Abstract Algebra Theory and Applications - Computer Science ...

19.1 EXTENSION FIELDS 329Proof. Let α ∈ E. Since [E : F ] = n, the elements1, α, . . . , α ncannot be linearly independent. Hence, there exist a i ∈ F , not all zero, suchthata n α n + a n−1 α n−1 + · · · + a 1 α + a 0 = 0.Therefore,p(x) = a n x n + · · · + a 0 ∈ F [x]is a nonzero polynomial with p(α) = 0.Remark. Theorem 19.6 says that every finite extension of a field F is analgebraic extension. The converse is false, however. We will leave it as anexercise to show that the set of all elements in R that are algebraic over Qforms an infinite field extension of Q.The next theorem is a counting theorem, similar to Lagrange’s Theoremin group theory. Theorem 19.6 will prove to be an extremely useful tool inour investigation of finite field extensions.Theorem 19.7 If E is a finite extension of F and K is a finite extensionof E, then K is a finite extension of F and[K : F ] = [K : E][E : F ].Proof. Let {α 1 , . . . , α n } be a basis for E as a vector space over F and{β 1 , . . . , β m } be a basis for K as a vector space over E. We claim that{α i β j } is a basis for K over F . We will first show that these vectors spanK. Let u ∈ K. Then u = ∑ mj=1 b jβ j and b j = ∑ ni=1 a ijα i , where b j ∈ Eand a ij ∈ F . Then(m∑ n∑)u = a ij α i β j = ∑ a ij (α i β j ).j=1 i=1i,jSo the mn vectors α i β j must span K over F .We must show that {α i β j } are linearly independent. Recall that a setof vectors v 1 , v 2 , . . . , v n in a vector space V are linearly independent ifc 1 v 1 + c 2 v 2 + · · · + c n v n = 0□implies thatc 1 = c 2 = · · · = c n = 0.