Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996

## § 6.2 Finite and algebraic extensions

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:

(1) u is algebraic over K;

(2) K(u) is a finite extension of K;

(3) u belongs to a finite extension of K.

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 u1, u2, . . . , un in F. Then the degree of

K ( u1, u2, . . . , un )

over K is at most the product of the degrees of ui 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.

## § 6.2 Solved problems

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(u2) = K(u). Solution

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