Page 298 - DMTH403_ABSTRACT_ALGEBRA

Unit 32: Solvability by Radicals

Theorem 1: Let K be a field of characteristic zero that contains all nth roots of unity, let a be an Notes
element of K, and let F be the splitting field of x -a over K. Then Gal(F/K) is a cyclic group whose
n
order is a divisor of n.
Theorem 2: Let p be a prime number, let K be a field that contains all pth roots of unity, and let
F be an extension of K. If [F:K] = |Gal(F/K)| = p, then F = K(u) for some u in F such that u is in
p
K.
Lemma: Let K be a field of characteristic zero, and let E be a radical extension of K. Then there
exists an extension F of E that is a normal radical extension of K.
Theorem 3: Let f(x) be a polynomial over a field K of characteristic zero. The equation
f(x) = 0 is solvable by radicals if and only if the Galois group of f(x) over K is solvable.
S is not solvable for n  5, and so to give an example of a polynomial equation of degree n that
n
is not solvable by radicals, we only need to find a polynomial of degree n whose Galois group
over Q is S .
n
Lemma: Any subgroup of S that contains both a transposition and a cycle of length 5 must be
5
equal to S itself.
5
Theorem 4: There exists a polynomial of degree 5 with rational coefficients that is not solvable
by radicals

Example: Let f(x) be irreducible over Q, and let F be its splitting field over Q. Show that if
Gal (F/Q) is abelian, then F = Q(u) for all roots u of f(x).
Solution: Since F has characteristic zero, we are in the situation of the fundamental theorem of
Galois theory. Because Gal (F/Q) is abelian, every intermediate extension between Q and F must
be normal. Therefore, if we adjoin any root u of f(x), the extension Q(u) must contain all other
roots of f(x), since it is irreducible over Q. Thus Q(u) is a splitting field for f(x), so Q(u) = F.

Example: Find the Galois group of x -1 over Q.
9
Solution: We can construct the splitting field F of x -1 over Q by adjoining a primitive 9th root
9
of unity to Q. We have the factorization
x -1 = (x -1)(x +x +1)
9
3
6
3
= (x-1)(x +x+1)(x +x +1).
2
6
3
Substituting x+1 in the last factor yields
(x+1) +(x+1) +1 = x +6x +15x + 21x +18x +9x+3.
2
6
5
4
6
3
3
This polynomial satisfies Eisenstein’s criterion for the prime 3, which implies that the factor
x +x +1 is irreducible over Q. The roots of this factor are the primitive 9th roots of unity, so it
3
6
follows that [F:Q] = 6. Gal (F/Q) is isomorphic to a subgroup of Z Since Z is abelian of order
×
×
9
9
6, it is isomorphic to Z . It follows that Gal (F/Q)  Z .
6 6
Comment: The Galois group of x -1 over Q is isomorphic to Z and so the Galois group is cyclic
×
n
n
of order (n) iff n = 2, 4, p , or 2p , for an odd prime p.
k
k
Example: Show that x -x +x -x+1 is irreducible over Q, and use it to find the Galois group
3
2
4
of x -1 over Q.
10
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