Abstract Algebra Theory and Applications - Computer Science ...

332 CHAPTER 19 FIELDSwhere each field F (α 1 , . . . , α i ) is algebraic over F (α 1 , . . . , α i−1 ).(3) ⇒ (1). LetE = 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 ). SinceF (α 1 , . . . , α i ) = F (α 1 , . . . , α i−1 )(α i )is simple extension and α i is algebraic over F (α 1 , . . . , α i−1 ), it follows that[F (α 1 , . . . , α i ) : F (α 1 , . . . , α i−1 )]is finite for each i. Therefore, [E : F ] is finite.□Algebraic ClosureGiven a field F , the question arises as to whether or not we can find a fieldE such that every polynomial p(x) has a root in E. This leads us to thefollowing theorem.Theorem 19.11 Let E be an extension field of F . The set of elements inE that are algebraic over F form a field.Proof. Let α, β ∈ E be algebraic over F . Then F (α, β) is a finite extensionof F . Since every element of F (α, β) is algebraic over F , α ± β, α/β, andα/β (β ≠ 0) are all algebraic over F . Consequently, the set of elements inE that are algebraic over F forms a field.□Corollary 19.12 The set of all algebraic numbers forms a field; that is, theset of all complex numbers that are algebraic over Q makes up a field.Let E be a field extension of a field F . We define the algebraic closureof a field F in E to be the field consisting of all elements in E that arealgebraic over F . A field F is algebraically closed if every nonconstantpolynomial in F [x] has a root in F .Theorem 19.13 A field F is algebraically closed if and only if every nonconstantpolynomial in F [x] factors into linear factors over F [x].