Definition of Finite Field: Addition and Multiplication Groups

  • Thread starter Thread starter Greg Bernhardt
  • Start date Start date
  • Tags Tags
    Field Finite
Click For Summary
Finite fields, known as Galois fields GF(p^n), are defined where p is a prime number, with addition and multiplication groups represented by Z(p)^n and Z(p^n-1) respectively. The multiplication groups of these fields are cyclic, and for p = 2, operations can be efficiently executed using bitwise exclusive or and shifting on computers. Elements in GF(p^n) are represented as polynomials with coefficients from {0, 1, ..., p-1}, with operations performed modulo a primitive polynomial of degree n. Subfields exist within GF(p^n) if m divides n, and the Wedderburn Theorem confirms that finite division rings are commutative, thus qualifying as finite fields. The historical context includes contributions from mathematicians like Galois and Gauss, who advanced the understanding of calculations in modular arithmetic.
Messages
19,851
Reaction score
10,885
Definition/Summary

All finite fields are known; they are the Galois fields GF(p^n), where p is a prime.

They have addition group Z(p)^n and multiplication group Z(p^n-1); their multiplication groups are cyclic.

If p = 2, then addition and multiplication can be done very fast by typical computer hardware using bitwise exclusive or and shifting.

Equations



Extended explanation

The finite fields GF(p) are {0, 1, ..., p-1} under addition and multiplication modulo p, which must be a prime number.

Finite fields GF(pn) for n > 1 can be described using polynomials in a variable x with coefficients having values in {0, 1, ..., p-1}.

Every element is a polynomial with a degree at most n-1.

Element addition is polynomial addition modulo p, while element multiplication is polynomial multiplication modulo p and a degree-n primitive or irreducible polynomial.

A primitive polynomial is one that cannot be factored in this construction of GF(pn). Primitive polynomials are not unique; there are
N(p,n) = \frac{1}{n} \sum_{m|n} \mu(m) p^{n/m}

monic ones, where μ is the Möbius mu function. That function is (-1)number of prime factors if they all have power 1, and 0 otherwise.

A finite field GF(pn) has subfield GF(pm) if m evenly divides n. If n is a prime, then GF(pn) only has only the trivial subfields, itself and GF(p).


* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
 
Physics news on Phys.org
E. H. Moore probably coined in 1893 the English term Galois field in honor of Évariste Galois, who has already calculated with certain imaginary numbers modulo ##p##.

The Wedderburn Theorem states that multiplication in a finite division ring is necessarily commutative. That is, finite division rings are always finite fields.

Gauss had already shown that one can count on numbers modulo a prime "as with rational numbers". Galois introduced imaginary numbers into calculation modulo ## p##, much like the imaginary unit ##\mathrm {i}## in the complex numbers. He was probably the first to consider field extensions of ##\mathbb{F}_ {p}## - although the abstract concept of fields was first introduced by Heinrich Weber in 1895 and Frobenius was the first to introduce it 1896 as extensions to finite structures.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
48
Views
4K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 26 ·
Replies
26
Views
852
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K