i.e.
deg(Φ
p
1
...p
t
(
x
p
m
1

1
1
...p
m
t

1
t
)) = (
p
1

1)
. . .
(
p
t

1)
p
m
1

1
1
. . . p
m
t

1
t
= deg(Φ
n
(
x
))
.
Furthermore, it is easy to see, because Φ
n
is always a monic polynomial,
and the arguments in both these cases have a leading coefficient of 1, both
the polynomials are monic.
Now, let
ζ
be a primitive
n
th root of unity. We claim that
ζ
p
m
1

1
1
...p
m
t

1
t
is a primitive
p
1
. . . p
t
th root of unity. This is because
(
ζ
p
m
1

1
1
...p
m
t

1
t
)
p
1
...p
t
=
ζ
p
m
1
1
...p
m
t
t
=
ζ
n
= 1
,
and for any
k
∈
N
, where
k < p
1
. . . p
t
, we have (
ζ
p
m
1

1
1
...p
m
t

1
t
)
k
6
= 1, as
p
m
1

1
1
. . . p
m
t

1
t
k < n
.
Hence, from the definition of the cyclotomic polynomials, it follows that
ζ
p
m
1

1
1
...p
m
t

1
t
is a root of Φ
p
1
...p
t
(
x
). However, this simply means that
ζ
is a root of Φ
p
1
...p
t
(
x
p
m
1

1
1
...p
m
t

1
t
).
Hence, Φ
p
1
...p
t
(
x
p
m
1

1
1
...p
m
t

1
t
) has
ζ
has a root, is monic, and has the same
degree as Irr(
ζ,
Q
, x
) = Φ
n
(
x
), and therefore we must have
Φ
n
(
x
) = Φ
p
1
...p
t
(
x
p
m
1

1
1
...p
m
t

1
t
)
.
2. Let
E
be an algebraic extension of
F
, and let
σ
:
E
→
E
be an
F
linear
embedding of
E
into itself. Show that
σ
is an automorphism.
Solution.
Let
σ
:
E
→
E
be the
F
linear embedding. Then,
σ
(
F
) =
F
⊆
σ
(
E
). Now, given any
a
∈
E
, since
E/F
is algebraic, there must be some
f
∈
F
[
x
] such that
f
(
a
) = 0. We apply
σ
to the coefficients of
f
, and ob
tain
σ
(
f
) =
f
, since
σ
(
F
) =
F
. Then,
f
(
σ
(
a
)) =
σ
(
f
)(
σ
(
a
)) =
σ
(
f
(
a
)) =
0. So,
σ
sends
a
to another root of
f
. Since
σ
is a homomorphism between
fields, it must necessarily be injective. Hence,
σ
permutes the roots of
f
.
In particular, there is a root
b
of
f
such that
σ
(
b
) =
a
.
Hence
σ
is surjective. Since
σ
is injective, it must be bijective, and hence
it is an automorphism.