A **field** is an algebraic structure that lets you do everything you’re used to from basic math: you can add and multiply elements, and addition and multiplication have the usual properties you’d expect. More formally, the elements of a field form an Abelian (commutative) group with respect to addition, the non-zero elements form an Abelian group with respect to multiplication, and multiplication distributes over addition.

The rational numbers are a field. So are the real numbers and complex numbers. But it may be surprising that there are **finite fields**. If *p* is a prime, then the integers mod *p* form a field. For example, the integers mod 7 are the numbers 0 through 6 with addition and multiplication defined as usual, except you take the remainder by 7.

All finite fields have *p*^{n} elements where *p* is prime and *n* is an integer at least 1. Conversely, for every number of the form *p*^{n} there is a field that size. Furthermore, all groups of a given size are isomorphic. The field with *p*^{n} elements is sometimes called the **Galois field **with that many elements, written *GF*(*p*^{n}).

The Galois fields of order *GF*(*p*) are simply the integers mod *p*. For *n* > 1, the elements of *GF*(*p*^{n}) are **polynomials** of degree *n*-1 with coefficients coming from *GF*(* p*). You add polynomials as you’d expect, but

**multiplication is a little different**. You pick an irreducible polynomial

*g*(

*x*) of degree n and define multiplication in

*GF*(

*p*

^{n}) to be the ordinary polynomial product except you take the remainder after dividing by

*g*(

*x*). If you chose a different irreducible polynomial

*g*(

*x*) you’ll get a different definition of multiplication and hence a different field, but all such fields will be isomorphic.

Finite fields arise in applications, such as coding theory. **Reed -Solomon codes**, for example, are defined in terms of finite fields. Finite fields also arise in **cryptography**, for example in elliptic curves over finite fields. In many applications, *p* = 2.

Finite fields cannot be algebraically complete. In that sense you can’t have a finite analog of the complex numbers.

## Example

Let’s look at *GF*(2^{3}), the field with 8 elements, i.e. 3-bit numbers. We choose *g*(*x*) = *x*³ + *x* + 1 as our irreducible polynomial.

The number 3 (11_{two}) corresponds to the polynomial *x* + 1. The number 5 (101_{two}) corresponds to *x*² + 1. The sum of 3 and 5 in this field corresponds to *x*² + *x*, which correspond to 6. (The constant terms cancel out because 1 + 1 = 0 mod 2.)

To compute the product of 5 and 6 in this field, we multiply (*x*² + 1)(*x*² + *x*). When we divide by *x*³ + *x* + 1, working mod 2, we get a remainder of *x* + 1, which corresponds to 3. So in this field, 5 times 6 equals 3.