文档介绍：page 1 of Solutions Chapters 6-10
Section
1. We have r1 =2,r2 =1,r3 =1sot1 =1,t2 =0,t3 = 1. The algorithm terminates in
one step after after subtraction of (X1 + X2 + X3)(X1X2X3). The given polynomial can be
expressedas e1e3.
2. We have r1 =2,r2 =1,r3 =0sot1 =1,t2 =1,t3 = 0. At step 1, subtract
(X1 + X2 + X3)(X1X2 + X1X3 + X2X3). The result is −3X1X2X3 +4X1X2X3 = X1X2X3.
By inspection (or by a secondstep of the algorithm), the given polynomial can be expressed
as e1e2 + e3.
3. Equation (1) follows upon taking σ1(h) outside the summation and using the lin-
ear dependence. Equation (2) is also a consequence of the linear dependence, because
σi(h)σi(g)=σi(hg).
4. By hypothesis, the characters are distinct, so for some h ∈ G we have σ1(h) = σ2(h).
Thus in (3), each ai is nonzero and
=0 ifi =1;
σ(h) −σ(h)
1 i =0 ifi =2.
This contradicts the minimality of r. (Note that the i = 2 case is important, since there is
no contradiction if σ1(h) −σi(h) = 0 for all i.)
5. By (), the Galois group consists√ of the identity alone. Since the identity fixes all
elements, the fixedfieldof G is Q( 3 2).
6. Since C = R[i], an R-automorphism σ of C is determined by its action on i. Since σ must
permute the roots of X2 + 1 by (), we have σ(i)=i or −i. Thus the Galois group has
two elements, the identity automorphism plex conjugation.
7. plex number z is plex conjugation if andonly if z is real, so the
fixedfieldis R.
Section
p
1. The right side is a subset of the left since both Ei and Ei+1 are containedin Ei+1.
∈ p
Since Ei is containedin the set on the right, it is enough to show that αi+1 Ei(Ei+1).
p
By hypothesis, αi+1 is separable over F , hence over Ei(αi+1). By Section , Problem 3,
∈ p ⊆ p
αi+1 Ei(αi+1) Ei(Ei+1).
p p
2. Apply Section , Problem 7, with E = F (E ) replacedby Ei+1 = Ei(Ei+1), to conclude
that Ei+1 is separable over Ei. By the induction hypothesis, Ei is separable over F .By
transitivity of separable