Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996

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§ 8.4 Solvability by Radicals

 
In most results in this section we will assume that the fields have characteristic zero, in order to guarantee that no irreducible polynomial has multiple roots. When we say that a polynomial equation is solvable by radicals, we mean that the solutions can be obtained from the coefficients in a finite sequence of steps, each of which may involve addition, subtraction, multiplication, division, or taking nth roots. Only the extraction of an nth root leads to a larger field, and so our formal definition is phrased in terms of subfields and adjunction of roots of xn-a for suitable elements a.

 
Definition 8.4.1. An extension field F of K is called a radical extension of K if there exist elements
u1, u2, ... , um in F and positive integers n1, n2, ... , nm such that

(i) F = K (u1, u2, ... , um), and

(ii) u1n1 is in K and uini is in K ( u1, ... , ui-1 ) for i = 2, ... , m .

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).

 
Proposition 8.4.2. Let F be the splitting field of xn - 1 over a field K of characteristic zero. Then Gal(F/K) is an abelian group.

 
Theorem 8.4.3. 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 xn-a over K. Then Gal(F/K) is a cyclic group whose order is a divisor of n.

 
Theorem 8.4.4. 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 up is in K.

 
Lemma 8.4.5. 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 8.4.6. 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.

 
Theorem 7.7.2 shows that Sn is not solvable for n 5, and so to give an example of a polynomial equation of degree n that is not solvable by radicals, we only need to find a polynomial of degree n whose Galois group over Q is Sn.

 
Lemma 8.4.7. Any subgroup of S5 that contains both a transposition and a cycle of length 5 must be equal to S5 itself.

 
Theorem 8.4.8. There exists a polynomial of degree 5 with rational coefficients that is not solvable by radicals.



§ 8.4 Solved problems

 
7. 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

8. Find the Galois group of x9-1 over Q.     Solution

9. Show that x4-x3+x2-x+1 is irreducible over Q, and use it to find the Galois group of x10-1 over Q.     Solution

10. Show that p(x) = x5-4x+2 is irreducible over Q, and find the number of real roots. Find the Galois group of p(x) over Q, and explain why the group is not solvable.     Solution

 


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