4Computational techniques

II Galois Theory


4.2 Trace, norm and discriminant
We are going to change direction a bit and look at traces and norms. These will
help us understand the field better, and perhaps prove some useful facts from
it. They will also lead to the notion of the discriminant, which is again another
tool that can be used to compute Galois groups, amongst many other things.
Definition (Trace). Let
K
be a field. If
A
= [
a
ij
] is an
n × n
matrix over
K
,
we define the trace of A to be
tr(A) =
n
X
i=1
a
ii
,
i.e. we take the sum of the diagonal terms.
It is a well-known fact that if B is an invertible n × n matrix, then
tr(B
−1
AB) = tr(A).
Hence given a finite-dimensional vector space
V
over
K
and
σ
:
V → V
a
K-linear map, then we can define the trace for the linear map as well.
Definition (Trace of linear map). Let
V
be a finite-dimensional vector space
over K, and σ : V → V a K-linear map. Then we can define
tr(σ) = tr(any matrix representing σ).
Definition (Trace of element). Let
K ⊆ L
be a finite field extension, and
α ∈ L
. Consider the
K
-linear map
σ
:
L → L
given by multiplication with
α
, i.e.
β 7→ αβ. Then we define the trace of α to be
tr
L/K
(α) = tr(σ).
Similarly, we can consider the determinant, and obtain the norm.
Definition (Norm of element). We define the norm of α to be
N
L/K
(α) = det(σ),
where σ is, again, the multiplication-by-α map.
This construction gives us two functions
tr
L/K
, N
L/K
:
L → K
. It is easy to
see from definition that tr
L/K
is additive while N
L/K
is multiplicative.
Example. Let
L/K
be a finite field extension, and
x ∈ K
. Then the matrix of
x is represented by xI, where I is the identity matrix. So
N
L/K
(x) = x
[L:K]
, tr
L/K
(x) = [L : K]x.
Example. Let
K
=
Q
,
L
=
Q
(
i
). Consider an element
a
+
bi ∈ Q
(
i
), and pick
the basis {1, i} for Q(i). Then the matrix of a + bi is
a −b
b a
.
So we find that tr
L/K
(a + bi) = 2a and N (a + bi) = a
2
+ b
2
= |a + bi|
2
.
In general, if
K
=
Q
and
L
=
Q
(
√
−d
) where
d >
0 is square-free, then
N
(
a
+
b
√
−d
) =
a
2
+
b
2
d
=
|a
+
b
√
−d|
2
. However, for other fields, the norm is
not at all related to the absolute value.
In general, computing norms and traces with the definition directly is not
fun. It turns out we can easily find the trace and norm of
α
from the minimal
polynomial of
α
, just like how we can find usual traces and determinants from
the characteristic polynomial.
To do so, we first prove the transitivity of trace and norm.
Lemma. Let L/F/K be finite field extensions. Then
tr
L/K
= tr
F/K
◦tr
L/F
, N
L/K
= N
F/K
◦ N
L/F
.
To prove this directly is not difficult, but involves some confusing notation.
Purely for the sake of notational convenience, we shall prove the following more
general fact:
Lemma. Let
F/K
be a field extension, and
V
an
F
-vector space. Let
T
:
V → V
be an F -linear map. Then it is in particular a K-linear map. Then
det
K
T = N
F/K
(det
F
T ), tr
K
T = tr
F/K
(tr
F
T ).
Taking
V
to be
L
and
T
to be multiplication by
α ∈ F
clearly gives the
original intended result.
Proof.
For
α ∈ F
, we will write
m
α
:
F → F
for multiplication by
α
map viewed
as a K-linear map.
By IB Groups, Rings and Modules, there exists a basis
{e
i
}
such that
T
is
in rational canonical form, i.e. such that
T
is block diagonal with each diagonal
looking like
0 0 ··· 0 a
0
1 0 ··· 0 a
1
0 1 ··· 0 a
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· 1 a
r−1
.
Since the norm is multiplicative and trace is additive, and
det
A 0
0 B
= det A det B, tr
A 0
0 B
= tr A + tr B,
we may wlog T is represented by a single block as above.
From the rational canonical form, we can read off
det
F
T = (−1)
r−1
a
0
, tr
F
T = a
r−1
.
We now pick a basis
{f
j
}
of
F
over
K
, and then
{e
i
f
j
}
is a basis for
V
over
K. Then in this basis, the matrix of T over K is given by
0 0 ··· 0 m
a
0
1 0 ··· 0 m
a
1
0 1 ··· 0 m
a
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· 1 m
a
r−1
.
It is clear that this has trace
tr
K
(m
a
r−1
) = tr
F/K
(a
r−1
) = tr
F/K
(tr
F
T ).
Moreover, writing n = [L : K], we have
det
K
0 0 ··· 0 m
a
0
1 0 ··· 0 m
a
1
0 1 ··· 0 m
a
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· 1 m
a
r−1
= (−1)
n(r−1)
det
K
m
a
0
0 0 ··· 0
m
a
1
1 0 ··· 0
m
a
2
0 1 ··· 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
a
r−1
0 0 ··· 1
= (−1)
n(r−1)
det
K
(m
a
0
)
= det
K
((−1)
r−1
m
a
0
)
= N
F/K
((−1)
r−1
a
0
)
= N
F/K
(det
F
T ).
So the result follows.
As a corollary, we have the following very powerful tool for computing norms
and traces.
Corollary. Let
L/K
be a finite field extension, and
α ∈ L
. Let
r
= [
L
:
K
(
α
)]
and let P
α
be the minimal polynomial of α over K, say
P
α
= t
n
+ a
n−1
t
n−1
+ ··· + a
0
.
with a
i
∈ K. Then
tr
L/K
(α) = −ra
n−1
and
N
L/K
(α) = (−1)
nr
a
r
0
.
Note how this resembles the relation between the characteristic polynomial
and trace/determinants in linear algebra.
Proof.
We first consider the case
r
= 1. Write
m
α
for the matrix representing
multiplication by
α
. Then
P
α
is the minimal polynomial of
m
α
. But since
deg P
α
=
n
=
dim
K
K
(
α
), it follows that this is also the characteristic polynomial.
So the result follows.
Now if
r
= 1, we can consider the tower of extensions
L/K
(
α
)
/K
. Then we
have
N
L/K
(α) = N
K(α)/K
(N
L/K(α)
(α)) = N
K(α)/K
(α
r
)
= (N
K(α)/K
(α))
r
= (−1)
nr
a
r
0
.
The computation for trace is similar.
It is also instructive to prove this directly. In the case
r
= 1, we can pick the
basis {1, α, α
2
, ··· , α
n−1
} of L over K. Then the multiplication map sends
1 7→ α
α 7→ α
2
.
.
.
α
n−1
7→ α
n
= −a
n−1
α
n−1
− ··· − a
0
So the matrix is just
A =
0 0 ··· −a
0
1 0 ··· −a
1
0 1 ··· −a
2
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· −a
n−1
The characteristic polynomial of this matrix is
det(tI − A) = det
t 0 ··· a
0
−1 t ··· a
1
0 −1 ··· a
2
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· t + a
n−1
By adding t
i
multiples of the ith row to the first row for each i, this gives
det(tI − A) = det
0 0 ··· P
α
−1 t ··· a
1
0 −1 ··· a
2
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· t + a
n−1
= P
α
.
Then we notice that for
r
= 1, in an appropriate choice of basis, the matrix
looks like
C =
A 0 ··· 0
0 A ··· 0
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· A
.
Theorem. Let
L/K
be a finite but not separable extension. Then
tr
L/K
(
α
) = 0
for all α ∈ L.
Proof.
Pick
β ∈ L
such that
P
β
, the minimal polynomial of
β
over
K
, is not
separable. Then by the previous characterization of separable polynomials, we
know p = char K > 0 with P
β
= q(t
p
) for some q ∈ K[t].
Now consider
K ⊆ K(β
p
) ⊆ K(β) ⊆ L.
To show
tr
L/K
= 0, by the previous proposition, it suffices to show
tr
K(β)/K(β
p
)
=
0.
Note that the minimal polynomial of
β
p
over
K
is
q
because
q
(
β
p
) = 0 and
q
is irreducible. Then [
K
(
β
) :
K
] =
deg P
β
=
p deg q
and
deg
[
K
(
β
p
) :
K
] =
deg q
.
So [K(β) : K(β
p
)] = p.
Now
{
1
, β, β
2
, ··· , β
p−1
}
is a basis of
K
(
β
) over
K
(
β
p
). Let
R
β
i
be the
minimal polynomial of β
i
over K(β
p
). Then
R
β
i
=
(
t − 1 i = 0
t
p
− β
ir
i = 0
,
We get the second case using the fact that
p
is a prime number, and hence
K
(
β
p
)(
β
i
) =
K
(
β
) if 1
≤ i < p
. So [
K
(
β
p
)(
β
i
) :
K
(
β
p
)] =
p
and hence the
minimal polynomial has degree p. Hence tr
K(β)/K(β
p
)
(β
i
) = 0 for all i.
Thus, tr
K(β)/K(β
p
)
= 0. Hence
tr
L/K
= tr
K(β
p
)/K
◦tr
K(β)/K(β
p
)
◦tr
L/K(β)
= 0.
Note that if L/K is a finite extension, and char K = 0, then
tr
L/K
(1) = [L : K] = 0.
So
tr
L/K
= 0. It is in fact true that all separable extensions have
tr
L/K
= 0, not
only when the field has characteristic 0.
Example. We want to show
3
√
3 ∈ Q
(
3
√
2
). Suppose not. Then we have
L
=
Q
(
3
√
3
) =
Q
(
3
√
2
), since both extensions of
Q
have degree 3. Then there
exists some a, b, c ∈ Q such that
3
√
3 = a + b
3
√
2 + c
3
√
2
2
.
We now compute the traces over Q. The minimal polynomials over Q are
P
3
√
3
= t
3
− 3, P
3
√
2
= t
3
− 2, P
3
√
4
= t
3
− 4.
So we have
tr
L/Q
(
3
√
3) = a tr
L/Q
(1) + b tr
L/Q
(
3
√
2) + c tr
L/Q
(
3
√
4).
Since the minimal polynomials above do not have coefficients in
t
2
, the traces of
the cube roots are zero. So we need a = 0. Then we are left with
3
√
3 = b
3
√
2 + c
3
√
4.
We apply the same trick again. We multiply by
3
√
2 to obtain
3
√
6 = b
3
√
4 + 2c.
We note that the minimal polynomial of
3
√
6 is t
3
− 6. Taking the trace gives
tr
L/Q
(
3
√
6) = b tr
L/Q
(
3
√
4) + 6c.
Again, the traces are zero. So c = 0. So we have
3
√
3 = b
3
√
2.
In other words,
b
3
=
3
2
,
which is clearly nonsense. This is a contradiction. So
3
√
3 ∈ Q(
3
√
2).
We can obtain another formula for the trace and norm as follows:
Theorem. Let
L/K
be a finite separable extension. Pick a further extension
E/L such that E/K is normal and
|Hom
K
(L, E)| = [L : K].
Write Hom
K
(L, E) = {φ
1
, ··· , φ
n
}. Then
tr
L/K
(α) =
n
X
i=1
φ
i
(α), N
L/K
(α) =
n
Y
i=1
φ
i
(α)
for all α ∈ L.
Proof.
Let
α ∈ L
. Let
P
α
be the minimal polynomial of
α
over
K
. Then there
is a one-to-one correspondence between
Hom
K
(K(α), E) ←→ Root
P
α
(E) = {α
1
, ··· , α
d
}.
wlog we let α = α
1
.
Also, since
|Hom
K
(L, E)| = [L : K],
we get
|Hom
K
(K(α), E)| = [K(α) : K] = deg P
α
.
Moreover, the restriction map
Hom
K
(
L, E
)
→ Hom
K
(
K
(
α
)
, E
) (defined by
φ 7→ φ|
K(α)
) is surjective and sends exactly [
K
(
α
) :
K
] elements to any particular
element in Hom
K
(K(α), E).
Therefore
X
φ
i
(α) = [L : K(α)]
X
ψ∈Hom
K
(K(α),E)
ψ(α) = [L : K(α)]
d
X
i=1
α
i
.
Moreover, we can read the sum of roots of a polynomial is the (negative of the)
coefficient of t
d−1
, where
P
α
= t
d
+ a
d−1
t
d−1
+ ··· + a
0
.
So
X
φ
i
(α) = [L : K(α)](−a
d−1
) = tr
L/K
(α).
Similarly, we have
Y
φ
i
(α) =
Y
ψ∈Hom
K
(K(α),E)
ψ(α)
[L:K(α)]
=
d
Y
i=1
α
i
!
[L:K(α)]
= ((−1)
d
a
0
)
[L:K(α)]
= N
L/K
(α).
Corollary. Let
L/K
be a finite separable extension. Then there is some
α ∈ L
such that tr
L/K
(α) = 0.
Proof. Using the notation of the previous theorem, we have
tr
L/K
(α) =
X
φ
i
(α).
Similar to a previous lemma, we can show that
φ
1
, ··· , φ
n
are “linearly indepen-
dent” over
E
, and hence
P
φ
i
cannot be identically zero. Hence there is some
α
such that
tr
L/K
(α) =
X
φ
i
(α) = 0.
Example. Let
K
=
F
q
⊆ L
=
F
q
n
, with
q
is a power of some prime number
p
.
By a previous theorem on finite fields, we know L/K is Galois and
Gal(L/K) =
Z
nZ
and is generated by the Frobenius φ = Fr
q
.
To apply the theorem, we had to pick an
E
such that
E/K
is normal and
Hom
K
(
L, E
) = [
L
:
K
]. However, since
L/K
is Galois, we can simply pick
E = L.
Then we know
tr
L/K
(α) =
X
ψ∈Gal(L/K)
ψ(α)
=
n−1
X
i=0
φ
i
(α)
= α + α
q
+ α
q
2
+ ··· + α
q
n−1
.
Similarly, the norm is
N
L/K
(α) =
n−1
Y
i=0
φ
i
(α) = α · α
q
· ··· · α
q
n−1
.
Recall that when solving quadratic equations
f
=
t
2
+
bt
+
c
, we defined the
discriminant as
b
2
−
4
c
. This discriminant then determined the types of roots of
f
. In general, we can define the discriminant of a polynomial of any degree, in a
scary way.
Definition (Discriminant). Let
K
be a field and
f ∈ K
[
t
],
L
the splitting field
of f over K. So we have
f = a(t − α
1
) ···(t − α
n
)
for some a, α
1
, ··· , α
n
∈ L. We define
∆
f
=
Y
i<j
(α
i
− α
j
), D
f
= ∆
2
f
= (−1)
n(n−1)/2
Y
i=j
(α
i
− α
j
).
We call D
f
the discriminant of f.
Clearly, D
f
= 0 if and only if f has no repeated roots.
Theorem. Let
K
be a field and
f ∈ K
[
t
],
L
is the splitting field of
f
over
K
.
Suppose D
f
= 0 and char K = 2. Then
(i) D
f
∈ K.
(ii)
Let
G
=
Gal
(
L/K
), and
θ
:
G → S
n
be the embedding given by the
permutation of the roots. Then
im θ ⊆ A
n
if and only if ∆
f
∈ K
(if and
only if D
f
is a square in K).
Proof.
(i) It is clear that D
f
is fixed by Gal(L/K) since it only permutes the roots.
(ii)
Consider a permutation
σ ∈ S
n
of the form
σ
= (
ℓ m
), and let it act on
the roots. Then we claim that
σ(∆
f
) = −∆
f
. (†)
So in general, odd elements in
S
n
negate ∆
f
while even elements fix it.
Thus, ∆
f
∈ K
iff ∆
f
is fixed by
Gal
(
L/K
) iff every element of
Gal
(
L/K
)
is even.
To prove (
†
), we have to painstakingly check all terms in the product. We
wlog
ℓ < m
. If
k < ℓ, m
. Then this swaps (
α
k
−α
ℓ
) with
α
k
−α
m
), which
has no effect. The
k > m
case is similar. If
ℓ < k < m
, then this sends
(
α
ℓ
− α
k
)
7→
(
α
m
− α
k
) and (
α
k
− α
m
)
7→
(
α
ℓ
− α
m
). This introduces two
negative signs, which has no net effect. Finally, this sends (
α
k
− α
m
) to
its negation, and so introduces a negative sign.
We will later use this result to compute certain Galois groups. Before that,
we see how this discriminant is related to the norm.
Theorem. Let
K
be a field, and
f ∈ K
[
t
] be an
n
-degree monic irreducible
polynomial with no repeated roots. Let
L
be the splitting field of
f
over
K
, and
let α ∈ Root
F
(L). Then
D
f
= (−1)
n(n−1)/2
N
K(α)/K
(f
′
(α)).
Proof.
Let
Hom
K
(
K
(
α
)
, L
) =
{φ
1
, ··· , φ
n
}
. Recall these are in one-to-one
correspondence with Root
f
(L) = {α
1
, ··· , α
n
}. Then we can compute
Y
i=j
(α
i
− α
j
) =
Y
i
Y
j=i
(α
i
− α
j
).
Note that since f is just monic, we have
f = (t − α
1
) ···(t − α
n
).
Computing the derivative directly, we find
Y
j=i
(α
i
− α
j
) = f
′
(α
i
).
So we have
Y
i=j
(α
i
− α
j
) =
Y
i
f
′
(α
i
).
Now since the φ
i
just maps α to α
i
, we have
Y
i=j
(α
i
− α
j
) =
Y
i
φ
i
(f
′
(α)) = N
K(α)/K
(f
′
(α)).
Finally, multiplying the factor of (−1)
n(n−1)/2
gives the desired result.
Example. Let
K
be a field with
char K
= 2
,
3. Let
f ∈ K
[
t
] have degree 3, say
f = t
3
+ bt + c
where we have gotten rid of the
t
2
term as in the first lecture. We further assume
f is irreducible with no repeated roots, and let L be the splitting field of f.
We want to compute the discriminant of this polynomial. Let
α ∈ Root
f
(
L
).
Then
β = f
′
(α) = 3α
2
+ b.
Then we can see
β = −2b −
3c
α
.
Alternatively, we have
α =
−3c
β + 2b
. (∗)
Putting (
∗
) into
α
3
+
bα
+
c
= 0, we find the minimal polynomial of
β
has
constant term −4b
3
− 27c
2
. This then gives us the norm, and we get
D
f
= −N
K(α)/K
(β) = −4b
3
− 27c
2
.
This is the discriminant of a cubic.
We can take a specific example, where
f = t
3
− 31t + 62.
Then
f
is irreducible over
Q
. We can compute
D
f
, and find that it is a square.
So the previous theorem says the image of the Galois group
Gal
(
L/K
) is a
subgroup of
A
3
. However, we also know
Gal
(
L/K
) has three elements since
deg f = 3. So we know Gal(L/K)
∼
=
A
3
.