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