map is surjective.
z = 1+
(1 + 8y)1/2
∞ X
1/2
:=
(8y)k
k
k=0
∞ X
1/2
=
4k (2y)k
k
=
=
k=0
∞
X
bk (2y)k
k=0
32
⇐⇒
where
1/2
4k
k
1/2(1/2 − 1) · · · · · (1/2 − k + 1)
=
4k
k!
(1/2)k · 1 · (−1) · · · · · (−2k + 3) k
=
4
k!
2k
= (−1)k−1 · 1 · 3 · · · · · (2k − 3) · .
k!
bk
=
k = ν2 (2k ) ≥ ν2 (k!) since ν2 (k!) P
≤ (number ofP
even number ≤ k ) + (number of number divisible
∞
∞
4 ≤ k ) + · · · ≤ b k2 c + b k4 c + · · · ≤ i=1 2ki = k2 i=0 21i = k2 1−1 1 = k2 2 = k . Hence ν2 (bk ) ≥ 0 and
2
since Z2 = {t ∈ Q2 |ν2 (t) ≥ 0} we have that bk ∈ Z2 .
∞
m
X
X
k k−n n
k
k
b
y
2
=
b
y
·
2
k
k
k2 k2
|{z}
k=n
k=n ∈Z2
|
{z
}
2
Now
by
∈Z2
2
−n
≤ 1·2
||a||2 = 2−ν2 (a) ≤ 1 for all a ∈ Z2 . Hence m 7→
k k
2
k=0 bk y 2 denes an element in Z2 . Then z = x.
where
P∞
Remark. For
p
odd
∼
=
2∗
∗
Z∗p /Z2∗
p → Fp /Fp
(reduction mod
p)
Pm
k=0 bk y
k k
2
is a Cauchy sequence
⇒ z :=
is an isomorphism (exercise)
Corollary 2.89. The map
∼
=
Q∗p /Q2∗
p →
dened by
Z/2Z ×
pν a 7→ ν, a
Z∗2 /Z2∗
p
where
(
Z/2Z × F∗p /F2∗
p
=
Z/2Z × (Z/8Z)∗ = Z/2Z × (Z/2Z)2 = (Z/2Z)3
a ∈ Z∗p is
p odd
p=2
an isomorphism.
∼
=
F , the map F ∗ → Z × R∗ dened
∼
∗
2∗ =
by p a 7→ ν, a where a ∈ R is a isomorphism. Hence F /F
→ Z/2Z × R∗ /R2∗ . Now (Z/8Z)∗ is
2
2
generated by 3, 5 and 3 = 5 = 1 mod 8, hence (Z/8Z) ∼
= Z/2Z ⊕ Z/2Z.
Proof. For any Discrete Valuation Ring
ν
∗
Corollary 2.90.
Z∗2 /Z2∗
2 = {1, 3, 5, 7}
Proposition 2.91. Let
4-dimensional
Q∗p /Q2∗
p .
p ∈ Z
R
and
with eld of fractions
Q∗2 /Q2∗
2 = {1, 3, 5, 7, 2, 6, 10, 14}.
be a prime.
regular quadratic form over
Then there is, up to isometry, a unique anisotropic
Qp .
This form has determinant
1
and represents all of
p = odd (exercise)
∗
2∗
p = 2: Consider all possible 2-dimensioanl forms h1, ai where a ∈ Q∗2 /Q2∗
2 . Set Da = {t ∈ Q2 /Q2 |t
represent h1, ai}
h1, ai Da ⊂ Q∗2 /Q2∗
2 = {1, 3, 5, 7, 2, 6, 10, 14}
h1, 1i
1, 2, 5, 10
h1, 2i
1, 2, 3, 6
h1, 3i
1, 3, 5, 7
h1, 5i
1, 5, 6, 14
h1, 6i
1, 6, 7, 10
h1, 7i
Hyperbolic
h1, 10i
1, 3, 10, 14
h1, 14i
1, 2, 7, 14
2
2
2
2
2
2
We check this table for h1, 1i: This represent 1, 2, 5, 10 because 1 = 1·1 +1·0 , 2 = 1 +1 , 5 = 2 +1
2
2
2
2
and 10 = 3 + 1 , and it does not represent 3, 7, 6, 14 because x + y ∈ {3, 7, 6, 14} has no solution
Proof.
33
Z2 since it has no solution mod 8 as x2 + y 2 ∈ {0, 1, 2} mod 8 since x2 , y 2 ∈ {0, 1} mod 8. If
x + y 2 = a ∈ {3, 7, 6, 14} has a solution in Q2 clearing denominators (multiplying with respect to 2n )
(∗) x2 + y 2 = at2 has a solution in Z2 and not all of x, y, t are divisible by 2.
in
2
t ∈ Z∗2 ⇒ t2 = 1 mod 8
Case 1.
2-t
Case 2.
t = 2u and 2 - x then x2 = 1 mod 8, |{z}
x2 +y 2 = 4u2 a has no solution mod 8 as y 2 ∈ {1, 0}
then
and
(∗)
has no solution
mod 8
1
mod 8.
h1, 1i
3, 7, 6, 14.
h1, 1i ∼
= h2, 2i ∼
= h5, 5i ∼
= h10, 10i h3, 3i ∼
= h7, 7i ∼
= h6, 6i ∼
= h14, 14i.
∼
∼
∼
e.g., h1, 1i = h2, 2i = h5, 5i = h10, 10i since h1, 1i represents 2,5 and 10 and the all have the same
determinant. Now h1, 1i h3, 3i = h3i·h1, 1i because h1, 1i represent 1, 2, 5, 10 but h3i·h1, 1i represents
∗
3, 6, 15 = 7, 30 = 14 ∈ Q∗2 /Q2∗
2 = Z/2Z × (Z/8Z) .
Let φ be a 4-dimensional anisotropic form over Q2 , φ = hd, . . . i, then ψ = hdi φ is also anisotropic
2
∗
2∗
∗
and represent d = 1 ∈ Q2 /Q2 . So ψ = h1, a, −b, −bci for some a, b, c ∈ Q2 . Rewrite this as
ψ = h1, ai ⊥ h−bi · h1, ci. If h1, ai and hbi · h1, ci represent a common element, then ψ represent 0
which contradicts the fact that ψ is anisotropic. Note also that a, c 6= 7 because h1, ai and h1, ci are
not hyperbolic. Therefore Da ∩ bDc = ∅
⇒
Da t bDc = Q∗2 /Q2∗
2 . We can use the table to see
Hence
does not represent
We also can check that
|Da |=|Dc |
Da ⊂ Q∗2 /Q2∗
2
is a subgroup.
Now
1 ∈ Da , Dc , 1 ∈
/ bDc
⇒
Dc subgroup
bDc ∩ Dc = ∅ ⇒ Dc t bDc =
∗
2∗
Da t bDc = Q∗2 /Q2∗
2 ⇒ Da = Dc ⇒ a = c ∈ Q2 /Q2 ⇒ ψ = h1, a, −b, −abi ⇒ det ψ = 1. Now
table
2
φ = d · ψ = hdi ψ = hd, da, −db, −dabi has determinant = 1⇒ every anisotropic 4-dimensional form
∗
2∗
has determinant 1. In particular h−1, a, −b, −abi is isotropic as it has determinant −1 6= 1 ∈ Q2 /Q2 ⇒
h−1, a, −b, −abi = h−1, 1, . . . i ⇒ ha, −b, −abi represent 1. Hence ψ = h1, a, −b, −abi = h1, 1, e, ei
since det ψ = 1. But e ∈
/ {3, 7, 6, 14} because otherwise he, ei = h7, 7i = h−1, −1iand ψ isotropic
table
⇒ e ∈ {1, 2, 5, 10}, he, ei ∼
= h1, 1i ⇒ ψ = h1, 1, 1, 1i.
table
ψ = h1, 1, 1, 1i is indeed anisotropic because otherwise h1, 1, 1, 1i ∼
= h1, −1, _, _i ⇒
2
2
2
∗
h1, 1, 1i represent −1 = 7 ∈ Q∗2 /Q2∗
but
x
+
y
+
z
=
7
has
no
solution
in
Q
because
x2 + y 2 + z 2
2
2
2
2
2
has no solution in Z2 (since no solution
mod 8). If x + y + z = 7 has a solution in Q2 then there
2
2
2
2
exists x + y + z = 7t for some x, y, z, t ∈ Z2 and not all of x, y, z, t are divisible by 2.
Let us check
t ∈ Z∗2 ⇒ t2 = 1 mod 8
x2 + y 2 + z 2 = 7
Case 1.
2-t
mod 8
Case 2.
2|t ⇒ t = 2u, u ∈ Z2 and one of x, y, z is not divisible by 2, say 2 - x⇒ x2 +y 2 +z 2 = 4·7·u2
2
2 2
2
has no solution mod 8 since x = 1 mod 8 and y , z ∈ {1, 0} mod 8 while 4·7u ∈ {0, 4}
mod 8.
If
then
contradiction since
has no solution
If
ψ = h1, 1, 1, 1i is anisotropic.Now h1, 1i, hence ψ , represents 1, 2, 5, 10 and ψ also represents
∼
−1 = 7 = 22 + 12 + 12 + 12 ∈ Q∗2 /Q2∗
2 . h−1i · ψ represents 1 and is anisotropic ⇒ h−1i · ψ = ψ ⇒ ψ =
∗
2∗
∼
h−1, −1, −1, −1i ≡ h7, 7, 7, 7i ⇒ ψ = hdi ψ = φ ∀d ∈ Q2 /Q2
Hence
table
Theorem 2.92. Let
1. Every
2.
p∈Z
5-dimensional
be a prime then:
inner product space over
Qp
is isotropic
I 3 (Qp ) = 0, I 2 (Qp ) = Z/2Z generated by the unique anisotropic form of dimension 4. I/I 2 (Qp ) =
W (Qp )
Q∗p /Q2∗
p , I(Qp ) = Z/2Z
Proof.
1.
ha1 , . . . , a5 i
anisotropic
⇒ ha1 , . . . , a4 i is anisotropic hence is the unique 4-dimensional
Q∗p /Q2∗
p , in particular ha1 , . . . , a4 i represents −a5 ⇒ ha1 , . . . , a5 i
anisotropic form representing all of
isotropic
I 2 (Qp ) = Z/2Z because let φ be the unique 42
dimensional anisotropic form over Qp then φ ∈ I because dim φ = 4 = 0 ∈ Z/2Z and 0 6= φ ∈ I
2
2
because discφ = det φ = 1 ⇒ 0 6= φ ∈ ker(disc) = I . If 0 6= ψ ∈ I is anisotropic, φ 6= ψ ⇒
dim ψ < 4, dim ψ = 0 mod 2 since ψ ∈ I, ⇒ dim ψ = 2 ⇒ ψ = ha, bi but 1 = discψ = −ab since
2. Now 1.
⇒ I 3 (Qp ) = 0
by Corollary 2.85.
34
2
= 1 ⇒ ψ = ha, −ai is hyperbolic, in particular not anisotropic ⇒ ψ = 0 ∈ W (Qp ).
I = {0, φ} = Z/2Z. The rest is true for any eld F with charF 6= 2
discI
Hence,
2
Theorem 2.93. Let
p
Case 1.
p∈Z
be a prime. Then the Witt groups of
Zp
and
Qp
are:
odd:
∼
=
W (Zp )
1
∂ ,∂
∼
=
W (Qp )
where
∂1, ∂2
/ W (Fp )
2
(reduction
mod p)
/ W (Fp ) ⊕ W (Fp )
are the rst and second residue homomorphism (∂
1
(ζ) = ∂ 2 (hpi ⊗ ζ))
p=2
Case 2.
W (Z2 )
W (Q2 )
∼
=
/ Z/8Z × Z/2Z
∼
=
/ Z/8Z × (Z/2Z)2
Proof. The case p is odd is left as an exercise.
∗
3
∼
p = 2: I 3 (Q2 ) = 0, I 2 (Q2 ) = Z/2Z, I/I 2 (Q2 ) = Q∗2 /Q2∗
2 = (Z/8Z) ×(Z/2Z) = (Z/2Z) , W (Q)/I =
3
2
2
2
Z/2Z. 0 = I ⊂ I ⊂ I ⊂ W (Q2 ). |W (Q2 )| = |W/I| · |I/I | · |I | = 2 · 8 · 2 = 32 ⇒ every element of
W (Q2 )
has order a power of
2.
We have:
• h1i ∈ W (Q2 ) has order 8 because 0 6= 4 h1i = h1, 1, 1, 1i as it is a generator of I 2 (Q2 ) = Z/2Z and
8 h1i = 4 h1i+4 h1i = 4 h1i+4 h−1i = 0 because h1, 1, 1, 1i = h−1, −1, −1, −1i = h−1i⊗h1, 1, 1, 1i
(both are anisotropic and there exists a unique anisotropic form of dimension 4) over Q2 .
• h1, 3i ∈ W (Q2 ) has order 2 because it represents −1 as 1+3·32 = 28 = 22 ·7 = 7 = −1 ∈ Q∗2 /Q2∗
2 .
=
0
∈
W
(Q
)
and
h1,
3i
=
6
0
∈
W
(Q
So h1, 3i ∼
h−1,
−3i
and
h1,
3i
+
h1,
3i
=
h1,
−1i
+
h3,
−3i
=
2
2)
| {z }
hyperbolic
since disc h1, 3i
= −3 = 5 6= 1 ∈ Q∗2 /Q2∗
2
• h1, 6i ∈ W (Q2 ) has order 2 because it represent −1 as −1 = 7 = 1 · 12 + 6 · 12 ∈ Q∗2 /Q2∗
2 . So
h1, 6i ∼
= h−1, −6i ⇒ h1, 6i + h1, 6i = h−1, −6i + h1, 6i = h1, −1i + h6, −6i = 0 ∈ W (Q2 ) and
0 6= h1, 6i ∈ W (Q2 ) because disc h1, 6i = −6 6= 1 ∈ Q∗2 /Q2∗
2
Z/8Z ⊕ Z/2Z ⊕ Z/2Z → W (Q2 ) dened by a, b, c 7→ a h1i + b h1, 3i + c h1, 6i is well
32. In order to show that the map is an isomorphism it suces to
show that it is injective. Assume 0 = a h1i + b h1, 3i + c h1, 6i ∈ W (Q2 ). Now b h1, 3i + c h1, 6i has order
≤ 2 ⇒ a h1i has order ≤ 2 ⇒ a = 4 mod 8 ⇒ a h1i = a0 h1, 1, 1, 1i with a0 ∈ Z/2Z. We compute the
Hence the map
dened. Both groups have order
discriminant
0
disc(a
h1, 1, 1, 1i + b h1, 3i + c h1, 6i)
0
=
(disc h1, 1, 1, 1i)a · disc(h1, 3i)b · disc(h1, 6i)c ∈ Q∗2 /Q2∗
2
=
1 · (−3)3 (−6)c = 5b · 10c ∈ Q∗2 /Q2∗
2 .
5 6= 10 ∈ Q∗2 /Q2∗
2 , hence, they linearly independent
(Z/2Z)3 ⇒ b, c = 0 ∈ Z/2Z ⇒ a0 h1, 1, 1, 1i = 0 ⇒ a0 = 0
map is injective. Hence W (Q2 ) ∼
= Z/8Z ⊕ Z/2Z ⊕ Z/2Z.
h1i , h1, 3i ∈ W (Z2 ) since 1, 3 ∈ Z∗2 , W (Z2 ) ,→ W (Q2 ) is
Now
(Z/2Z) h1, 3i ⊂ W (Z2 ) ⇒ |W (Z2 )| ≥ 8 · 2 = 16.
(Q2 )|
W (Z2 ) ⊂ ker(∂ 2 ) ⇒ |W (Z2 )| ≤ | ker(∂ 2 )| = |W
|W (F2 )| =
(Z/2Z) h1, 3i = W (Z2 ).
Lemma 2.94 (Denition). Set
F2 -vector space in Q∗2 /Q2∗
=
2
h1, 1, 1, 1i 6= 0 ∈ W (Q2 ) ⇒ the
in the
since
injective, it follows that
(Z/8Z) h1i ⊕
∂2
Also W (Z2 ) → W (Q2 ) W (F2 ) is zero.
32
2 = 16 ⇒ |W (Z2 )| = 16. Hence (Z/8Z) h1i ⊕
Q∞ = R, p = ∞ = innite prime. For p ∈ Z ∪ {∞} prime there
is a unique quaternion algebra over Qp that doesn't split (i.e., M2 (Qp )). Therefore, Hasse(V ) =
35
(
[A]
[Qp ]
∈ Br(Qp )
,
∈ Br(Qp )
where
inner product space
For
a, b ∈ Q∗p ,
V
A is a division quaternion algebra.
The Hasse symbol
hp (V ) for a symmetric
over
Qp is dened by
(
−1 if Hasse(V ) does not split (6= [Qp ] ∈ Br(Qp ))
hp (V ) =
1
if Hasse(V ) = [Qp ] ∈ Br(Qp )
the Hilbert Symbol is:
(a, b)p = hp (ha, bi) =

1
if
−1
if
a,b
Qp a,b
Qp
splits
does not split
Qp . If p = ∞
Br(Q∞ ) = Br(R) = {R, H}, so H is the unique non-split quaternion algebra over R.
c,d
a,b
If p < ∞: (
Qp ), ( Qp ) M2 (Q2 ) ⇐⇒ ha, b, −ab, −1i , hc, d, −cd, −1i h1, 1, −1, −1i (all forms
2
are in I (Qp ) ∼
= Z/2Z as disc = 0 for them) ⇐⇒ ha, b, −ab, −1i , hc, d, −cd, −1i are both the unique
anisotropic 4-dimensional form over Qp ⇐⇒ ha, b, −ab, −1i ∼
= hc, d, −cd, −1i h1, 1, −1, −1i ⇐⇒
c,d
1,1
∼
( a,b
(
)
)
(
)
=
M
(Q
)
.
=
2
p
Qp
Qp
Qp
Proof. We need to justify that there exists a unique non-split quaternion algebra over
then
Hilbert Reciprocity Law. Let
all but nitely many primes
V be a symmetric
Q inner product space over Q.
p ∈ Z ∪ {∞}. And p∈Z∪{∞} prime hp (V ) = 1
Then
hp (V ) = 1
for
is a product of Hilbert Symbols (a, b)p , it suces to show claim for V = ha, bi
Q
(a,
b)p = 1 ∀a, b ∈ Q∗p . To show (a, b)p = 1, using bilinearity of Hilbert symbol
p∈Z∪{∞}
Q
(ab, c) = (a, c)p (b, c)p , we just need to show (a, b)p = 1 for a, b prime or ±1. In this case, express
(a, b)p in terms of Legendre symbol which mean the proof is a consequence of Quadratic Reciprocity.
Q hp (V )
Proof. Since
and thus
(Details are left as an exercise)
Corollary 2.95. Let
V, W be inner product spaces over Q.
hp (W ) ∀p ∈ Z ∪ {∞} prime, p 6= q . Then hq (V ) = hq (W )
Q
Q
Proof.
p∈Z∪{∞} hp (V ) = 1 =
p∈Z∪{∞} hp (W )
Let
q ∈ Z ∪ {∞}
be a prime. If
hp (V ) =
We will need this theorem:
Theorem 2.96 (Dirichlet). Let
form
a + nb, n ∈ Z,
a, b ∈ Z
be integers with
gcd(a, b) = 1,
then the set of integers of the
contains innitely many primes.
Proof. This theorem is beyond the scope of this module
Strong Hasse principle for quadratic forms over
Q
is isotropic if and only if
V
is isotropic over
R
and
Q. A symmetric
Qp ∀p ∈ Z prime.
inner product space
Remark. The Theorem says: A homogeneous quadratic polynomial has a non-trivial zero in
only if it has a non-trivial zero in
R
and
Qp ∀p ∈ Z
V
Q
over
if and
prime.
Proof. ⇒: Is clear.
⇐: We assume Dirichlet Theorem. We use induction on
n = dimQ V
n = 1:
Every 1-dimensional inner space is anisotropic (over any eld)
n = 2:
A
2-dimensional form V is isotropic over Q (any eld of characteristic not 2) ⇐⇒ V
Q, i.e., V ∼
= H ⇐⇒ V ∼
= H over R and Qp ∀p ∈ Z prime ⇐⇒ V is isotropic
over R and Qp ∀p ∈ Z prime
n = 3 : V isotropic over Q ⇐⇒ V ∼
= h1, −1, − det V i over R
= h1, −1, − det V i over Q ⇐⇒ V ∼
and Qp ∀p prime (Weak Hasse Principle) ⇐⇒ V isotropic over R and Qp ∀p primes.
hyperbolic over
n = 4: Write V = hd1 , d2 , d3 , d4 i with di ∈ Z \ {0} square free, d = det V square free. Let
P = {2} ∪ {p ∈ Z prime : p|d1 . . . d4 } < ∞. Write Vp for V ⊗Q Qp . Now Vp is isotropic by
assumption, ⇒ Vp ∼
= h1, −1i ⊥ hap , −ap di (over Qp ) with ap ∈ Z \ {0} square free.
36
•
p∈
/ P we can assume that ap ∈ Z∗p and a∞ = 1. Otherwise if p = ∞ replace (V, β)
with (V, −β), and if ∞ =
6 p∈
/ P we would have ap = pbp (as ap ∈ Zp \ {0} square free).
Therefore, 0
= ∂p2 V = ∂p2 h1, −1, pbp , −pbp di = hbp , −bp di ⇒ disc hbp , −bp di = d =
If
p-d1 ...d4
∼ ∗ 2∗
∼
∼
1 ∈ F∗p /F2∗
p = Zp /Zp and d is a square in Qp ⇒ hap , −ap di = hap , −ap i = h1, −1i over
Qp and we can even assume ap = 1
∗
2∗
• There
exists q ∈ Z prime such that a := qπ = ap ∈ Qp /Qp ∀p ∈ P where π =
Q
p∈P,ν(ap )=1 p (Note ν(ap ) = 0 or 1 as ap ∈ Zp is square free). To justify existence
∗
of q note that ap = πup , up ∈ Zp . By the Chinese Remainder Theorem Z Z/8Z ×
Q
∗
p∈P,p6=2 Z/pZ is surjective. So there exists an integer r such that r = a2 ∈ (Z/8Z) ⊂
Z/8Z and
Q r = up ∈ Z/pZ for p ∈ P \ {2}. In fact, any integer of the form r + ns with
s = 23 p∈P\{2} p can be chosen instead of r. Since the ap 's are units, it follows that
s and r are relatively prime. By Dirichlet's theorem on existence on innitely many
primes in an arithmetic progression, we can choose r = q a prime. By construction
a = qπ = ap ∈ Q∗p /Q2∗
p .
∼
Claim: V = h1, −1i ⊥ ha, −adi over Q (in particular, V isotropic over Q as it contains H)
Proof of claim: By the weak Hasse principle it suces to show that
h1, −1, a, −adi
Case 1.
Case 2.
over
Qp ∀p ∈ Z ∪ {∞}
(Vp =) h1, −1, ap , −ap di =
prime.
∼ ha, −adi since, by construction of a, we have
p ∈ P : We have hap , −ap di =
for
p
∈
P
.
a = ap ∈ Q∗p /Q2∗
p
p ∈
/ P and p 6= q, ∞: One checks that ∂ 1 and ∂ 2 agree: ∂ 1 hap , −ap di =
hap , −ap di = ha, −adi = ∂ 1 ha, −adi ∈ W (Fp ) because p does not divide a, ap , d
and over Fp quadratic forms are classied by rank and determinant. Further, we
2
2
have ∂ hap , −ap di = 0 = ∂ ha, −adi ∈ W (Fp ) because p does not divide a, ap , d.
∼
And so ⇒ ha, −adi = hap , −ap di over Qp .
p6=2
p = ∞: ha, −adi = ha∞ , −a∞ di over R = Q∞ because a∞ = 1 and a > 0.
q : Over Qq the forms h1, −1, a, −adi and V have the same rank (= 4), determinant d and Hasse invariant (by Hilbert reciprocity, as both are isometric over Qp ,
p 6= q , and thus have same Hasse symbol over Qp , p 6= q ) ⇒ h1, −1, a, −adi ∼
=V
over Qq .
n ≥ 5: Choose an orthogonal sum decomposition V ∼
= U ⊥ W with dim U = 2 and dim W =
n − 2 ≥ 3. Want to nd a non-degenerate subspace of V of dimension less than n which is
isotropic over R and Qp ∀p. Then by induction hypothesis, this subspace is isotropic over
Q, hence V is isotropic over Q. If (U or) W is isotropic over R and Qp ∀p then by induction
(U or)W is isotropic over Q (then so is V and we are done). Hence suppose W anisotropic
over some Qp . Let P = {p ∈ Z ∪ {∞} prime|W anisotropic over Qp } (6= ∅). P is a nite
set because dim W ≥ 3 and ha, b, ci (a, b, c ∈ Z) is isotropic over Qp (p 6= 2) if and only if
ha, b, ci ∼
= h1, −1, −abci over Qp , but if p - a, b, c and p 6= 2, ∞ then ha, b, ci ∼
= h1, −1, −abci
2
2
1
over Qp . (This holds because if p 6= 2, ∞ then ∂ LHS = 0 = ∂ RHS, ∂ LHS= ha, b, ci =
h1, −1, −abci = ∂ 1 RHS, recall W (Qp ) ∼
= W (Fp ) ⊕ W (Fp ) and over Fp quadratic forms
Case 3.
Case 4.
∂ 1 ,∂ 2
are classied by rank and determinant)
(V, β), q(x) = β(x, x), q isotropic over Qp ∀p ∈ Z ∪ {∞}
∀p ∈ Z ∪ {∞} there exists 0 6= up ∈ U ⊗ Qp and 0 6= wp ∈ W ⊗ Qp such that
q(up ) + q(wp ) = 0
∗
2∗
Claim: There exists u ∈ Z \ {0} such that q(u) = q(up ) ∈ Qp /Qp for all p ∈ P
Then the claim ⇒ Qu ⊥ W ⊂ V has dimension n−1 and is isotropic over R and Qp ∀p prime
because W isotropic over Qp ∀p ∈
/ P and Qu ⊥ W isotropic over p ∈ P as q(u) + q(wp ) =
0 ∀p ∈ P . So by induction hypothesis the subspace Qu ⊥ W is isotropic over Q ⇒ V is
isotropic over Q
2
2
Justication of the claim: Now q = ax + by with a, b ∈ Z \ {0} so up = (xp , yp ) ∈ Zp × Zp .
Let
q
be the quadratic form of
prime. Hence
Case 1.
∞∈
/ P , then plp ξp = q(up ) = ax2p +byp2 ∈ Zp \{0} where ξp ∈ Z∗p and
Q
lp ∈ Z≥0 . By the Chinese Remainder Theorem the map Z p∈P Z/plp +3 =
Q
lp +3
is surjective. Hence there exists x, y ∈ Z such that x = xp , y =
p∈P Zp /p
Assume rst
37
yp mod plp +3 for p ∈ P . and Set u = (x, y). Then q(u) = ax2 + by 2 =
ax2p + byp2 ∈ Zp /plp +3 implies that ax2 + by 2 = plp ep with ep = ξp mod p3 . Now
ξp ∈ Z∗p implies that ξp = ep ∈ Q∗p /Q2∗
p due to the fact that
• ep = ξp mod p3
∗
2∗
• Z∗p /Z2∗
mod p (p
p = Fp /Fp
∗
2∗
3
∗
• Z2 /Z2 = (Z/2 Z)
odd)
lp
lp
∗
2∗
2
2
ξp = ep ∈ Q∗p /Q2∗
p ⇒ p ξp = p ep ∈ Qp /Qp ∀p ∈ P ⇒ ax + by =
2
∗
2∗
∗
2∗
+ byp ∈ Qp /Qp ⇒ q(u) = q(up ) ∈ Qp /Qp ∀p ∈ P
∗
2∗
If ∞ ∈ P , Q∞ = R. Then q(u∞ ) ∈ R /R
either > 0 or < 0, by replacing q with
−q we can assume q(u∞ ) > 0. Let u∞ = (x∞ , y∞ ) ∈ R × R, q(u∞ ) = ax2 + by 2
not both a, b < 0 since q(u∞ ) > 0, so without loss of generality we can assume
a > 0. Choose x, y as in the rst case such that moreover x2 > − ab y 2 then
2
2
q(u) = q(up ) ∈ Q∗p /Q2∗
p ∀p ∈ P \ {∞} as above. Furthermore q(u) = ax + by >
2
2∗
0 ⇒ q(u) = q(u∞ ) ∈ R /R .
Now
ax2p
Case 2.
This ends the proof of the claim.
Denition 2.97. Let
q
•
positive denite if
•
negative denite if
•
indenite if
q
0
over
q
is said to be
q(x) > 0 ∀x 6= 0.
q(x) < 0 ∀x 6= 0.
is neither positive nor negative denite.
Corollary 2.98. Let
represents
be a rational (or integral) quadratic form. Then
q
be a rational quadratic form of dimension
≥ 5.
If
q
is indenite, then it
Q.
q represents 0 over R and Qp for all p ∈ Z
⇒ qR = h1, −1i ⊥ . . . so q represent 0 over R. Since dimension of q≥ 5 ⇒ q
all p prime because every 5 dimensional form over Qp is isotropic.
Proof. By the strong Hasse principle, we nee to see that
prime. Since
represent
0
q
indenite
over
Qp
for
2.9
Integral quadratic forms
Recall
W (Z) → Z.
∼
=
sgn
Denition 2.99. A symmetric inner product space over
•
even (or of type II) if
•
odd (or of type I) if
•
Remark.
•
If
q
β(x, x) ∈ Z
∃x ∈ V
is even
such that
Z
then
is called:
∀x ∈ V
β(x, x) ∈ Z
A symmetric inner product space over
is a quadratic form over
Z, (V, β)
Z
is odd.
is also called unimodular lattice
β(x, y) = q(x + y) − q(x) − q(y) and β(x, x) = 2q(X). So the
Z are precisely the inner product spaces that come
even symmetric inner product spaces over
from regular quadratic forms.
Lemma 2.100. Let
x ∈ V, x 6= 0
(V, β) be an indenite
β(x, x) = 0.
symmetric inner product space over
Z
then there exists
such that
W (Z) ,→ W (Q) is generated by h1i and h−1i. (V, β) indenite
⇒ VQ = V ⊗Z Q = h1, −1i ⊥ . . . isotropic ⇒ ∃x ∈ VQ , x 6= 0 such that β(x, x) = 0. But
V ⊂ VQ , x = ny for some n ∈ Z \ {0}, y ∈ V ⇒ β(y, y) = β(nx, nx) = n2 β(x, x) = 0 and 0 6= y ∈ V
Proof. Recall that the image of
Theorem 2.101. Let
Then
(V, β)
(V, β)
be an odd (i.e., type I) indenite symmetric inner product space over
has an orthogonal basis. In particular
(V, β) ∼
= m h1i ⊥ n h−1i
38
over
Z.
Z.
Proof. Claim:
(V, β) ∼
=
0
1
1
odd
⊥ (V 0 , β 0 ).
V : For k ∈ Z we have
0
1
0
1 k+1
∼
⇒
= h1, −1i .
1 2k + 1
−1
1
k
{z
}
|
The theorem follows from claim by induction on dimension
0
1
1
2k + 1
=
1
k+1
1
1
k
0
det=−1
h1i ⊥ V 0 or h−1i ⊥ V 0 is indenite and both are odd and
0
have dimension less that V , and V = h±1i ⊥ (h∓1i ⊥ V )
To prove the claim: We know (V, β) indenite
⇒
∃x ∈ V with x 6= 0 and β(x, x) = 0.
From this the theorem follows as we have
previous lemma
(V, β) inner product space over Z ⇒ V = Zn . So x = (a1 , . . . , an ) ∈ Zn , we can assume
d = gcd(a1 , . . . , an ) = 1 (otherwise replace x with xd ). We can extend x to a Z basis x1 = x, x2 , . . . , xn
n
of V = Z because φ : V /Zx → (V ⊗Z Q)/Qx is injective as y = (c1 , . . . , cn ) ∈ ker φ then ∃r, s ∈ Z
with r 6= 0 and ry = sx ⇒ rci = sai for all i = 1, . . . , n, since gcd(a1 , . . . , an ) = 1 there exists
P
P
P
P
b1 , . . . , bn ∈ Z such that
ai bi = 1. So r ci bi = s ai bi = s ⇒ s = rt where t =
ci bi , hence
ry = sx ⇒ ry = rtx ⇒ y = tx ⇒ y = 0 ∈ V /Zx. Hence our map is indeed injective. Now V /Zx is
Now
r6=0
p
⊗Z Q
= Qn−1 ⇒ V /x is a free Z-module ⇒ V V /x
Z-module, submodule of V Qx
σ : V /Zx → V (pσ = 1)⇒ V = x ⊕
im
σ . Hence x can be extended to a Z-basis
|{z}
a nitely generated
has a section
∼
=V /x∼
=Zn−1
x1 = x, x2 , . . . , xn
of
V.
(
Let
y1 , . . . , yn
be the dual basis of
is non-degenerated
β
∼
=
⇒ β : V → Hom(V, Z) ∼
= Zn
Now
(V, β)
Theorem 2.102. If
divisible by
8.
Proof. Let
(V, β)
odd
(V, β)
⇒
i.e.,
β(xi , yj ) =
has a
Z
basis
1 i=j
0 i=
6 j
e1 , . . . , en ,
which exists because
where
(ei )(
P
β
αj xj ) = αi
k ∈ {1, . . . ,
n} such that
k , yk ) is odd. If β(y1 , y1 )
β(y
non−degenerate
0
1
0
1
is odd then β|(Zx1 +Zy1 ) =
⇒ (V, β) ∼
⊥
(Zx1 ⊕ Zy1 )⊥ .
=
1 odd
1 odd
0
1
If β(y1 , y1 ) even and β(yk , yk ) odd for k 6= 1 then β|(Zx1 +Z(y1 +yk )) =
⇒ (V, β) ∼
=
1 odd
non−degenerate
0
1
⊥
(Zx1 ⊕ Z(y1 + yk ))⊥
1 odd
,
yi ↔ ei .
x1 , . . . , x n ,
there exists
is an even symmetric inner product space over
Z
then its signature is
Z. Then V /2V = V ⊗Z F2
x →
7 β(x, x) is linear because
be an arbitary symmetric inner product space over
is a symmetric inner product space over
F2 . But over F2
β(x + y, x + y) = β(x, x) + 2β(x, y) + β(y, y) ∈ F2 . As V /2V
| {z }
the map
is non degenerate, there exists a unique
=0
u ∈ V /2V such that β(u, x) = β(x, x) mod 2 ∀x ∈ V. If u, u0 ∈ V are two lifts of u ∈ V /2V then
u0 = u + 2v for some v ∈ V , and β(u0 , u0 ) = β(u, u) + 4(β(u, v) + β(v, v)) = β(u, u) ∈ Z/8Z because
|
{z
}
=0 mod 2
β(u, v) = β(v, v) ∈ F2 by denition of u. Set φ(V ) := β(u, u) ∈ Z/8Z for any lift u of u ∈ V /2V . We
have seen that φ(V ) does not depend on the lift u of u. From the denition of φ we have φ(V ⊥ W ) =
φ(V )+φ(W ) and φ(h1i) = 1, φ(h−1i) = −1, so φ : W (Z) → Z/8Z : V 7→ φ(V ) is a well dened map. If
(V, β) is even then β(x, x) = 0 mod 2 ∀x ∈ V and we can choose u = 0 ⇒ u = 0 ⇒ φ(V ) = 0 ∈ Z/8Z.
Since φ(h1i) = 1 ⇒ φ : Z · h1i = W (Z) → Z/8Z is the signature. Now, if (V, β) is even then β(x, x) = 0
mod 2 ∀x ∈ V and u = 0 ⇒ we can choose u = 0 ⇒ φ(V ) = 0 ∈ Z/8Z. This implies that signature of
any even symmetric inner product space over Z is divisible by 8.
Corollary 2.103. Every even positive denite inner product space over
Proof. If
M
is positive denite then
Theorem 2.104. Let
⇐⇒ M, N
M, N
Z
has rank divisible by
8.
rk M = sgn M .
be indenite symmetric inner product spaces over
have the same rank, signature and type (odd or even).
39
Z.
Then
M ∼
= N
M, N are odd then M, N have orthogonal basis, by Theorem 2.101,then
M, N are even, we do not have the time to prove this in this course.
Proof. If
If
the theorem follows.
Example. Of even positive denite inner product spaces over
Z.
n
R
be equipped with standard Euclidean inner product h(x1 , . . . , xn ), (y1 , . . . , yn )i =
Pn
n
∼ k
i=1 xi yi . If M ⊆ R is a nitely generated Z-submodule, then M = Z for some k ≤ n. Restricting
h, iRn to M denes a symmetric bilinear form β(x, y) = hx, yi ∈ R on M with values in R. Assume
rkZ M = n = dimR Rn ⇒ Rn /M compact Riemanian manifold. Vol(Rn /M ) = volume of parallelepiped spanned by a Z-basis of M . If we let A = (b1 , . . . , bn ) where b1 , . . . , bn is a Z-basis of M then
s
p
n
det(AT A) = det (hbi , bj i)i,j=1,...,n ⇒ a nitely generated M ⊂ Rn of
Vol(R /M ) = | det A| =
|
{z
}
General Remark: Let
biliner form matrix of M
rkZ M = n
denes a (positive denite) inner product space over
• hx, yi ∈ Z ∀x, y ∈ M ,
•
Vol(R
n
Z
if and only if:
and
/M ) = 1
In fact every possible denite inner product space
M ⊗Z R ⊃ M
and
βR ∼
= h1, . . . , 1i
| {z }
over
R
since
(M, β)
(M, β)
over
Z
Rn ∼
=
arises in that way, because
is positive denite.
n
E4m ⊆ R4m be the Z-submodule (m ∈ Z≥1 ) generated by ei + ej (i, j = 1, . . . , 4m)
+ e2 + · · · + e4m ) where ei = (0, . . . , 0, 1, 0, . . . , 0) is the standard basis vector of R4m . Then
Lemma 2.105. Let
1
and 2 (e1
↑
i
E4m
is a symmetric inner product space over
Z
of rank
4m
which is even (respectively odd) if
m
is
even (respectively odd)
• E4m ⊂ R4m nitely generated Z-submodule ⇒ E4m free Z-module, i.e., E4m ∼
= Zk . Now
rkZ E4m = dimQ E4m ⊗Z Q = 4m because ei + ej , i, j = 1, . . . , 4m and 21 (e1 + · · · + e4m ) span Qn .
(Note that this contains 2ei , i = 1, . . . , m by setting i = j ).
(
2 i 6= j
• hx, yi ∈ Z ∀x, y ∈ E4m . (check for x, y generators of E4m ). E.g., hei + ej , ei + ej i =
,
4 i=j
1
1
1
2 (e1 + · · · + e4m ), 2 (e1 + · · · + e4m ) = 4 4m = m ⇒ E4m is even if and only if m even.
Proof.
•
M ⊂ N ⊂ Rn of
rank n Z-submodule, then R /M R /N covering with |N/M | sheets because N/M acts freely
n
n
n
n
n
on R /M with quotient R /N . So |N/M | · Vol(R /N ) = Vol(R /M ) ∀M ⊂ N ⊂ R rk = n
Z-submodules.
We are left to check it is non-degenerate. We will use the following trick: If
n
n
/E4m ) = 1. Let E 0 ⊂ E4m be the Z1
0
submodule generated by ei + ej , i, j = 1, . . . 4m. Then E4m /E is generated by ξ = (e1 + · · · +
2
0
0
0
0
e4m ) ∈
/ E , and 2ξ ∈ E so 2ξ = 0 ∈ E4m /E . Therefore E4m /E = Z/2Z ⇒ 2Vol(R4m /E4m ) =
4m
Vol(R
/E 0 ). But notice E 0 ⊂ Z4m where Z4m is generated by e1 , . . . , e4m . Now Z4m /E 0 is
generated by e1 because ei = ei + e1 − e1 ∀i = 2, . . . , 4m. Now e1 ∈
/ E 0 but 2e1 = e1 + e1 ∈ E 0 ⇒
4m
0
4m
0
4m
4m
Z /E = Z/2Z ⇒ Vol(R /E ) = 2Vol(R /Z ) ⇒ Vol(R4m /E 4m ) = Vol(R4m /Z4m ) =
1 ⇒ E4m is non-degenerate.
Now we prove
Corollary 2.106.
Fact. For all
E4m
E8m
is non-degenerate, i.e., Vol(R
4m
8m
is an even positive denite symmetric inner product space of rank
n ∈ Z≥0 , {symmetric
inner product spaces over
Z
of given rank
n}/isometry
is a nite
set.
Example.
rankn
number of even positive denite inner products space over
Z
8
16
24
32
1
2
24
Unknown
E8
representative
40
E16 , E8 ⊥ E8
Niemeier
(1968)
≥ 10