Abstract Algebra Theory and Applications - Computer Science ...

372 CHAPTER 21 GALOIS THEORYThe problem is that some of the a i ’s may be in E but not in F . We mustshow that this is impossible.Suppose that at least one of the a i ’s is in E but not in F . By rearrangingthe α i ’s we may assume that a 1 is nonzero. Since any nonzero multiple of asolution is also a solution, we can also assume that a 1 = 1. Of all possiblesolutions fitting this description, we choose the one with the smallest numberof nonzero terms. Again, by rearranging α 2 , . . . , α n+1 if necessary, we canassume that a 2 is in E but not in F . Since F is the subfield of E that is fixedelementwise by G, there exists a σ i in G such that σ i (a 2 ) ≠ a 2 . Applyingσ i to each equation in the system, we end up with the same homogeneoussystem, since G is a group. Therefore, x 1 = σ i (a 1 ) = 1, x 2 = σ i (a 2 ), . . .,x n+1 = σ i (a n+1 ) is also a solution of the original system. We know thata linear combination of two solutions of a homogeneous system is also asolution; consequently,x 1 = 1 − 1 = 0x 2 = a 2 − σ i (a 2 ).x n+1 = a n+1 − σ i (a n+1 )must be another solution of the system. This is a nontrivial solution becauseσ i (a 2 ) ≠ a 2 , and has fewer nonzero entries than our original solution. Thisis a contradiction, since the number of nonzero solutions to our originalsolution was assumed to be minimal. We can therefore conclude that a 1 =· · · = a n+1 = 0. □Let E be an algebraic extension of F . If every irreducible polynomial inF [x] with a root in E has all of its roots in E, then E is called a normalextension of F ; that is, every irreducible polynomial in F [x] containing aroot in E is the product of linear factors in E[x].Theorem 21.13 Let E be a field extension of F . Then the following statementsare equivalent.1. E is a finite, normal, separable extension of F .2. E is a splitting field over F of a separable polynomial.3. F = E G for some finite group of automorphisms of E.Proof. (1) ⇒ (2). Let E be a finite, normal, separable extension of F . Bythe Primitive Element Theorem, we can find an α in E such that E = F (α).