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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 *n*th roots.
Only the extraction of an *n*th root leads to a larger field,
and so our formal definition is phrased in terms of subfields
and adjunction of roots of
x^{n}-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

u_{1}, u_{2}, ... , u_{m} in F
and positive integers
n_{1}, n_{2}, ... , n_{m}
such that

**(i)**F = K (u_{1}, u_{2}, ... , u_{m}), and**(ii)**u_{1}^{n1}is in K and u_{i}^{ni}is in K ( u_{1}, ... , u_{i-1}) for i = 2, ... , m .

**Proposition 8.4.2.**
Let F be the splitting field of
x^{n} - 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 *n*th roots of unity,
let a be an element of K,
and let F be the splitting field of x^{n}-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 *p*th 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^{p} 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 S_{n}
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 S_{n}.

**Lemma 8.4.7.**
Any subgroup of S_{5}
that contains both a transposition and a cycle of length 5
must be equal to S_{5} 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 x^{9}-1 over **Q**.
*Solution*

**9.**
Show that x^{4}-x^{3}+x^{2}-x+1
is irreducible over **Q**,
and use it to find the Galois group of x^{10}-1 over **Q**.
*Solution*

**10.**
Show that p(x) = x^{5}-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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