finite field

finite field

A

finite field

(also called a

Galois field

) is a field that has finitely many

elements

.
The number of elements in a finite field is sometimes called the

order

of the field.
We will present some basic facts about finite fields.

1

Size of a finite field

Theorem 1.1

.

A finite field

F

has positive characteristic

p

>

0

for some prime

p

. The

cardinality

of

F

is

p

n

where

n

:=

[

F

:

F

p

]

and

F

p

denotes
the

prime subfield

of

F

.

Proof.

The

characteristic

of

F

is

positive

because otherwise the

additive

subgroup

generated by

1

would be an

infinite subset

of

F

. Accordingly, the prime subfield

𝔽

p

of

F

is

isomorphic

to
the field

ℤ

/

p

⁢

ℤ

of

integers

mod

p

. The integer

p

is prime since otherwise

ℤ

/

p

⁢

ℤ

would have

zero divisors

. Since the field

F

is an

n

–dimensional

vector space

over

𝔽

p

for some finite

n

, it is set–isomorphic to

𝔽

p

n

and thus has cardinality

p

n

.
∎

2

Existence of finite fields

Now that we know every finite field has

p

n

elements, it is natural
to ask which of these actually arise as cardinalities of finite
fields. It turns out that for each prime

p

and each

natural number

n

, there is essentially exactly one finite field of size

p

n

.

Lemma 2.1

.

In any field

F

with

m

elements, the

equation

x

m

=

x

is satisfied by all elements

x

of

F

.

Proof.

The result is clearly true if

x

=

0

. We may therefore assume

x

is not zero. By

definition

of field, the set

F

×

of nonzero elements of

F

forms a group under

multiplication

. This set has

m

-

1

elements, and by Lagrange’s

theorem

x

m

-

1

=

1

for any

x

∈

F

×

, so

x

m

=

x

follows.
∎

Theorem 2.2

.

For each prime

p

>

0

and each natural number

n

∈

N

, there
exists a finite field of cardinality

p

n

, and any two such are
isomorphic.

Proof.

For

n

=

1

, the finite field

𝔽

p

:=

ℤ

/

p

⁢

ℤ

has

p

elements, and any
two such are isomorphic by the map sending

1

to

1

.

In general, the

polynomial

f

⁢

(

X

)

:=

X

p

n

-

X

∈

𝔽

p

⁢

[

X

]

has

derivative

-

1

and thus is

separable

over

𝔽

p

. We claim that the

splitting field

F

of this polynomial is a finite field of size

p

n

. The field

F

certainly contains the set

S

of roots of

f

⁢

(

X

)

. However, the set

S

is

closed under

the field

operations

, so

S

is itself a field. Since splitting fields are

minimal

by
definition, the containment

S

⊂

F

means that

S

=

F

. Finally,

S

has

p

n

elements since

f

⁢

(

X

)

is separable, so

F

is a field of
size

p

n

.

For the uniqueness part, any other field

F

′

of size

p

n

contains a

subfield

isomorphic to

𝔽

p

. Moreover,

F

′

equals the splitting field of
the polynomial

X

p

n

-

X

over

𝔽

p

, since by Lemma

2.1

every element of

F

′

is a root of this polynomial, and all

p

n

possible roots of the polynomial are accounted for in this way. By the uniqueness of
splitting fields up to

isomorphism

, the two fields

F

and

F

′

are
isomorphic.
∎

Note: The proof of Theorem

2.2

given here, while standard
because of its

efficiency

, relies on more

abstract algebra

than is

strictly

necessary. The reader may find a more concrete

presentation

of this and many other results about finite fields
in

[

1

, Ch. 7]

.

Corollary 2.3

.

Every finite field

F

is a

normal extension

of its prime subfield

F

p

.

Proof.

This follows from the fact that

field extensions

obtained from
splitting fields are normal extensions.
∎

3

Units in a finite field

Henceforth, in light of Theorem

2.2

, we will write

𝔽

q

for the unique (up to isomorphism) finite field of cardinality

q

=

p

n

. A fundamental step in the investigation of finite fields is the
observation that their

multiplicative groups

are cyclic:

Theorem 3.1

.

The multiplicative group

F

q

*

consisting of nonzero elements of
the finite field

F

q

is a

cyclic group

.

Proof.

We begin with the

formula

∑

d

∣

k

ϕ

⁢

(

d

)

=

k

,

(1)

where

ϕ

denotes the

Euler totient function

. It is proved as
follows. For every

divisor

d

of

k

, the cyclic group

C

k

of size

k

has exactly one cyclic subgroup

C

d

of size

d

. Let

G

d

be
the subset of

C

d

consisting of elements of

C

d

which have the
maximum possible order (

http://planetmath.org/OrderGroup

) of

d

. Since every element of

C

k

has

maximal order

in the subgroup of

C

k

that it

generates

, we see that
the sets

G

d

partition

the set

C

k

, so that

∑

d

∣

k

|

G

d

|

=

|

C

k

|

=

k

.

The

identity

(

1

) then follows from the observation that the
cyclic subgroup

C

d

has exactly

ϕ

⁢

(

d

)

elements of maximal order

d

.

We now prove the theorem. Let

k

=

q

-

1

, and for each divisor

d

of

k

, let

ψ

⁢

(

d

)

be the number of elements of

𝔽

q

*

of order

d

. We claim that

ψ

⁢

(

d

)

is either zero or

ϕ

⁢

(

d

)

. Indeed, if
it is nonzero, then let

x

∈

𝔽

q

*

be an element of order

d

, and
let

G

x

be the subgroup of

𝔽

q

*

generated by

x

. Then

G

x

has
size

d

and every element of

G

x

is a root of the polynomial

x

d

-

1

. But this polynomial cannot have more than

d

roots in a field, so
every root of

x

d

-

1

must be an element of

G

x

. In particular,
every element of order

d

must be in

G

x

already, and we see that

G

x

only has

ϕ

⁢

(

d

)

elements of order

d

.

We have proved that

ψ

⁢

(

d

)

≤

ϕ

⁢

(

d

)

for all

d

∣

q

-

1

. If

ψ

⁢

(

q

-

1

)

were 0, then we would have

∑

d

∣

q

-

1

ψ

⁢

(

d

)

<

∑

d

∣

q

-

1

ϕ

⁢

(

d

)

=

q

-

1

,

which is impossible since the first sum must equal

q

-

1

(because
every element of

𝔽

q

*

has order equal to some divisor

d

of

q

-

1

).
∎

A more

constructive proof

of Theorem

3.1

, which actually
exhibits a

generator

for the cyclic group, may be found
in

[

2

, Ch. 16]

.

Corollary 3.2

.

Every

extension

of finite fields is a

primitive

extension.

Proof.

By Theorem

3.1

, the multiplicative group of the

extension field

is cyclic. Any generator of the multiplicative group of the extension field also algebraically generates the extension field over the

base field

.
∎

4

Automorphisms of a finite field

Observe that, since a splitting field for

X

q

m

-

X

over

𝔽

p

contains all the roots of

X

q

-

X

, it follows that the field

𝔽

q

m

contains a subfield isomorphic to

𝔽

q

. We will show
later (Theorem

4.2

) that this is the only way that extensions of
finite fields can arise. For now we will construct the

Galois group

of
the field extension

𝔽

q

m

/

𝔽

q

, which is normal by
Corollary

2.3

.

Theorem 4.1

.

The Galois group of the field extension

F

q

m

/

F

q

is a cyclic
group of size

m

generated by the

q

th

power

Frobenius map

Frob

q

.

Proof.

The fact that

Frob

q

is an element of

Gal

⁡

(

𝔽

q

m

/

𝔽

q

)

, and
that

(

Frob

q

)

m

=

Frob

q

m

is the identity on

𝔽

q

m

, is

obvious

. Since the extension

𝔽

q

m

/

𝔽

q

is normal and of degree

m

, the group

Gal

⁡

(

𝔽

q

m

/

𝔽

q

)

must have size

m

, and we will
be done if we can show that

(

Frob

q

)

k

, for

k

=

0

,

1

,

…

,

m

-

1

, are distinct elements of

Gal

⁡

(

𝔽

q

m

/

𝔽

q

)

.

It is enough to show that none of

(

Frob

q

)

k

, for

k

=

1

,

2

,

…

,

m

-

1

, is the

identity map

on

𝔽

q

m

, for then we will have
shown that

Frob

q

is of order exactly equal to

m

. But, if any
such

(

Frob

q

)

k

were the identity map, then the polynomial

X

q

k

-

X

would have

q

m

distinct roots in

𝔽

q

m

, which is
impossible in a field since

q

k

<

q

m

.
∎

We can now use the

Galois correspondence

between subgroups of the
Galois group and intermediate fields of a field extension to
immediately classify all the intermediate fields in the extension

𝔽

q

m

/

𝔽

q

.

Theorem 4.2

.

The field extension

F

q

m

/

F

q

contains exactly one intermediate
field isomorphic to

F

q

d

, for each divisor

d

of

m

, and no
others. In particular, the subfields of

F

p

n

are precisely the
fields

F

p

d

for

d

∣

n

.

Proof.

By the

fundamental theorem of Galois theory

, each intermediate field
of

𝔽

q

m

/

𝔽

q

corresponds to a subgroup of

Gal

⁡

(

𝔽

q

m

/

𝔽

q

)

. The latter is a cyclic group of order

m

, so
its subgroups are exactly the cyclic groups generated by

(

Frob

q

)

d

, one for each

d

∣

m

. The

fixed field

of

(

Frob

q

)

d

is the set of roots of

X

q

d

-

X

, which forms a
subfield of

𝔽

q

m

isomorphic to

𝔽

q

d

, so the result
follows.

The subfields of

𝔽

p

n

can be obtained by applying the above
considerations to the extension

𝔽

p

n

/

𝔽

p

.
∎

References

1

Kenneth Ireland & Michael Rosen,

A Classical

Introduction

to Modern

Number Theory

, Second Edition

,
Springer–Verlag, 1990 (GTM

84

).

2

Ian Stewart,

Galois Theory

, Second Edition

,
Chapman & Hall, 1989.

Title

finite field

Canonical

name

FiniteField

Date of creation

2013-03-22 12:37:50

Last modified on

2013-03-22 12:37:50

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

16

Author

yark (2760)

Entry type

Definition

Classification

msc 12E20

Classification

msc 11T99

Synonym

Galois field

Related topic

AlgebraicClosureOfAFiniteField

Related topic

IrreduciblePolynomialsOverFiniteField

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