and complete solution attached

Question 5:


1  i 3 1  i 3 1  i 3 
 1  i 3

, 
,
,
 . Then the
2
2
2
2 




The set of roots of x 4  x 2  1 is 

 1  i 3 
 . It can be observed that L  F i 3 . Moreover,

2



 

splitting filed is equal to L  F 
because of the identity i 3 

 
 1 is F  i 3  . Now, it is left to

1  i 3 1  i 3

, it can be written as F i 3  L . Putting
2
2

 

together we have L  F i 3 , so the splitting field of x 4  x 2

 

1  i 3
. But this is true since  is of the form of
2
1
1
a  bi 3 where a    F and b   F .
2
2

prove F i 3  F   where  

Hence, the desired result follows as F   is proved to be the splitting field of x 4  x 2  1 .
Question 6:
a) Let L be the splitting field. According to Euler’s formula, the roots of f  x   x4  1 are

1 
 1
i /4
, i
ei /4 , e3i /4 , e5i /4 , e7i /4 , which are simplified to be 
 . Since the powers of e
2
 2





can generate the other three zeros of f  x  , it follows that F ei /4  L and by adding or









 i and  ei /4   











  2   F  2  : F  and

subtracting the roots, we have i, 2  L . But since ei /4

2

3

1
i
, we

2
2

have F i, 2  F ei /4 and therefore i, 2  F ei / 4 . By minimality, we can conclude









F i, 2  L . Finally, since  F i, 2 : F    F i, 2 : F

 
F





but F i, 2 

, it can be concluded that we have extensions on the right-hand side





of degree 2 and hence  F i, 2 : F   4 . Thus, it implies that





L : F   4 .

b) Let L be the splitting field. The roots of x 6  1 are i,  ,  5 where   ei /6 . Hence, the





splitting field L of x 6  1 over F is F i,  ,  5  F  i,   . Moreover,  





form. Thus, F  i,    F  i, 2  i   F i, 3 . Now,  F



3 i
 in Cartesian
2 2

 3  : F   2 because

3  F and

 3  does not contain i because it’s a
real extension of F . Thus,  F  3, i  : F  3   1 and  F  3, i  F  3   2 because




x  1 F  3   x  is satisfied by i . Hence,  F  3, i  F  3   2 . Finally  F  i,   : F   4


has the minimal polynomial x 2  3  0 . Moreover, F

2

and hence the degree of spl...