MATH42122 Galois Theory.
Solutions IV.
A.Isomorphisms and Monomorphisms
1. We have
(0) + 0 = (0) = (0 + 0) = (0) + (0)
and hence 0 = (0) by cancelation in the additive group of E . Also,
(1) 1 = (1) = (1 1) = (1) (1),
and hence 1 = (1) by cancelation in t
MATH42122
Three hours
THE UNIVERSITY OF MANCHESTER
GALOIS THEORY
XX XXX 2013
X.00 X.00
Answer the ONE question in Section A (40 marks in total). Answer TWO of the THREE questions
in Section B (40 marks in total). If more then TWO questions from Section B
MATH42122 Galois Theory.
Solutions III.
1. (a) (Q( n 5 : Q) = n. This is because the minimum polynomial of n 5
over Q is xn 5. Indeed, this polynomial is irreducible by Eisensteins
Criterion, and it has n 5 as a zero.
(b) (Q( 2 + 3) : Q) = 4. Indeed, we h
MATH42122/62122
Three hours
THE UNIVERSITY OF MANCHESTER
GALOIS THEORY
XX May 2012
X.00 X.00
Answer the ONE question in Section A (40 marks)
and
TWO of the THREE questions in Section B (20 marks each).
The total number of marks on the paper is 80. A furth
MATH42122 Galois Theory.
Solutions I.
1. Let a, b, c, d Q, then
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 E,
(a + b 2) = a b 2 E,
(a + b 2)(c + d 2) = (ac + 2bd) + (ad + bc) 2 E,
and if a + b 2 = 0, then 1/(a + b 2) E as
(a + b 2)1 =
a2
a
b
+ 2
2.
2
2
MATH42122/MATH62122 Galois Theory
Coursework 2010/2011
TEST - SOLUTIONS
1. Since 7 = 1 and x7 1 = (x 1)f (x) where f (x) is the polynomial
given in Question 1, the minimum polynomial of is the irreducible polynomial f (x). Hence the degree of the simple a
MATH42122 Galois Theory.
1. The roots of f are 4 2 and i 4 2. Hence
Solutions VIII.
4
4
4
Q( 2, i 2) = Q( 2, i) = Q(, i)
is a splitting eld of f inside C.
2. The splitting eld E can be obtained from Q by two consecutive simple
extension, namely by rst adj
MATH42122 Galois Theory.
Solutions IX.
1. (a) The roots of f are i with i = 0, 1, 2, 3, 4 where = 5 2 and =
exp( 2i ). Hence E = Q(, , 2 , 3 , 4 ) = Q(, ) is a splitting
5
eld of f inside C.
(b) The eld E can be obtained from Q by two consecutive simple e
MATH42122 Galois Theory.
Solutions VI.
1. (a) The ve complex roots of x5 1 are precisely the elements , 2 , 3 , 4
and 5 = 1. But x5 1 = (x4 + x3 + x2 + x + 1)(x 1) and hence
the rst four powers of are the four roots of x4 + x3 + x2 + x + 1.
Consequently,