SOLUTIONS TO PROBLEMS 3.5, 3.9.5, 3.11, 3.13.5
KATE THOMPSON
3.5 Let
p
be an odd prime number, and let
u
∈
Z
be a quadratic nonresidue modulo
p
.
Then
[
Q
×
p
:
Q
×
2
p
] = 4, and a set of coset representatives is 1
, p, u, pu
.
Solutions:
First, we intend to use Hensel’s Lemma for this proof; given that, we feel obligated
to state which version we will use.
Lemma.
(Hensel’s Lemma) Let
(
K, v
)
be a Henselian (e.g.
complete) valued field with valuation
ring
R
and residue field
k
. Let
f
∈
R
[
t
]
. If there exists an
α
∈
R
such that
v
(
f
(
α
))
>
2
v
(
f
(
α
))
,
then there exists a
β
∈
R
with
f
(
β
) = 0
and
v
(
α

β
)
> v
(
f
(
α
))
.
Let
P
(
X
) =
X
2

a
for
a
a
p
adic integer. It is clear that there is a root in
Z
p
only if
v
(
a
) =
v
(
x
2
) = 2
v
(
x
)
is even. Then, by dividing
a
by an appropriate power of
p
2
m
, we can reduce to the case
v
(
a
) = 0, or
a
∈
Z
×
p
.
Now we claim that if
a
∈
Z
×
p
is a
p
adic unit and if there exists
α
such that
α
2
≡
a
(mod
p
Z
p
), then
a
is a square of an element in
Z
×
p
. To see this, apply Hensel’s lemma to
X
2

a
. Since
p
= 2 and
a
∈
Z
×
p
, we are guaranteed to have 2
α
≡
0 (mod
p
) so we’re fine.
Next, we claim that
x
∈
Q
p
is a square if and only if
x
=
p
2
n
y
2
for
n
∈
Z
, and
y
∈
Z
×
p
a
p

adic unit.
One direction is obvious.
The second direction is also fairly obvious–just refer to the
previous case, and also note that
p
is itself NOT a square in
Q
p
(since if it had a square root, then
that square root would have valuation
1
2
). We can conclude then that
Q
×
p
:
Q
×
2
p
= 4, and is given
by
{
1
, p, u, pu
}
for
u
a quadratic nonresidue in
Z
×
p
. Referring to the text below this Exercise in the
course notes, we can moreover conclude that
Q
×
p
/
(
Q
×
2
p
)
∼
=
Z
/
2
Z
×
Z
/
2
Z
3.9.5 A very special and important quadratic form is
q
H
(
x
1
, x
2
) =
x
1
x
2
, the socalled
hyperbolic plane
.
a) Let
K
be any field of characteristic different from 2. Give an explicit change of variables that
diagonalizes
q
H
.
Solutions:
Short answer: let
P
=
1

1
2

1
2

1
Why does this definitely work? Since the characteristic of our field is not 2, we can represent
the quadratic form
x
1
x
2
as the following symmetric matrix:
M
=
0
2

1
2

1
0
Thus, using matrix theory, we claim that
P
works. To confirm, we compute:
1
2

1

1
2

1
0
2

1
2

1
0
1

1
2

1
2

1
=
2

2
2

1
2

2

2

1
1

1
2

1
2

1
=
2

1
0
0

2

1
Last, note that since the characteristic of a field is NOT 2, 2

1
exists.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
b) Show by brute force that
q
H
cannot be diagonalized over
F
2
.
See part (c) for a less forceful method.
c) Show that
q
H
cannot be diagonalized over any field of characteristic 2.
Solutions:
Suppose that it can be diagonalized. This means that for some
a, b, c, d
∈
F
,
ad

bc
=
0, we have
a
c
b
d
0
1
0
0
a
b
c
d
=
e
0
0
f
Working this out, we see that we have:
a
c
b
d
0
1
0
0
a
b
c
d
=
0
a
0
b
a
b
c
d
=
ac
ad
bc
bd
This means that
ad
=
bc
= 0; however, that would contradict our assumption that
ad

bc
= 0.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '11
 Staff
 Number Theory, Quadratic form, bilinear form, Clifford algebra, symmetric bilinear form, Hensel

Click to edit the document details