Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-V
137
element of
defines a permutation of the roots of f(x) that lie in K.
as σbg= α α V α ∈k so σf = σ . Hence
f(x)
is one-to-one so
See Proof of the theorem : Topics in Algebra, by Hersteim.
Theorem 10.
The general polynomial of degree
Note:
Proof:
are roots of f(x). Sinc there are only finitely many roots of
defines a permutation of those roots of f(x) that lie in K).
is not solvable by radicals.
is called general polynomial of degree over k.
If F(a 1
, a 2
, -------a n
) is the field of rational functions in the n variables a 1
, a 2
, -------a n,
then the Galois group of
n n−1 the polynomial f ( x) = x + a1x + − − − − −a
over F(a
n
1
, a 2
, -------a n
) is S n
, the symmetric group of degree
n (see thu 5.6.3, Hersteins Topics in Algebra). But S n
is not solvable group when n ≥ 5 . Hence by thu 9, f(x)
is not solvable by radicals over F(a 1
, a 2
, -------a n
) when .
Summary of basic results, questions and examples:
1. Let F be a subfield of a field K. K may be regarded as a vector space over F. If is a finite dimensional
vector space, we call K a finite extension of F. If the dimension of the vector space K is n, we say
G ne
Θ f
σ
bg ∈
Ga xσ n
k
∈ jG K n
( ≥ K i
5 2 −1
that n-1 K is an extension of degree n over F. We write
∴ 1, c, Q) c
e= , x− k2 jj− + : −a kQ
−j 1, x−=
f,
bg 2a c+ i
−= −0 − − − σac nf −bg
1 ax +
i ha
= n
, 0a i
∈b σk
fgbg
cσ ai h= 0 fcσ
bg ai
h=
0
[K : F] = n
2. Let
This is read, "the degree of K over F is equal to n."
n. Then n elements
be algebraic over F and let p(x) be the minimal polynomial of e over F. Let degree of p(x) be
are linearly independent over F and generate the smallest
field F(c) which contains F and e. Now F(e) is a vector space of dimension n over the field F. Hence the
degree of F(c) over F is equal to the degree of the minimal polynomial of e over F.
bg
Example 1.
F c : F = deg I F,
C
Q 2
rr
b g
e j : Q = deg of irreducible
polynomial p(x) = x 2 –2 over Q = deg Irr eF
, 2j.
.
3. If K is a finite extension of F and K = F(a 1
, a 2
, -------a n
), then a 1
, a 2
, -------a n
have to be algebraic over
F.
This is a consequence of important theorem: