Abstract Algebra Theory and Applications - Computer Science ...

19.1 EXTENSION FIELDS 331This example shows that it is possible that some extension F (α 1 , . . . , α n ) isactually a simple extension of F even though n > 1.Example 9. Let us compute a basis for Q( 3√ 5, √ 5 i), where √ 5 is thepositive√square √ root of 5 and 3√ 5 is the real cube root of 5. We know that5 i /∈ Q(35 ), so[Q( 3√ 5, √ 5 i) : Q( 3√ 5 )] = 2.It is easy to determine that {1, √ 5i } is a basis for Q( 3√ 5, √ 5 i) over Q( 3√ 5 ).We also know that {1, 3√ 5, ( 3√ 5 ) 2 } is a basis for Q( 3√ 5 ) over Q. Hence, abasis for Q( √ 5, 3√ 5 ) over Q is{1, √ 5 i, 3√ 5, ( 3√ 5 ) 2 , ( 6√ 5 ) 5 i, ( 6√ 5 ) 7 i = 5 6√ 5 i or 6√ 5 i}.Notice that 6√ 5 i is a zero of x 6 + 5. We can show that this polynomial isirreducible over Q using Eisenstein’s Criterion, where we let p = 5. Consequently,Q ⊂ Q( 6√ 5 ) ⊂ Q( 3√ 5, √ 5 i).But it must be the case that Q( 6√ 5 i) = Q( 3√ 5, √ 5 i), since the degree ofboth of these extensions is 6.Theorem 19.10 Let E be a field extension of F . Then the following statementsare equivalent.1. E is a finite extension of F .2. There exists a finite number of algebraic elements α 1 , . . . , α n ∈ E suchthat E = F (α 1 , . . . , α n ).3. There exists a sequence of fieldsE = F (α 1 , . . . , α n ) ⊃ F (α 1 , . . . , α n−1 ) ⊃ · · · ⊃ F (α 1 ) ⊃ F,where each field F (α 1 , . . . , α i ) is algebraic over F (α 1 , . . . , α i−1 ).Proof. (1) ⇒ (2). Let E be a finite algebraic extension of F . Then E is afinite dimensional vector space over F and there exists a basis consisting ofelements α 1 , . . . , α n in E such that E = F (α 1 , . . . , α n ). Each α i is algebraicover F by Theorem 19.6.(2) ⇒ (3). Suppose that E = F (α 1 , . . . , α n ), where every α i is algebraicover F . ThenE = F (α 1 , . . . , α n ) ⊃ F (α 1 , . . . , α n−1 ) ⊃ · · · ⊃ F (α 1 ) ⊃ F,