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Proposition 6.2.1.
Let F be an extension field of K and let
u be an element of F that is algebraic over K.
If the minimal polynomial of u over K has degree n,
then K(u) is an n-dimensional vector space over K.
Definition 6.2.2.
Let F be an extension field of K.
If the dimension of F as a vector space over K is finite,
then F is said to be a
finite
extension of K.
The dimension of F as a vector space over K is called the
degree
of F over K, and is denoted by [F:K].
Proposition 6.2.3.
Let F be an extension field of K and let
u be an element of F.
The following conditions are equivalent:
Theorem 6.2.4.
Let E be a finite extension of K and
let F be a finite extension of E.
Then F is a finite extension of K, and
[F:K] = [F:E][E:K].
Corollary 6.2.5.
Let F be a finite extension of K.
Then the degree of any element of F is a divisor of [F:K].
Corollary 6.2.6.
Let F be an extension field of K, with algebraic elements
u_{1},
u_{2},
. . . ,
u_{n} in F.
Then the degree of
K ( u_{1}, u_{2}, . . . , u_{n} )
over K is at most the product of the degrees of u_{i} over K, for 1 i n.
Corollary 6.2.7.
Let F be an extension field of K.
The set of all elements of F that are algebraic over K forms a subfield of F.
Definition 6.2.8.
An extension field F of K is said to be
algebraic
over K if each element of F is algebraic over K.
Proposition 6.2.9.
Every finite extension is an algebraic extension.
Example 6.2.3. (Algebraic numbers)
Let Q* be the set of all complex numbers u in C
such that u is algebraic over Q.
Then Q* is a subfield of C by Corollary
6.2.7, called the
field of algebraic numbers.
1.
Let F be a field generated over the field K by u and
v of relatively prime degrees m and n, respectively, over K.
Prove that [F:K] = mn.
Solution
2. Let F K be an extension field, and let u be an element of F. Show that if [K(u):K] is an odd number, then K(u^{2}) = K(u). Solution
Solutions to the problems | Forward to §6.3 | Back to §6.1 | Up | Table of Contents