Cryptography Reference
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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
( 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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