E    
  D  
    T

Encyclopaedia of DesignTheory: a Galois field

Here is the Galois field GF(8) with 8 elements.

Its prime field is the binary field GF(2)={0,1}. The polynomial x³+x+1 is irreducible over GF(2), so we adjoin a root of this polynomial, an element a satisfying a³=a+1. The eight elements of the field are

0, 1, a, a+1, a², a²+1, a²+a, a²+a+1.

We now calculate successive powers of a:

a0 1
a1 a
a2 a²
a3 a+1
a4 a²+a
a5 a²+a+1
a6 a²  +   1

Finally we have a7=1. The element a is a primitive root, since every non-zero element of the field occurs among the powers of a.

This table functions as a "table of logarithms" in the field. We can easily multiply elements using the rule ai·aj=ai+j and reducing the exponent modulo 7 if necessary. We can add elements in the form given in the second column, using the rule that x+x=0 for any element x.

For example,

(a²+1)+(a²+a+1) = a
(a²+1)·(a²+a+1) = a6·a5 = a11 = a4 = a²+a

Exercise

Construct GF(8) using the irreducible polynomial x³+x²+1 instead, and show that it is isomorphic to the above field.

Table of contents | Glossary | Topics | Bibliography | History

Peter J. Cameron
7 August 2002