Mandelbrot Set and Fermat’s Last Theorem
115
This theorem, while it does restrict the size of a finite field, does not say that one
will exist for a particular power of a prime, nor does it specify how many finite fields
can exist of a particular order. The answers to these questions can be deduced from
the following theorem.
Theorem 2.3 Let L be a field with characteristic p and prime subfield K . Then L
is the splitting field for F(x) = x
P
n − x iff L has P
n elements.
Proof Suppose that L is the splitting field for F(x) = x
P
n − x over K . Since
( f (x), f
(x)) = 1, the roots of f (x) are distinct and so L has at least p
n elements.
Consider the subset
E = {P ∈ L|P
p
n = P}
of L. Clearly E contains p
n elements since it consists of the roots of f (x). Suppose
that p, β ∈ E; then (Pβ)
p
n = (P)
p
n (β)
p
n = Pβ and hence, Pβ in E. Also,
(P + β)
p
n =
p
n
i
P
i
β
p
n −i
= P
p
n + β
p
n = P + β
Since p | C( p
n
, i) for 0 < i < p
n , and hence ( p + β) in E. The existence of additive
and multiplicative inverses is easy to show, so E is a subfield of L and also a splitting
field for f (x). Thus by 2.1 E = L and L contains p
n elements.
Suppose now that L contains p
n . The multiplicative group of L, which we will
denote by L
∗ , forms a group of order p
n
− 1 and hence the order of any element of
L
∗ divides p
n
− 1. Thus P
p
n = P for all P in L
∗ and the relation is trivially true for
P = 0. Thus f (x) splits in L.
Theorem 2.4 G F( p
n
)∗ is cyclic
Proof The multiplicative group G F( p
n
)∗is, by definition, abelian and of order p
n
−
1. If p
n
− 1 = P
e 1
1 , . . . , p
e k
k , then, factoring G F( p
n
)∗ into a direct product of its
Sylow subgroups, we have G F( p
n
)∗ = S( p 1 ) × . . . × S( p k ), where S( p i ) is the
Sylow subgroup of order ( p i )
e i . The order of every element in S( p i ) is a power of
p i and let a i in S( p i ) have the maximal order, say( p i )
e
i , e
i ≤ e i , for i = 1, . . . , k.
Since ( p i , p j ) = 1, i not equal j, the element a = a 1 a 2 . . . a k has maximal order m =
( p 1 )
e
−1
. . . (p k )
e
k in G F( p
n
)∗. Furthermore, every element of G F( p
n
)∗ satisfies the
polynomial x
m
− 1, implying that m ≥ p
n
− 1. Since a ∈ G F( p
n
)∗ has order m, m
divides p
n
− 1 and so, m = p
n
− 1. Thus the element a is a generator and G F( p
n
)
∗
is cyclic.
A generator of G F( p
n
)∗ is called a primitive element of G F( p
n
).
The following theorem has some useful consequences.
Theorem 2.5 Over any field K , (x
m
− 1) | (x
n
− 1) iff m divides n
Proof If n = qm + r , with r < m, then by direct computation
115
This theorem, while it does restrict the size of a finite field, does not say that one
will exist for a particular power of a prime, nor does it specify how many finite fields
can exist of a particular order. The answers to these questions can be deduced from
the following theorem.
Theorem 2.3 Let L be a field with characteristic p and prime subfield K . Then L
is the splitting field for F(x) = x
P
n − x iff L has P
n elements.
Proof Suppose that L is the splitting field for F(x) = x
P
n − x over K . Since
( f (x), f
(x)) = 1, the roots of f (x) are distinct and so L has at least p
n elements.
Consider the subset
E = {P ∈ L|P
p
n = P}
of L. Clearly E contains p
n elements since it consists of the roots of f (x). Suppose
that p, β ∈ E; then (Pβ)
p
n = (P)
p
n (β)
p
n = Pβ and hence, Pβ in E. Also,
(P + β)
p
n =
p
n
i
P
i
β
p
n −i
= P
p
n + β
p
n = P + β
Since p | C( p
n
, i) for 0 < i < p
n , and hence ( p + β) in E. The existence of additive
and multiplicative inverses is easy to show, so E is a subfield of L and also a splitting
field for f (x). Thus by 2.1 E = L and L contains p
n elements.
Suppose now that L contains p
n . The multiplicative group of L, which we will
denote by L
∗ , forms a group of order p
n
− 1 and hence the order of any element of
L
∗ divides p
n
− 1. Thus P
p
n = P for all P in L
∗ and the relation is trivially true for
P = 0. Thus f (x) splits in L.
Theorem 2.4 G F( p
n
)∗ is cyclic
Proof The multiplicative group G F( p
n
)∗is, by definition, abelian and of order p
n
−
1. If p
n
− 1 = P
e 1
1 , . . . , p
e k
k , then, factoring G F( p
n
)∗ into a direct product of its
Sylow subgroups, we have G F( p
n
)∗ = S( p 1 ) × . . . × S( p k ), where S( p i ) is the
Sylow subgroup of order ( p i )
e i . The order of every element in S( p i ) is a power of
p i and let a i in S( p i ) have the maximal order, say( p i )
e
i , e
i ≤ e i , for i = 1, . . . , k.
Since ( p i , p j ) = 1, i not equal j, the element a = a 1 a 2 . . . a k has maximal order m =
( p 1 )
e
−1
. . . (p k )
e
k in G F( p
n
)∗. Furthermore, every element of G F( p
n
)∗ satisfies the
polynomial x
m
− 1, implying that m ≥ p
n
− 1. Since a ∈ G F( p
n
)∗ has order m, m
divides p
n
− 1 and so, m = p
n
− 1. Thus the element a is a generator and G F( p
n
)
∗
is cyclic.
A generator of G F( p
n
)∗ is called a primitive element of G F( p
n
).
The following theorem has some useful consequences.
Theorem 2.5 Over any field K , (x
m
− 1) | (x
n
− 1) iff m divides n
Proof If n = qm + r , with r < m, then by direct computation