ABSTRACT
This section applies field theory and group theory to prove that there is no general method to solve fifth degree equations using only field operations and extractions of roots. We start by defining a splitting field for a polynomial f over a field
F , which is a minimal field extension of F where f factors into linear polynomials. There is always a splitting field and a splitting field is a finite extension of F (Theorem 45.1). Furthermore, any two splitting fields for a given f ∈ F [x] are isomorphic (Theorem 45.2). Splitting fields have the surprising property that if K is a splitting field over F for f , and g ∈ F [x] is irreducible over F with a root in K, then g also factors into linear polynomials in K (Theorem 45.3). An extension field K of F is a normal extension of F if whenever
an irreducible f ∈ F [x] has one root in K, then it splits in K. If F has characteristic zero, then K is a finite normal extension of F if and only if K is a splitting field for some irreducible polynomial in F [x] (Theorem 45.6). Furthermore, the roots of an irreducible polynomial over F are all distinct in its splitting field (Theorem 45.4), as long as the field is of characteristic zero. Another important property of fields of characteristic zero is this: any finite extension is actually simple (Theorem 45.5). The section digresses a bit with a chapter on finite fields where we use
the idea of splitting field to prove that every finite field of characteristic p (prime) has pn elements (Theorem 46.1), and, indeed, a unique such field exists of order pn, for every p and n. We denote this Galois field by GF (pn). Also, GF (pm) is a subfield of GF (pn) if and only if m divides n. If E is an extension field of F , then the set of automorphisms of
E that fix elements in F forms a group, denoted by Gal(E|F ) and called the Galois group of the field E over F . We then explore the relationship between the order of these Galois groups and the degrees of the field extensions and discover that these counts are equal in the
∈ F [x] an ducible polynomial, and α ∈ K\F a root of f . Suppose that ϕ ∈ Gal(F (α)|F ). Then ϕ is entirely determined by ϕ(α). Furthermore, ϕ(α) must be a root of f in K, and so
|Gal(F (α)|F )| ≤ deg(f) = [F (α) : F ].
