Abstract Algebra Theory and Applications - Computer Science ...

368 CHAPTER 21 GALOIS THEORYof G(E/F ). However, σ r cannot be the identity for 1 ≤ r < k; otherwise,x prk − x would have p m roots, which is impossible.□Example 3. We can now confirm that the Galois group of Q( √ 3, √ 5 )over Q in Example 2 is indeed isomorphic to Z 2 × Z 2 . Certainly the groupH = {id, σ, τ, µ} is a subgroup of G(Q( √ 3, √ 5 )/Q); however, H must be allof G(Q( √ 3, √ 5 )/Q), since|H| = [Q( √ 3, √ 5 ) : Q] = |G(Q( √ 3, √ 5 )/Q)| = 4.Example 4. Let us compute the Galois group off(x) = x 4 + x 3 + x 2 + x + 1over Q. We know that f(x) is irreducible by Exercise 19 in Chapter 15.Furthermore, since (x − 1)f(x) = x 5 − 1, we can use DeMoivre’s Theoremto determine that the roots of f(x) are ω i , where i = 1, . . . , 4 andω = cos(2π/5) + i sin(2π/5).Hence, the splitting field of f(x) must be Q(ω). We can define automorphismsσ i of Q(ω) by σ i (ω) = ω i for i = 1, . . . , 4. It is easy to check thatthese are indeed distinct automorphisms in G(Q(ω)/Q). Since[Q(ω) : Q] = |G(Q(ω)/Q)| = 4,the σ i ’s must be all of G(Q(ω)/Q). Therefore, G(Q(ω)/Q) ∼ = Z 4 since ω isa generator for the Galois group.Separable ExtensionsMany of the results that we have just proven depend on the fact that apolynomial f(x) in F [x] has no repeated roots in its splitting field. It isevident that we need to know exactly when a polynomial factors into distinctlinear factors in its splitting field. Let E be the splitting field of a polynomialf(x) in F [x]. Suppose that f(x) factors over E asf(x) = (x − α 1 ) n 1(x − α 2 ) n2 · · · (x − α r ) nr =r∏(x − α i ) n i.We define the multiplicity of a root α i of f(x) to be n i . A root withmultiplicity 1 is called a simple root. Recall that a polynomial f(x) ∈ F [x]i=1