10.1.2
We know that Q is a field. Q is a group under addition, therefore so is Q2.
Let a,b be in Q, v=(x,y),u=(z,t) be in Q2. Then a(x,y):=(ax,ay) is in Q2
(i) a(v+u)=a((x,y)+(z,t))=a(x+z,y+t)=(ax+az,ay+at)=(ax,ay)+(az,at)= a(x,y)+a(z,t)=av+au
(ii) (a+b)v=(ax+bx,ay+by)=(ax,ay)+(bx,by)=av+bv.
(iii) a(bv)=a(bx,by)=(abx,aby)=ab(x,y)=(ab)v.
(iv) 1v=1(a,b)=(1a,1b)=(a,b)=v.
So Q2 is a vector space over Q.
10.1.9 
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let a= ( - \/ 2 - \/ 3 )/ \/ 2 , b=1, c=1.
Then it is easy to check that
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 a\/ 2 + b(\/ 2 + i) + c(\/ 3 - i) = 0.

(a,b,c are non-zero)
10.1.10
Suppose {v} is linearly dependent. Then there is a c in F s.t. cv=0. F is a field, so there must be a an inverce d in F s.t. cd=dc=1. Then v=dcv=d0=0. This contradicts to the assumption that v!=0, so {v} must be linearly independent.
10.2.1
Let A be the intersection. We need to show that 0 is in A, if a,b are in A, then a+b, ab, -a, a-1 are there. I will show only the first two, the others can be done similarly. Suppose 0 is in all of Ei's, so it must be in their intersection A.
a,b are in A. This means a and b are in Ei for every i in I. Ei is a field, therefor a+b is in Ei for every i in I. This means that a+b must be in the intersection of the Ei's. So a+b is in A.
10.2.2
F(u2) by definition is the smallest subfield of K that contains u2 and F, so it is enough to show that F(u) contains F and u2. F(u) is a field and contains u, so it must contain u2. F(u) contains F by definition.
10.2.4
4 and 1-i are in Q(1-i) , therefor 3+i=4-(1-i) is also in Q(1-i). This means Q(3+i) is a subfield of Q(1-i). The same way Q(1-i) is a subfield of Q(3+i), therefor Q(1-i)=Q(3+i).
10.2.5a 
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             3/    
Let x = 1 + \/  2    .
Then (x-1)3=2. So x3 - 3x2 + 3x1 -3 = 0. This means x is algebraic over Q.
10.2.9 
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\/Pi  satisfies x2 - Pi=0, so 
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\/Pi is algebraic over Q(Pi).