Page 272 - DMTH403_ABSTRACT_ALGEBRA

Unit 28: Galois Theory

Notes
p
r
m
 (x) = x . To compute the order of in Gal(F/K), first note that  is the identity since  (x) =
m
x p rm = x p n  x for all x  F. But  cannot have lower degree, since this would give a polynomial
with too many roots. It follows that  is a generator for Gal(F/K).
28.2 Repeated Roots

In computing the Galois group of a polynomial, it is important to know whether or not it has
repeated roots. A field F is called perfect if no irreducible polynomial over F has repeated roots.
This section includes the results that any field of characteristic zero is perfect, and that any finite
field is perfect.
In the previous section, we showed that the order of the Galois group of a polynomial with no
repeated roots is equal to the degree of its splitting field over the base field. The first thing in this
section is to develop methods to determine whether or not a polynomial has repeated roots.
Let f(x) be a polynomial in K[x], and let F be a splitting field for f(x) over K. If f(x) has the
factorization f(x) = (x – r ) m 1 ... (x – r ) m t over F, then we say that the root r has multiplicity m . i
t
1
i
If m = 1, then r is called a simple root.
i
i
Let f(x)  K[x], with f(x) =  t k=0 a x . The formal derivative f’(x) of f(x) is defined by the formula
k
k
f'(x) =  t k=0 ka x , where ka denotes the sum of a added to itself k times. It is not difficult to
k-1
k
k
k
show from this definition that the standard differentiation formulas hold. Proposition shows
that the polynomial f(x)  K[x] has no multiple roots iff it is relatively prime to its formal
derivative f’(x). Proposition shows that f(x) has no multiple roots unless char(K) = p  0 and f(x)
has the form f(x) = a + a x + a x + ... + a x .
p
np
2p
0
n
2
1
A polynomial f(x) over the field K is called separable if its irreducible factors have only simple
roots. An algebraic extension field F of K is called separable over K if the minimal polynomial
of each element of F is separable. The field F is called perfect if every polynomial over F is
separable.
Theorem states that any field of characteristic zero is perfect, and a field of characteristic
p > 0 is perfect if and only if each of its elements has a pth root in the field. It follows immediately
from the theorem that any finite field is perfect.
To give an example of a field that is not perfect, let p be a prime number, and let K = Z . Then in
p
the field K(x) of rational functions over K, the element x has no pth root. Therefore, this rational
function field is not perfect.
The extension field F of K is called a simple extension if there exists an element u  F such that
F = K(u). In this case, u is called a primitive element. Note that if F is a finite field, then Theorem
shows that the multiplicative group F is cyclic. If the generator of this group is a, then it is easy
x
to see that F = K(a) for any subfield K. Theorem shows that any finite separable extension is a
simple extension.
28.3 The Fundamental Theorem

Here we study the connection between subgroups of Gal(F/K) and fields between K and F. This
is a critical step in proving that a polynomial is solvable by radicals if and only if its Galois
group is solvable.

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