Fraleigh, *A First Course in Abstract Algebra*, Addison-Wesley, Reading, 1976.

Herstein, *Topics in Algebra*, Xerox, Lexington, 1975.

Lang, *Algebra*, Addison-Wesley, Reading, 1971.

It will be assumed that you are familiar with the definition of a field, the definition and
basic
properties of vector spaces and the fact that the integers modulo a prime p form a field (a
finite
one), which we will denote by **GF**(p) (standing for the Galois Field of order p).

If L is an extension of K (denoted L > K), and þ is an element of L but not an
element of K, then the smallest field containing both K and þ will be denoted by
K(þ) and is a subfield of L. Similarly, the smallest field containing K and the elements
þ_{1},þ_{2}, ... ,þ_{n} in L > K will be written as K(þ_{1},þ_{2}, ...
,þ_{n}).
Any extension field L of K can be viewed as a vector space over K and the dimension of this
vector space is called the* degree* of L over K, and is denoted by [L:K]. If the vector
space is
finite dimensional we say that L is a *finite extension* of K. If L is a finite extension of
K and M
is a finite extension of L, then [M:K] = [M:L][L:K].

Denote by K[x] the ring of polynomials over K in the variable x. K[x] is a principal ideal
domain (an ideal is a subring that is closed under multiplication, a principal ideal is an ideal
generated by a single element, and a principal ideal domain is a commutative ring with unity
all
of whose ideals are principal). A polynomial in K[x] is said to be* monic* if the
coefficient of the
highest power in x is unity. It is* irreducible* if it is not the product of two nonscalar
polynomials
in K[x]. If f(x) is an irreducible polynomial in K[x], then any zero (i.e., root) of f(x) is not in
K
and so there is a smallest extension field L of K that contains it. Furthermore, L is isomorphic
to the quotient field K[x]/<f(x)>, where <f(x)> denotes the principal ideal of K[x]
generated by f(x). If the irreducible polynomial f(x) is of degree n, then [L:K] = n.
Furthermore,
if þ is the zero of f(x) in question, then L = K(þ) and the elements
1,þ,þ^{2},..., þ^{n-1} form a basis of L over K.

Before we continue, let us illustrate these ideas with a well-known example. The field
**C** of
complex numbers is usually described in one of two ways, either as the set of numbers {a +
bi}
where a and b are real numbers and i = sqrt(-1), or if you prefer not to mention the imaginary
number i, **C** can be described as the set of all pairs (a,b) of real numbers where
addition of two
pairs is the usual componentwise addition and multiplication of two pairs is defined by
(a,b)(c,d)
= (ac - bd, ad + bc). The second version is of course viewing **C** as a vector space of
dimension
2 over the real numbers **R**. Let us see how we would construct **C** starting
with the subfield **R**.
Now K =** R** and K[x] is the ring of all polynomials with real coefficients. The
polynomial x2 +
1 is monic since the coefficient of x2 is 1 and it is irreducible over** R** since it cannot
be factored
into polynomials with only real coefficients. The principal ideal <x^{2}+1> consists of all
polynomials that have x^{2}+1 as a factor. The quotient field structure,
**R**[x]/<x^{2}+1> is obtained
by taking each polynomial in **R**[x] and dividing it by x^{2}+1. The polynomials that
have the same
remainder after division form equivalence classes, which are the elements of the quotient
field.
The different possible remainders are the polynomials a + bx, where a,b in **R**, and we
identify
the equivalence classes with these remainders. The association a + bx iff a + bi clearly
shows
that the quotient field and **C** are isomorphic. Now, let þ be a zero of x^{2} + 1,
i.e. þ^{2}
+ 1 = 0 , so þ = sqrt(-1) = i which is not an element of** R**. {1,i} forms a basis
for** C** over
**R** since every element of **C** can be written as a(1) + b(i) with a,b in
**R** and 1 and i are easily seen
to be linearly independent. Using this basis, we can identify the elements of **C** with
their
coefficient vectors, i.e. a + bi iff (a,b) to get the second representation as a vector space of
dimension 2 (notice the highest power of x in the irreducible polynomial).

Let f(x) be in K[x], then the smallest field containing K and all the zeros of f(x) is called
the *splitting field* of f(x) over K. We may call it "the" splitting field due to the
following
theorem:

**Theorem II.1.1** -* Let f(x) be an irreducible polynomial in K[x], then a splitting
field for f(x) over
K exists and any two such splitting fields are isomorphic*.

If f(x) is a polynomial in K[x] of degree n, then its splitting field over K is at most of
degree
n! and this bound may or may not be obtained, depending on the polynomial and the field.
We
now consider some elementary properties of polynomials over a field. If f(x) in K[x] is given
by

f(x) = ß_{n} x^{n} + ß_{n-1} x^{n-1} + ... + ß_{1} x + ß_{0} ,
where ß_{i}
in K

then the formal derivative of f(x), denoted by f'(x) is the polynomial

f'(x) = nß_{n} x^{n-1} + (n-1)ß_{n-1} x^{n-2} + ... +
ß_{1}

which is a polynomial of degree at most n - 1. Note that the formal derivative of a
polynomial
may be zero even though the polynomial is not a constant, for example if K = **GF**(3),
where 3
= 0, the polynomial f(x) = x^{3} + 2 has a zero formal derivative. A polynomial f(x) in K[x] is
said
to have a zero þ of *multiplicity* m in some extension field L of K if m is the
largest
positive integer for which

(x-þ)^{m} | f(x)

in L[x], where the vertical bar indicates division with no remainder. The zeros of an
irreducible
polynomial f(x) in K[x] in the splitting field for f(x) over K are called *conjugates*. Let
f(x), g(x)
in K[x] be two polynomials such that deg f > deg g. By the Euclidean algorithm there
exist
two polynomials q(x) and r(x) such that

f(x) = q(x)g(x) + r(x), where deg r < deg g.

By repeated application of the algorithm, the greatest common divisor (or gcd) d(x) of f(x)
and
g(x) (denoted by (f(x), g(x)) ) can be expressed as

d(x) = (f(x),g(x)) = a(x)f(x) + b(x)g(x), for some a(x),b(x) in K[x].

For further reference we collect some elementary properties of polynomials in the following theorem.

**Theorem II.1.2** -* Let f(x), g(x) in K[x] and let L be any extension of K.
Then:
(i) if (f(x), g(x)) = d(x) in K[x] then (f(x), g(x)) = d(x) in L[x].
(ii) f(x) | g(x) in K[x] iff f(x) | g(x) in L[x].
(iii) f(x) has a multiple zero iff (f(x), f'(x)) is not 1. *

It is significant that two polynomials have a common root in some extension field if they
have
a common divisor over the original field. The question of multiplicities of roots of irreducible
polynomials can be settled precisely. If f(x) in K[x], char K = 0 and f(x) is irreducible, then
f(x)
cannot have multiple zeros. If char K = p, then an irreducible polynomial f(x) having multiple
zeros must be of the form f(x) = g(x^{p} ) for some polynomial g(x) in K[x]. In this case each
zero
of f(x) has the same multiplicity. It can be shown that if f(x) is an irreducible polynomial
over
a finite field, then it has only simple (multiplicity one) zeros.

**Theorem II.1.3** - *A polynomial f(x) in K[x] of degree n has at most n zeros in
any extension
of K.
*

We consider now the concept of field isomorphism, which will be useful in the
investigation
of finite fields. An* isomorphism* of the field K_{1} onto the field K_{2} is a one-to-one
onto map that
preserves both field operations, i.e.,

µ(þ + ß) = µ(þ) + µ(ß),
µ(þß)
= µ(þ)µ(ß) for all þ,ß in K1 .

An *automorphism* of K is an isomorphism of K onto itself. The set of all
automorphisms of a
field forms a group under composition. If µ_{1},µ_{2},...,µ_{n} are distinct
isomorphisms of
K_{1}onto K_{2}, then these isomorphisms are linearly independent over K_{2} in the sense that
if

Sum a_{i} µ_{i} (b) = 0 with a_{i} in K_{2}

for all b in K_{1} , then all the a_{i} must = 0. In particular, distinct automorphisms of a field are
linearly independent.

A field automorphism leaves the elements of the prime subfield fixed. The automorphism may leave a larger subfield fixed. More generally, if

K' = { a in K | µ_{i} (a) = a, i = 1,...,n }

then K' is a subfield of K, called the* fixed field* of K with respect to the
automorphisms µ_{i}.
It can be shown that [K:K'] >= n. We will be interested in subgroups of the group of all
automorphisms that fix certain subfields. We denote the group of all automorphisms of a field
L by G(L) and the subgroup of G(L) that fixes all elements of the subfield K of L by G(L/K).
It is important to note that the fixed field of G(L/K) may properly contain K. It is easily
shown
that G(L/K) is a subgroup of G(L), conversely, if H is any subgroup of G(L), then the set of
elements of L fixed by H is a subfield of L.

**Theorem II.2.1** - *Any finite field with characteristic p has p ^{n} elements for some
positive integer
n.*

Proof: Let L be the finite field and K the prime subfield of L. The vector space of L over K
is
of some finite dimension, say n, and there exists a basis þ_{1},þ_{2}, ... ,þ_{n}
of L
over K. Since every element of L can be expressed uniquely as a linear combination of the
þ_{i} over K, i.e., every a in L can be written as, a = Sum ß_{i} þ_{i} , with
ß_{i}
in K, and since K has p elements, L must have p^{n} elements. ¶

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 II.2.2** - *Let L be a field with characteristic p and prime subfield K. Then
L is the
splitting field for iff L has
p ^{n} elements.*

Proof: Suppose that L is the splitting field for 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 = { þ in L | þ^{pn} = þ }

of L. Clearly E contains p^{n} elements since it consists of the roots of f(x). Suppose that
þ,ß in E; then (þß)^{pn} = (þ)^{pn} (ß)^{pn} =
þß and hence, þß in E. Also,

since p | C(p^{n}, i) for 0 < i < p^{n} , and hence, (þ + ß ) 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 Thm II.1.1 E = L and L contains p^{n} elements.

Suppose now that L contains p^{n} elements. 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 þ^{pn} = þ for all þ in L* and the relation is trivially true
for þ = 0.
Thus f(x) splits in L. ¶

Now for some important corollaries.

**Corollary II.2.3** - *There is a unique (up to isomorphism) field of order p ^{n}.*

Proof: Since f(x) splits over any field with this many elements and by Thm II.1.1 splitting fields are unique, this result follows. ¶

**Corollary II.2.4** - **GF**(p^{n})* is the splitting field for over* **GF**(p).

Proof: This is just a restatement of Thm II.2.2 and Cor II.2.3. ¶

**Corollary II.2.5** - *For any prime p and integer n,*** GF**(p^{n})
*exists*.

Proof: By Thm II.1.1 the splitting field exists and by Cor II.2.4 it is GF(p^{n}). ¶

The following important theorem is useful in establishing the subfield structure of the Galois Fields among other things.

**Theorem II.2.6** - **GF**(p^{n})* *is cyclic*.

Proof: The multiplicative group **GF**(p^{n})* is, by definition, abelian and of order p^{n}
- 1. If
, then, factoring **GF**(p^{n})* into a direct product of its Sylow subgroups, we
have

**GF**(p^{n})* = S(p_{1} ) × ... × S(p_{k})

where S(p_{i} ) is the Sylow subgroup of order (p_{i})^{ei} . 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 **GF**(p^{n})*. Furthermore every
element of **GF**(p^{n})* satisfies the
polynomial x^{m} -1, implying that m >= p^{n} -1. Since a in **GF**(p^{n})* has order
m, m divides p^{n}
-1 and so, m = p^{n} -1. Thus the element a is a generator and **GF**(p^{n})* is cyclic.
¶

A generator of **GF**(p^{n})* is called a *primitive element* of
**GF**(p^{n}).

The following theorem has some useful consequences.

**Theorem II.2.7** -* Over any field K, (x ^{m} - 1) | (x^{n} - 1 ) iff m | n*.

Proof: If n = qm + r, with r < m, then by direct computation

It follows that (x^{m} - 1) | (x^{n} - 1) iff x^{r} - 1 = 0, i.e., r = 0. ¶

**Corollary II.2.8** - *For any prime integer p, (p ^{m} - 1) | (p^{n} - 1) iff m |
n.*

Proof: Basically the same as that of the theorem, do for homework.***********

**Theorem II.2.9** - **GF**(p^{m} ) is a subfield of ** GF**(p^{n} )* iff m |
n.*

Proof: Suppose **GF**(p^{m} ) is a subfield of **GF**(p^{n} ); then** GF**(p^{n} )
may be interpreted as a vector
space over **GF**(p^{m} ) with dimension, say, k. Hence, p^{n} = p^{km} and m | n.

Now suppose m | n, which from the previous theorem and its corollary implies that
(x^{pm - 1} - 1) | (x^{pn - 1} - 1). Thus every zero of x^{pm} - x that is in**
GF**(p^{m}) is also a zero of x^{pn} - x
and hence in **GF**(p^{n}). It follows that **GF**(p^{m} ) is contained in
**GF**(p^{n}). Notice that there is
precisely one subfield of **GF**(p^{n}) of order p^{m}, otherwise x^{pm} - x would have
more than p^{m} roots.
¶

Although we will not prove it, the automorphism group of a finite field is cyclic. The
standard generator of this group is the so-called* Frobenius automorphism* defined for a
finite field
of characteristic p as the map x --> x^{p} for all x in **GF**(p^{n} ).

(Homework: Prove that this map is a field automorphism)**************

- x
^{2} - x
^{2}+ 1 - x
^{2}+ 2 - x
^{2}+ x - x
^{2}+ x + 1 - x
^{2}+ x + 2 - x
^{2}+ 2x - x
^{2}+ 2x + 1 - x
^{2}+ 2x + 2

- (x + 1)(x + 1) = x
^{2}+ 2x + 1 - (x + 1)(x + 2) = x
^{2}+ 2 - (x + 2)(x + 2) = x
^{2}+ x + 1

x

þ

and so þ has order 4 and does not generate the cyclic group of order 8, i.e., þ is not a primitive element. On the other hand, consider µ a root of the polynomial x

- µ
^{1}=**µ** - µ
^{2}=**2µ + 1** - µ
^{3}= µ(2µ + 1) = 2µ^{2}+ µ = 2(2µ + 1) + µ =**2µ + 2** - µ
^{4}= 2µ^{2}+ 2µ = µ + 2 + 2µ =**2** - µ
^{5}=**2µ** - µ
^{6}= 2µ^{2}=**µ + 2** - µ
^{7}= µ^{2}+ 2µ = 2µ + 1 + 2µ =**µ + 1** - µ
^{8}= µ^{2}+ µ = 2µ + 1 + µ =**1**

- x
^{3}+ 1 - x
^{3}+ x + 1 - x
^{3}+ x^{2}+ 1 - x
^{3}+ x^{2}+ x + 1

- µ
^{1}= µ - µ
^{2}= µ^{2} - µ
^{3}= µ + 1 - µ
^{4}= µ^{2}+ µ - µ
^{5}= µ^{2}+ µ + 1 - µ
^{6}= µ^{2}+ 1 - µ
^{7}= 1

- þ
^{1}= þ - þ
^{2}= þ^{2} - þ
^{3}= þ^{2}+ 1 - þ
^{4}= þ^{2}+ þ + 1 - þ
^{5}= þ + 1 - þ
^{6}= þ^{2}+ þ - þ
^{7}= 1

Dickson,

This is a reprint of what had been the only source on finite fields. It is fairly difficult reading now since the notation and style are very old (the original book was written in 1900), but it deserves to be mentioned for its significance in the development of modern algebra. Only the first half of the book deals with finite fields per se, the rest is devoted to the automorphism groups of these fields.

Another place to look for finite fields is in any book on algebraic coding theory, since this theory builds on vector spaces over finite fields these books usually devote some time to them.

Berlekamp, *Algebraic Coding Theory*, McGraw-Hill, New York 1968.

Blake & Mullin, *An Introduction to Algebraic and Combinatorial Coding Theory*, Academic
Press, 1976.

Pless,* Introduction to the Theory of Error-Correcting Codes*, Wiley, 1982.

A recent tome is devoted to finite fields; it tends to be encyclopedic but is a good
source.

Lidl & Niederreiter,* Finite Fields*, Vol. 20 in the Encyclopedia of Mathematics and its
Applications, Addison-Wesley, 1983. [Reprinted by Cambridge University Press, 1987]

The same authors have also published a book on applications of finite fields which is
more
of a text than the above cited volume.

Lidl & Niederreiter,* Introduction to Finite Fields and their Applications*, Cambridge
University
Press, 1986.

A very readable account of the theory of finite fields is contained in

McEliece,* Finite Fields for Computer Scientists and Engineers*, Kluwer Academic Publishers,
1987.