Ciro Santilli @cirosantilli  40

There is a Turing machine that halts for every member of the language with the answer yes, but does not necessarily halt for non-members.

Non-examples: cs.stackexchange.com/questions/52503/non-recursively-enumerable-languages

Subset of recursively enumerable language as explained at: difference between recursive language and recursively enumerable language.

This is the classic result of formal language theory, but there is too much slack between context free and context sensitive, which is PSPACE (larger than NP!).

By Noam Chomsky.

A good summary table that opens up each category much more can be seen e.g. at the bottom of en.wikipedia.org/wiki/Automata_theory under the summary thingy at the bottom entitled "Automata theory: formal languages and formal grammars".

The opposite of from first principles.

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 $9=3\times 3$. Basically, for order $p^n$, we take:For a worked out example, see: GF(4).

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.