Page 299 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Solution: We can construct the splitting field F of x -1 over Q by adjoining a primitive 10th root
10
of unity to Q. We have the factorization
x -1 = (x -1)(x +1)
5
10
5
= (x-1)(x +x +x +x+1) (x+1)(x -x +x -x+1).
3
4
4
3
2
2
Substituting x-1 in the last factor yields
(x-1) -(x-1) +(x-1) -(x-1)+1
2
4
3
= (x -4x +6x -4x+1) - (x -3x +3x-1) + (x -2x+1) - (x-1) + 1
2
4
3
2
2
3
= x -5x +10x -10x+5.
2
3
4
This polynomial satisfies Eisenstein’s criterion for the prime 5, which implies that the factor
x -x +x -x+1 is irreducible over Q.
4
2
3
The roots of this factor are the primitive 10th roots of unity, so it follows that [F:Q] = 4. The proof
of Theorem 1 shows that Gal (F/Q)  Z and so the Galois group is cyclic of order 4.
×
10
Example: Show that p(x) = x -4x+2 is irreducible over Q, and find the number of real
5
roots. Find the Galois group of p(x) over Q, and explain why the group is not solvable.
Solution: The polynomial p(x) is irreducible over Q since it satisfies Eisenstein’s criterion for
p = 2. Since p(-2) = -22, p(-1) = 5, p(0) = 2, p(1) = –1, and p(2) = 26, we see that p(x) has a real root
between -2 and -1, another between 0 and 1, and a third between 1 and 2. The derivative
p’(x) = 5x -4 has two real roots, so p(x) has one relative maximum and one relative minimum,
4
and thus it must have exactly three real roots. It follows as in the proof of Theorem 2 that the
Galois group of p(x) over Q is S , and so it is not solvable.
5
Self Assessment
1. Let F be splitting field of x – 1 over a field K of characteristic .................. then G(F/K) is an
n
abelian group.
(a) 1 (b) 0
(c) –1 (d) –2
2. Let K be a field of characteristic zero and let  be a .................. of K. Thus there exists an
extension of F of  that is normal radical extension.
(a) radical extension (b) solvable group
(c) Galois group (d) finite element
3. There exists a polynomial of degree .................. with rational co-efficients that is not solvable
by radical.
(a) 4 (b) 5
(c) 6 (d) 7

4. Any subgroup of S that contains both a transposition and cycle of length .................. must
5
be equal to S itself.
5
(a) 4 (b) 5
(c) 3 (d) 6

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