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Finite fields are one of the few examples of an algebraic structure where one can classify everything completely.

We saw earlier how to make a finite field.

**Theorem**

*Let p be a prime and* *f*(*x*) *an irreducible polynomial of degree k in* **Z**_{p}[*x*]. *Then* **Z**_{p}[*x*]/ < *f*(*x*) > *is a field with p ^{k} elements.*

**Proof**

Note that a polynomial is irreducible if it cannot be written as a product of polynomials of lower degree.

As in the last section we can choose as coset representatives polynomials of the form *a*_{0} + *a*_{1}*x* + *a*_{2}*x*^{2} + ... + *a*_{k-1}*x*^{k-1} and so we get a ring of order *p*^{k}.

As in **Z**_{n} we use the Euclidean algorithm to find the inverse of an element.

Let *g*(*x*) be a coset representative of an element of the field. Since *f* (*x*) is irreducible it is not divisible by any lower degree polynomial and so the *gcd*(*g*(*x*), *f* (*x*)) = 1. Then by the Euclidean algorithm 1 = *a*(*x*)*g*(*x*) + *b*(*x*)*f* (*x*) for some polynomials *a*(*x*), *b*(*x*). Then *a*(*x*) is a coset representative for the inverse of *g*(*x*).

Here are some results I don't have time to prove.

**Important fact**

*For every prime p and positive integer k there is an irreducible polynomial of degree k in* **Z**_{p}[*x*]. *Hence there is a field of order p ^{k}.*

**Remarks**

- In fact there are
*lots*of such polynomials. For example, take*p*= 3. Then with*k*= 2 there are 3*monic*(leading coefficient 1) irreducible polynomials out of 9; with*k*= 3 there are 8 out of 27; with*k*= 5 there are 48 out of 243; ... - It is easy to tell if a quadratic or cubic is irreducible since if it were not it would have a linear factor and so the polynomial would have a root. That is easy to check! However, higher degree polynomials are harder. For example, the polynomial
*x*^{4}+ 1 ∈**Z**_{2}[*x*] has no root, but is reducible since it is (*x*^{2}+ 1)^{2}.

*Any two fields of order p ^{k} are isomorphic as rings.*

**Example**

We saw in the last section that the map defined by 1 ↦ 1 and *x* ↦ *y* + 1 is a ring isomorphism from **Z**_{3}[*x*]/ < *x*^{2} +1 > to **Z**_{3}[*y*]/ < *y*^{2}+ 2*y* +2 > and this is the unique field of order 9.

Following Exercises 2 Qu 6 we look at the the subring of a finite field generated by 1 and deduce that 1 has a prime number *p* for its additive order and that hence every non-zero element has this same order.

**Definitions**

This prime number is called the **characteristic** of the field.

The subfield generated by 1 is called the **prime subfield** and is isomorphic to **Z**_{p} .

**Remark**

If the element 1 does not have a finite order (in which case the field is not finite) we say the characteristic of the field is 0. In a field with characteristic 0 the element 1 generates a subfield isomorphic to the rational numbers **Q**.

**Theorem**

*Any field is a vector space over any subfield.*

**Proof**

it is easy to verify the axioms.

If *E* is a subfield of *F* then we can choose a basis for *F* over *E*. The number of elements in the basis is called the **dimension** of *F* over *E* and is written [*E*: *F*].

In particular, if *F* is finite with |*F*| = *p*^{k} it is a field of dimension *k* over its prime subfield.

Now let *r* be any element not in the prime subfield of the field *F* of size *p*^{k}. Then consider the elements 1, *r*, *r*^{2}, *r*^{3}, ... Eventually this list will stop producing linearly independent vectors and then one has a set {1, *r*, *r*^{2} , ... , *r*^{m-1} } of independent vectors and a polynomial *f*(*r*) = *a*_{0} + *a*_{1}*r* + *a*_{2}*r*^{2} + ... + *a*_{m}*r*^{m} = 0 with coefficients in the prime field **Z**_{p} .

The subspace spanned by this set is a subfield and has size *p*^{m}.

This polynomial f is called the **minimal polynomial** of the element *r* and since the field is a vector space over this subfield we must have *m* divides *k*.

A important result about finite fields is:

**The cyclic group theorem**

*The multiplicative group of a finite field is cyclic.*

**Proof**

In Exercises 4 Qu 6 you should have already noticed that the multiplicative group *U*_{n} of **Z**_{n} is cyclic if *n* is prime.

The main result we need to prove this in general is:

**Lemma**

*A polynomial of degree k with coefficients in a field can have at most k roots.*

**Proof of lemma**

If a polynomial *f*(*x*) over a field has a root *α* then divide *f*(*x*) by *x* - *α* to get *f*(*x*) = (*x* - *α*)*q*(*x*) + *r*. Since *f*(*α*) = 0 it follows that *r* = 0 and the polynomial is divisible by *x* - *α*. Hence the polynomial has a linear factor for each root and so cannot have more that *deg*(*f*) roots.

Before we prove our theorem we look at a particular case.

Consider the multiplicative group of the field with 9 elements. This abelian group has order 8 and so is one of *C*_{8} , *C*_{4} × *C*_{2} or *C*_{2} × *C*_{2} × *C*_{2} . If it were not *C*_{8} then any element *r* would satisfy *r*^{4} = 1. But then the polynomial *x*^{4} - 1 would have too many roots.

The general proof is similar. The multiplicative group of a field with order *p*^{k} has order *p*^{k} - 1 and suppose we have some prime *q* dividing this. Then the elements of the multiplicative group whose order is a power of *q* form a subgroup of order *q*^{s} (say). If this group is not cyclic then every element *r* of this subgroup satisfies *r*^{s-1} = 1 and then the polynomial *x*^{s-1} - 1 would have too many roots.

So the multiplicative group is a direct product of cyclic groups corresponding to various primes and hence is cyclic.

In particular we have now proved the result mentioned above:

*If n is prime, the group of units of* **Z**_{n} *is cyclic. *

**Remark**

The fact that the multiplicative group of *GF*(*p*^{k}) is cyclic makes it very conveneient for calculation. We can choose a particular multiplicative generator *z* (say) and then all the other elements are powers of this. So multiplication is easy. To calculate the sum of two elements *z*^{m} and *z*^{n} (say) observe that *z*^{m} + *z*^{n} = *z*^{m}(1 + *z*^{n-m}) and so we need only store a table which writes 1 + *z*^{r} as a power of *z*.

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