10.4.4
By the Fundamental Theorem of algebra every polynomial over R splits in C.
If all the roots of f are in R, f splits in R, and therefore its splitting field is R.
Suppose f has a root a+bi in C which is not in R (so b!=0). f splits in C, so the splitting field
K of f is contained in C. But the splitting field must contain R and a+bi, so
R(a+bi) is contained in K and K is contained in C.
a and a+bi are in R(a+bi), so bi is in R(a+bi). b!=0, so i is in R(a+bi). This means
R(i) is contained in R(a+bi). But R(i)=C, so R(a+bi)=C, and we get that K=C.
10.4.9
Suppose F={a1,...,an} is algebraicly closed. WLOG a1!=0.
Then f(x)=a1+(x-a1)(x-a2)...(x-an) must have all its root in F, so at least one
of a1,...,an is a root for f. But for every i, f(ai)=a1!=0, which is a contradiction.
This means F can not be algebraicly closed.
10.5.10
Assume p(x) is separable, and let K be the splitting field of p(x) over F.
Then there are distinct u1,...,un in K and a c!=0 in F s.t. p(x)=c(x-u1)...(x-un).
So p'(x)=c(x-u2)...(x-un)+...+c(x-u1)...(x-u(n-1)).
p'(u1)=c(u1-u2)...(u1-un)!=0 since ui's are distinct, [which follows from
separability of p(x)], and so p'(x)!=0.
Assume p'(x)!=0. Then deg(p'(x)) is bigger than 0 and less than deg p(x). This
means p(x) does not divide p`(x). Also p(x) is irreducible, so (p(x),p'(x))
must be one. Now apply Lemma 10.16 to get that p is separable.
10.5.12b
The minimal polynomials of sqrt(3) and i are x^2-3 and x^2+1.
x^2-3=(x-sqrt(3))(x+sprt(3)), and x^2+1=(x-i)(x+i).
From the proof of the theorem if we choose
c!=(+/- sprt(3) - sqrt(3))/(i +/- i), for example 1, then
Q(sqrt(3),i)=Q(sqrt(3)+i).
10.6.5
F is a subfield of K, so char(F)=char(K) (1F=1K, and n1F=n1K for every n).
By theorem 10.23 F has order p^a where p=char(F), a=[F:Zp] and
K has order p^b where b=[K:Zp].
[K:Zp]=[K:F][F:Zp] so b=na. This means order of K is p^(na)=(p^a)^n=q^n.
11.1.4
| i | t | a | b |
i | i | t | a | b |
t | t | i | b | a |
a | a | b | i | t |
b | b | a | t | i |
11.1.5
GalFK is isomorphic to a subgroup of
Sn. By Lagrange's theorem
order(GalFK) | order(Sn), which is n!
So order(GalFK) | n!.