Finite field arithmetic

Arithmetic in a finite field is different from standard arithmetic. All operations must always yield results that remain within the field. There are a limited number of numbers in the finite field; all of operations performed in the finite field result in a number within that field. Finite fields are used in a variety of applications, including cryptography, such as the Rijndael encryption algorithm.

While each finite field is itself not infinite, there are an infinite number of different finite fields.

Contents

1 Notation 2 Addition and subtraction 3 Multiplication 3.1 Multiplicative inverse 3.2 Primitive finite fields 4 Program examples 5 Rijndael's finite field 6 External links

Notation

Although elements of a finite field can be expressed in numerical form (i.e., hexadecimal, binary, etc.), it is often found convenient to express them as polynomials, with each term in the polynomials representing the bits in the elements' binary expressions. For example, the following are equivalent representations of the same value of a characteristic 2 finite field:

Hexadecimal: {53}

Binary: {01010011}

Polynomial: x⁶ + x⁴ + x + 1

The exponents in the polynomial notation serve as "tags", making it possible to keep track of each bit's value throughout arithmetical manipulation, without the need for zero-value placeholders or alignment of digits into columns. If hexadecimal or binary notation is used, braces ( "{" and "}" ) or similar devices are commonly used to indicate that the value is an element of a field.

Addition and subtraction

Addition and subtraction are performed by adding or subtracting two of these polynomials together. Any term in a polynomial can only have a value zero or one.

In a finite field with characteristic 2, addition and subtraction are identical, and are accomplished using the XOR operator. Thus,

Hexadecimal: {53} + {CA} = {99}

Binary: {01010011} + {11001010} = {10011001}

Polynomial: (x⁶ + x⁴ + x + 1) + (x⁷ + x⁶ + x³ + x) = x⁷ + x⁴ + x³ + 1

Notice that each of the numbers being added in the example have a x⁶ in them. x⁶ + x⁶ would be 2x⁶. But, since each coefficient needs to be mod 2, this becomes 0x⁶, and then gets dropped from the answer.

Here is a table with both the normal algebraic sum and the characteristic 2 finite field sum of a few polynomials:

p1	p2	p1 + p2 (normal algebra)	p1 + p2 in GF(2n)
x3 + x + 1	x3 + x2	2x3 + x2 + x + 1	x2 + x + 1
x4 + x2	x6 + x2	x6 + x4 + 2x2	x6 + x4
x + 1	x2 + 1	x2 + x + 2	x2 + x
x3 + x	x2 + 1	x3 + x2 + x + 1	x3 + x2 + x + 1

Note that a characteristic 2 finite field is sometimes called a GF(2ⁿ) Galois field.

Multiplication

Multiplication in a finite field is multiplication modulo the irreducible polynomial used to define the finite field. (I.e., it is multiplication followed by division using the irreducible polynomial as the divisor—the remainder is the product.) The symbol "•" may be used to denote multiplication in a finite field.

For example, if the irreducible polynomial used is f(x) = x⁸ + x⁴ + x³ + x + 1 (the irreducible polynomial used in Rijndael), then {53} • {CA} = {01} because

(x⁶ + x⁴ + x + 1)(x⁷ + x⁶ + x³ + x) =

x¹³ + x¹² + x⁹ + x⁷ + x¹¹ + x¹⁰ + x⁷ + x⁵ + x⁸ + x⁷ + x⁴ + x² + x⁷ + x⁶ + x³ + x =

x¹³ + x¹² + x⁹ + x¹¹ + x¹⁰ + x⁵ + x⁸ + x⁴ + x² + x⁶ + x³ + x =

x¹³ + x¹² + x¹¹ + x¹⁰ + x⁹ + x⁸ + x⁶ + x⁵ + x⁴ + x³ + x² + x

and

x¹³ + x¹² + x¹¹ + x¹⁰ + x⁹ + x⁸ + x⁶ + x⁵ + x⁴ + x³ + x² + x modulo x⁸ + x⁴ + x³ + x + 1 (= 11111101111110 mod 100011011) = 1, which can be demonstrated through long division (shown using binary notation, since it lends itself well to the task):

                  111101
100011011)11111101111110
          100011011     
           1110000011110
           100011011    
            110110101110
            100011011   
             10101110110
             100011011  
              0100011010
              000000000 
               100011010
               100011011
                00000001

(The elements {53} and {CA} happen to be multiplicative inverses of one another since their product is 1.)

Multiplication in this particular finite field can also be done as follows:

• Take two eight-bit numbers, a and b, and an eight-bit product p. The reason for eight bits is because this finite field can only have eight elements in the polynomial.
• Set the product to zero.
• Make a copy of a and b, which we will simply call a and b in the rest of this algorithm
• Run the following loop eight times:
  1. If the low bit of b is set, exclusive or the product p by the value of a
  2. Keep track of whether the high (leftmost) bit of a is set to one
  3. Rotate a one bit to the left, discarding the high bit, and making the low bit have a value of zero
  4. If a's hi bit had a value of one prior to this rotation, exclusive or a with the hexadecimal number 0x1b (27 in decimal). 0x1b corresponds to the irreducible polynomial x⁸ + x⁴ + x³ + x + 1.
  5. Rotate b one bit to the right, discarding the low bit, and making the high (leftmost) bit have a value of zero.
• The product p now has the product of a and b

Multiplicative inverse

The multiplicative inverse for n in a finite field can be calculated a number of different ways:

• By multiplying n by every number in the field until the product is one. This is a Brute-force search

• By using the Extended Euclidean algorithm

• By making a logarithm table of the finite field, and performing subtraction in the table. In a logarithm table, subtraction is the same as division.

Primitive finite fields

A finite field is considered a primitive finite field if the number 2 is a generator for the finite field. In other words, if repeated multiplications by 2 results in the entire finite field being traversed before the number has a value of one, the field is a primitive finite field. As it turns out, the GF(2⁸) finite field with the reducing polynomial x⁸ + x⁴ + x³ + x + 1 is not a primitive; the smallest generator in this field is 3. The GF(2⁸) finite field with the reducing primitive polynomial x⁸ + x⁴ + x³ + x² + 1, however, is a primitive field.

Primitive finite fields are used, for example, by Linear feedback shift registers.

Program examples

Here is some C code which will add, subtract, and multiply numbers in a characteristic 2 (GF(2ⁿ)) finite field with eight terms and the irreducible polynomial f(x) = x⁸ + x⁴ + x³ + x + 1:

/* Add two numbers in a GF(2^8) finite field */
unsigned char gadd(unsigned char a, unsigned char b) {
	return a ^ b;
}

/* Subtract two numbers in a GF(2^8) finite field */
unsigned char gsub(unsigned char a, unsigned char b) {
	return a ^ b;
}

/* Multiply two numbers in the GF(2^8) finite field defined 
 * by the polynomial x^8 + x^4 + x^3 + x + 1 */
unsigned char gmul(unsigned char a, unsigned char b) {
	unsigned char p = 0;
	unsigned char counter;
	unsigned char hi_bit_set;
	for(counter = 0; counter < 8; counter++) {
		if((b & 1) == 1) 
			p ^= a;
		hi_bit_set = (a & 0x80);
		a <<= 1;
		if(hi_bit_set == 0x80) 
			a ^= 0x1b; /* x^8 + x^4 + x^3 + x + 1 */
		b >>= 1;
	}
	return p;
}

Rijndael's finite field

Rijndael uses a characteristic 2 finite field with 8 terms, which can also be called a GF(2⁸) Galois field. Rijndael's finite field uses the following reduction polynomial for multiplication:

x⁸ + x⁴ + x³ + x + 1.

External links

• A description of Rijndael's finite field (http://www.samiam.org/galois.html)