Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

132
ADVANCED ABSTRACT ALGEBRA
precisely the roots of f(x), and it has exactly
Now we prove the beautiful result given below:
Theorem 7.
elements.
The multiplicative group of non-zero elements of a finite field is cyclic.
Proof:
Let k be a finite field of p n *
n
elements. k = k − ( 0).
So k * = p − 1 = m say. Let a ∈ k
* be of maximal
order, say m 1
i.e. o(a) = m 1
. Now we use the following result: (Let G be a finite abelian group. Let a ∈G
be an element of maximal order. Then order of every element of G is a divisor of this order of a).
By above result, each element of
satisfies f ( x) x m 1
= − 1 . Since k is a field, so there are at most m 1
roots of f(x), hence m ≤ m 1 . But m1 ≤ m, so m = m 1 , and = m. Therefore k * = < a > implies the
result.
Algebraically Closed field:
A field k is said to be algebraically closed, if every polynomial
in K.
Example (Fundamental Theorem of Algebra):
of +ve degree has a root
Every nonconstant polynomial with complex coefficients has a complex root i.e. splits into linear
factors.
Automorphism of extension:
Let K be an extension of the field k.
Define ψ : K → K
aC
Gk
pf
Ψ
K
such that
V
bg bg bg
Ψ ab = Ψ a • Ψ b
Ψ is 1–1 and onto
and (C) = C V
Then is k–automorphism of an extension field K.
The group of all k–automorphisms of K is called the Galois group of the field extension K. This
group is denoted by
.
Galois extension:
An extension K of the field k is called Galois extension if
1. K is algebraic extension of k.
2. The fixed field of
is k i.e.
= k