4Computational techniques
II Galois Theory
4.1 Reduction mod p
The goal of this chapter is to see what happens when we reduce a polynomial
f ∈ Z[t] to the corresponding polynomial
¯
f ∈ F
p
[t].
More precisely, suppose we have a polynomial
f ∈ Z
[
t
], and
E
is its splitting
field over
Q
. We then reduce
f
to
¯
f ∈ F
p
[
t
] by reducing the coefficients mod
p
,
and let
¯
E be the splitting field of
¯
f over F
p
.
The ultimate goal is to show that under mild assumptions, there is an
injection
Gal(E/F
p
) → Gal(E/Q).
To do this, we will go through a lot of algebraic fluff to obtain an alternative
characterization of the Galois group, and obtain the result as an easy corollary.
This section will be notationally heavy. First, in the background, we have a
polynomial
f
of degree
n
(whose field we shall specify later). Then we will have
three distinct set of variables, namely (
x
1
, ··· , x
n
), (
u
1
, ··· , u
n
), plus a
t
. They
will play different roles.
–
The
x
i
will be placeholders. After establishing our definitions, we will then
map each x
i
to α
i
, a root of our f.
– The u
i
will stay as “general coefficients” all the time.
– t
will be the actual variable we think our polynomial is in, i.e. all polyno-
mials will be variables in
t
, and
u
i
and
x
i
will form part of the coefficients.
To begin with, let
L = Q(x
1
, ··· , x
n
)
F = Q(e
1
, ··· , e
n
).
where
x
i
are variables and
e
i
are the symmetric polynomials in the
x
1
, ··· , x
n
.
We have seen that Gal(L/F )
∼
=
S
n
.
Now let
B = Z[x
1
, ··· , x
n
]
A = Z[e
1
, ··· , e
n
].
It is an exercise on example sheet 4 to show that
B ∩ F = A. (∗)
We will for now take this for granted.
We now add it new variables
u
1
, ··· , u
n
, t
. We previously mentioned that
S
n
can act on, say
L
[
u
1
, ··· , u
n
, t
] by permuting the variables. Here there are two
ways in which this can happen — a permutation can either permute the
x
i
, or
permute the u
i
. We will have to keep this in mind.
Now for each σ ∈ S
n
, we define the linear polynomial
R
σ
= t − x
σ(1)
u
1
− ··· − x
σ(n)
u
n
.
For example, we have
R
(1)
= t − x
1
u
1
− ··· − x
n
u
n
.
As mentioned, an element
ρ ∈ S
n
can act on
R
ρ
in two ways: it either sends
R
σ
7→ R
ρσ
or R
σ
7→ R
σρ
−1
.
It should be clear that the first action permutes the
x
i
. What the second
action does is permute the
u
i
. To see this, we can consider a simple case where
n = 2. Then the action ρ acting on R
(1)
sends
t − x
1
u
1
− x
2
u
2
7→ t − x
ρ
−1
(1)
u
1
− x
ρ
−2
(2)
u
2
= t − x
1
u
ρ(1)
− x
2
u
ρ(2)
.
Finally, we define the following big scary polynomial:
R =
Y
σ∈S
n
R
σ
∈ B[u
1
, ··· , u
n
, t].
We see that this is fixed by any permutation in
σ ∈ S
n
under both actions.
Considering the first action and using (∗), we see that in fact
R ∈ A[u
1
, ··· , u
n
, t].
This is since if we view
R
as a polynomial over
B
in the variables
u
1
, ··· , u
n
, t
,
then its coefficients will be invariant under permuting the
x
i
. So the coefficients
must be a function of the e
i
, i.e. lie in A.
With these definitions in place, we can focus on a concrete polynomial.
Let K be a field, and let
f = t
n
+ a
1
t
n−1
+ ··· + a
n
∈ K[t]
have no repeated roots. We let E be the splitting field of f over K. Write
Root
f
(E) = {α
1
, ··· , α
n
}.
Note that this is the setting we had at the beginning of the chapter, but with an
arbitrary field K instead of Q and F
p
.
We define a ring homomorphism
θ
:
B → E
by
x
i
7→ α
i
. This extends to a
ring homomorphism
θ : B[u
1
, ··· , u
n
, t] → E[u
1
, ··· , u
n
, t].
Note that the ring homomorphism
θ
send
e
i
7→
(
−
1)
i
a
i
. So in particular, if
we restrict the homomorphism
θ
to
A
, then the image is restricted to the field
generated by
a
i
. But we already have
a
i
∈ K
. So
θ
(
A
) =
K
. In particular, since
R ∈ A[u
1
, ··· , u
n
, t], we have
θ(R) ∈ K[u
1
, ··· , u
n
, t].
Now let
P
be an irreducible factor of
θ
(
R
) in
K
[
u
1
, ··· , u
n
, t
]. We want to say
each such irreducible polynomial is related to the Galois group
G
=
Gal
(
E/K
).
Since
f
has no repeated roots, we can consider
G
as a subgroup of
S
n
, where
the elements of
G
are just the permutations of the roots
α
i
. We will then show
that each irreducible polynomial corresponds to a coset of G.
Recall that at the beginning, we said
S
n
can act on our polynomial rings
by permuting the
x
i
or
u
i
. However, once we have mapped the
x
i
to the
α
i
and focus on a specific field,
S
n
as a whole can no longer act on the
α
i
, since
there might be non-trivial relations between the
α
i
. Instead, only the subgroup
G ≤ S
n
can act on α
i
. On the other hand, S
n
can still act on u
i
.
Recall that
R
is defined as a product of linear factors
R
σ
’s. So we can find a
subset Λ ⊆ S
n
such that
P =
Y
σ∈Λ
R
σ
.
We will later see this Λ is just a coset of the Galois group G.
Pick σ ∈ Λ. Then by definition of P ,
R
σ
| P
in
E
[
u
1
, ··· , u
n
, t
]. Now if
ρ ∈ G
, then we can let
ρ
act on both sides by
permuting the
x
i
(i.e. the
α
i
). This does not change
P
because
P
has coefficients
in K and the action of G has to fix K. Hence we have
R
ρσ
| P.
More generally, if we let
H =
Y
ρ∈G
R
ρσ
∈ E[u
1
, ··· , u
n
, t],
then
H | P
by the irreducibility of P .
Since
H
is also invariant under the action of
G
, we know
H ∈ K
[
u
1
, ··· , u
n
, t
].
By the irreducibility of P , we know H = P . Hence, we know
Λ = Gσ.
We have thus proved that the irreducible factors of
θ
(
R
) in
K
[
u
1
, ··· , u
n
, t
] are
in one-to-one correspondence with the cosets of
G
in
S
n
. In particular, if
P
corresponds to G itself, then
P =
Y
τ∈G
R
τ
.
In general, if
P
corresponds to a coset
Gσ
, we can let
λ ∈ S
n
act on
P
by
permuting the u
i
’s. Then this sends
P =
Y
ρ∈G
R
ρσ
7→ Q =
Y
ρ∈G
R
ρσλ
−1
.
So this corresponds to the coset
Gσλ
−1
. In particular,
P
=
Q
if and only if
Gσ
=
Gσλ
−1
. So we can use this to figure out what permutations preserve an
irreducible factor. In particular, taking σ = (1), we have
Theorem.
G = {λ ∈ S
n
: λ preserves the irreducible factor corresponding to G}. (†)
This is the key result of this chapter, and we will apply this as follows:
Theorem. Let
f ∈ Z
[
t
] be monic with no repeated roots. Let
E
be the splitting
field of
f
over
Q
, and take
¯
f ∈ F
p
[
t
] be the obvious polynomial obtained by
reducing the coefficients of
f
mod
p
. We also assume this has no repeated roots,
and let
¯
E be the splitting field of
¯
f.
Then there is an injective homomorphism
¯
G = Gal(
¯
E/F
p
) → G = Gal(E/Q).
Moreover, if
¯
f
factors as a product of irreducibles of length
n
1
, n
2
, ··· , n
r
, then
Gal(f) contains an element of cycle type (n
1
, ··· , n
r
).
Proof. We apply the previous theorem twice. First, we take K = Q. Then
θ(R) ∈ Z[u
1
, ··· , u
n
, t].
Let
P
be the irreducible factor of
θ
(
R
) corresponding to the Galois group
G
.
Applying Gauss’ lemma, we know P has integer coefficients.
Applying the theorem again, taking
K
=
F
p
. Denote the ring homomorphism
as
¯
θ
. Then
¯
θ
(
R
)
∈ F
p
[
u
1
, ··· , u
n
, t
]. Now let
Q
be the irreducible factor
¯
θ
(
R
)
corresponding to
¯
G.
Now note that
θ
(
R
(1)
)
| P
and
¯
θ
(
R
(1)
)
| Q
, since the identity is in
G
and
¯
G
.
Also, note that
¯
θ
(
R
) =
θ(R)
, where the bar again denotes reduction mod
p
. So
Q |
¯
P .
Considering the second action of
S
n
(i.e. permuting the
u
i
), we can show
¯
G ⊆ G, using the characterization (†). Details are left as an exercise.
This is incredibly useful for computing Galois groups, as it allows us to
explicitly write down some cycles in Gal(E, Q).