Cryptography Reference
In-Depth Information
the desired homomorphism from
K
C
. Since a homomorphism of fields is
injective,
K
is isomorphic to its image under this homomorphism. Therefore,
K
is isomorphic to a subfield of
C
.
→
The proposition also holds, with a similar proof, if the transcendence degree
of
K
is at most the cardinality of the real numbers, which is the cardinality
of a transcendence basis of
C
.
If
α ∈
C
is algebraic over
Q
,then
f
(
α
) = 0 for some nonzero polynomial
with rational coecients. Let Aut(
C
) be the set of field automorphisms of
C
and let
σ ∈
Aut(
C
). Then
σ
(1) = 1, from which it follows that
σ
is the
identity on
Q
. Therefore,
0=
σ
(
f
(
α
)) =
f
(
σ
(
α
))
,
so
σ
(
α
) is one of the finitely many roots of
f
(
X
). The next result gives a
converse to this fact.
PROPOSITION C.7
Let α
∈
C
. If the set
{σ
(
α
)
| σ ∈ Aut
(
C
)
},
where σ runs through all automorphisms of
C
, is finite, then α is algebraic
over
Q
.
PROOF
Suppose
α
is transcendental. There is a transcendence basis
S
of
C
with
α
∈
S
.Then
C
is algebraic over the field
F
generated by
Q
and
S
.
The map
σ
:
F
F
α −→ α
+1
β −→ β
−→
when
β ∈ S, β
=
α
defines an automorphism of
F
. By Proposition C.5,
σ
canbeextendedto
amap
σ
:
C
→
C
. We want to show that
σ
is an automorphism, which
means that we must show that
σ
is surjective. Let
y ∈
C
.Since
y
is algebraic
over
F
, there is a nonzero polynomial
g
(
X
) with coecients in
F
such that
g
(
y
) = 0. Let
g
σ
−
1
denote the result of applying
σ
−
1
to all of the coecients
of
g
(notethatweknow
σ
−
1
exists on
F
because we already know that
σ
is
an automorphism of
F
). For any root
r
of
g
σ
−
1
,wehave
0=
σ
g
σ
−
1
(
r
)
=
g
(
σ
(
r
))
.
Therefore,
σ
maps the roots of
g
σ
−
1
to roots of
g
. Since the two polynomials
havethesamenumberofroots,
σ
gives a bijection between the two sets of
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