Ciro Santilli @cirosantilli 37

A convenient notation for the elements of $GF(n)$ of prime order is to use integers, e.g. for $GF(7)$ we could write:which makes it clear what is the additive inverse of each element, although sometimes a notation starting from 0 is also used:

For fields of prime order, regular modular arithmetic works as the field operation.

For non-prime order, we see that modular arithmetic does not work because the divisors have no inverse. E.g. at order 6, 2 and 3 have no inverse, e.g. for 2:we see that things wrap around perfecly, and 1 is never reached.

For non-prime prime power orders however, we can find a way, see finite field of non-prime order.

Ciro Santilli tried to add this example to Wikipedia, but it was reverted, so here we are, see also: Section "Deletionism on Wikipedia".

This is a good first example of a field of a finite field of non-prime order, this one is a prime power order instead.

$4=2^2$, so one way to represent the elements of the field will be the to use the 4 polynomials of degree 1 over GF(2):

Note that we refer in this definition to anther field, but that is fine, because we only refer to fields of prime order such as GF(2), because we are dealing with prime powers only. And we have already defined fields of prime order easily previously with modular arithmetic.

Over GF(2), there is only one irreducible polynomial of degree 2:

Addition is defined element-wise with modular arithmetic modulo 2 as defined over GF(2), e.g.:

Multiplication is done modulo $X^2+X+1$, which ensures that the result is also of degree 1.

For example first we do a regular multiplication:

Without modulo, that would not be one of the elements of the field anymore due to the $1X^2$!

So we take the modulo, we note that:and by the definition of modulo:which is the final result of the multiplication.

TODO show how taking a reducible polynomial for modulo fails. Presumably it is for a similar reason to why things fail for the prime case.