Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-V
131
As F
= p dΘ F ≅ Z p i, each α i ∈ F can be chosen p ways. Hence that total number of ways in which an
element in k can be defined in p n ways. So
Theorem 6.
1. Let k be a finite field with p n elements. Then k is the splitting of the polynomial x
p n − x
subfield of k.
2. Two finite fields are isomorphic
.
they have the same number of elements.
over the prime
3. Let k be a finite field with p n elements. Then each subfield of k has p m elements for some divisor m of
n. Conversely, for each +ve divisor m of n
a unique subfield of k with
elements.
4. V prime p and V positive integer n, ∃ a field with
Proof:
elemetns.
1. Θ k has
elements, then k*, the multiplicative group of k has p n –1 elements. Hence for any
∗
−
x ∈k ⊂ k, x pn 1
= 1 , so p
x n = x V
. The polynomial
has atmost p n roots
and so its roots must be precisely the elements of k. Hence k is the splitting field of f xbgover the
prime subfield of k.
2. is the corollary of (1). Let and k 2
be two finite fields with p n elements, containing prime subfields
F 1
and F 2
respectively. But F1 ≅ Z
p
≅ F2
. By (1), k 1
and k 2
are splitting fields x
p n − x over isomorphic
⇔∃
mk x
p m
n ∈ =
kp.
n p
f n
1
≅ k2 ⇔ k 1 = k2
.
n
pn p
xf( ( x) ) = x −1 n
−
x1
. fields F 1
and F 2.
Hence from unit (IV),
p m 3. Let F 1
be the prime subfield of k. Let k 1
be a subfield of k. Then
n = k: F1 = k: k1 k1: F1 k1:
F1
n
Let k1: F1
= m,
so any subfield k 1
of k must have p m elements such that m n .
–1 is a divisor of p n − 1 and so q(x) =
Conversely, suppose
x pm − 1
− 1 is a divisor of
for some positive integer m. Then
p
As k is the splitting field of x n − x = xf ( x) over F 1
. We know
p
{ } n
is a subfield of k and
that a ∈ k:
a = a
has distinct roots. So k must contain all p m
distinct roots of xg(x). Hence these roots form a subfield of k. Moreover, any other subfield with p m elements
p
must be a splitting field of xg( x) = x m
− x.
Hence there exists unique subfield of k with p m elements.
p
4. Let k be the splitting field of f ( x) = x n
− x over its prime subfield isomorphic to Z p . Now
n
n
e j d i
p
p −1
x − x = x x − 1 and p × p n − 1 . p
So it is easy to see that f ( x) = x n
− x has distinct roots.
p
{ a ∈ k:
a n
= a} is a subfield of k and so set of all roots of f(x) is a subfield of k. Hence k consists of