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:

$GR(7)={−3,−2,−1,0,1,2,3}$

$GR(7)={0,1,2,3,4,5,6}$

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.

$0×2=01×2=22×2=43×2=04×2=25×2=4$

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

There's exactly one field per prime power, so all we need to specify a field is give its order, notated e.g. as $GF(n)$.

Every element of a finite field satisfies $x_{order}=x$.

It is interesting to compare this result philosophically with the classification of finite groups: fields are more constrained as they have to have two operations, and this leads to a much simpler classification!

As per classification of finite fields those must be of prime power order.

Video "Finite fields made easy by Randell Heyman (2015)" at youtu.be/z9bTzjy4SCg?t=159 shows how for order $9=3×3$. Basically, for order $p_{n}$, we take:For a worked out example, see: GF(4).

- each element is a polynomial in $GF(p)[x]$, $GF(p)[x]$, the polynomial ring over the finite field $GF(p)$ with degree smaller than $n$. We've just seen how to construct $GF(p)$ for prime $p$ above, so we're good there.
- addition works element-wise modulo on $GF(p)$
- multiplication is done modulo an irreducible polynomial of order $n$

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):

- 0X + 0
- 0X + 1
- 1X + 0
- 1X + 1

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:

$X_{2}+X+1$

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

$(1X+0)+(1X+1)=(1+1)X+(0+1)=0X+1$

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:

$(1X+0)×(1X+1)=(1×1)X_{2}+(1×1)X+(0×1)X+(0×1)=1X_{2}+1X+0$

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.

$1X_{2}+1X+0=1(X_{2}+X+1)+(0X+1)$

$(1X_{2}+1X+0)mod(X_{2}+X+1)=(0X+1)$

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.

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