Page 300 - DMTH403_ABSTRACT_ALGEBRA

Unit 32: Solvability by Radicals

32.3 Summary Notes

An extension field F of K is called a radical extension of K if there exist elements

u , u , ... , u in F and positive integers n , n , ... , n such that
2
m
1
1
2
m
(i) F = K (u , u , ... , u ), and
1
m
2
(ii) u is in K and u is in K ( u , ... , u ) for i = 2, ... , m .
n
n
1
i-1
i i
1 1
For a polynomial f(x) in K[x], the polynomial equation f(x) = 0 is said to be solvable by
radicals if there exists a radical extension F of K that contains all roots of f(x).
Let F be the splitting field of x - 1 over a field K of characteristic zero. Then Gal(F/K) is an
n

abelian group.
Let K be a field of characteristic zero that contains all nth roots of unity, let a be an 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.
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 K.
p
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.
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
n that is not solvable by radicals, we only need to find a polynomial of degree n whose
Galois group over Q is S .
n
Any subgroup of S that contains both a transposition and a cycle of length 5 must be equal
 5
to S itself.
5
There exists a polynomial of degree 5 with rational coefficients that is not solvable by

radicals
32.4 Keywords
Radical Extension: An extension field F of K is called a radical extension of K if there exist
elements u , u , ... , u in F and positive integers n , n , ... , n such that
1
m
2
2
m
1
(i) F = K (u , u , ... , u )
1
m
2
Solvable by Radicals: For a polynomial f(x) in K[x], the polynomial equation f(x) = 0 is said to be
solvable by radicals if there exists a radical extension F of K that contains all roots of f(x).
32.5 Review Questions
1. We know that the cyclotomic polynomial
p
x  1 p 1 p 2
 p (x)  x 1  x   x   ... x 1



is irreducible over  for every prime p. Let w be a zero  (x), and consider the field ().
p
(a) Show that ,  ,..., are distinct zeros of  (x), and conclude that they are all the
2
p-1
p
zeros of  (x).
p
LOVELY PROFESSIONAL UNIVERSITY 293