Fields - Elliptic Curves: Number Theory and Cryptography - page 482

rational coe cients, that is, if it is not algebraic over Q . A set of elements S =
{ K (with i running through some (possibly infinite) index set I ) is called

algebraically dependent if there are n distinct elements α 1 ,...,α n of S ,for
some n

α i }⊆

1, and a nonzero polynomial f ( X 1 ,...,X n ) with rational coe cients
such that f ( α 1 ,...,α n ) = 0. The set S is called algebraically independent
if it is not algebraically dependent. This means that there is no polynomial
relation among the elements of S . A maximal algebraically independent subset
of K is called a transcendence basis of K .The transcendence degree
of K over Q is the cardinality of a transcendence basis (the cardinality is
independent of the choice of transcendence basis). If every element of K
is algebraic over Q , then the transcendence degree is 0. The transcendence
degree of C over Q is infinite, in fact, uncountably infinite.
Let K be a field of characteristic 0, and choose a transcendence basis S .Let
F be the field generated by Q and the elements of S . The maximality of S
implies that every element of K is algebraic over F . Therefore, K can be ob-
tained by starting with Q , adjoining algebraically independent transcendental
elements, then making an algebraic extension.
Let K and L be fields and let f : K → L be a homomorphism of fields. We
always assume f maps 1 ∈ K to 1 ∈ L .Then f is injective. One way to see
this is to note that if 0 = x ∈ K ,then1= f ( x ) f ( x − 1 )= f ( x ) f ( x ) − 1 ;since
f ( x ) has a multiplicative inverse, it cannot be 0.
The following result is very useful. It is proved using Zorn's Lemma (see
[71]).

≥

PROPOSITION C.5
Let K and L be fields. Assume that L is algebraically closed and that there
is a field homomorphism

f : K −→ L.

f : K f restricted to K is f.

Then there is a homomorphism → L such that

Proposition C.5 has the following nice consequence.

COROLLARY C.6
Let K be a field of characteristic 0. Assume that K has finite transcendence
degree over Q . Then there is a homomorphism K → C . Therefore, K can be

regarded as a subfield of C .

PROOF Choose a transcendence basis S = {α 1 ,...,α n } of K and let F be
the field generated by Q and S .Since C has uncountable transcendence degree
over Q ,wecanchoose n algebraically independent elements τ 1 ,...,τ n ∈ C .

Define f : F C by making f the identity map on Q and sett in g f ( α j )= τ j

for all j . The proposition says that f canbeex te ndedto f : F → C .Since
K is an algebraic extension of F ,wehave K ⊆ F . Restricting

→

f to K yields