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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!

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.

A function:

A vector field with a bilinear map into itself, which we can also call a "vector product".

Note that the vector product does not have to be neither associative nor commutative.

Examples: en.wikipedia.org/w/index.php?title=Algebra_over_a_field&oldid=1035146107#Motivating_examples

An algebra over a field where division exists.

There are 3: real numbers, complex numbers and quaternions.

Notably, the octonions are not associative.

An elliptic curve is defined by numbers $a$ and $b$. The curve is the set of all points $(x,y)$ of the real plane that satisfy the Equation 1. "Definition of the elliptic curves"

Equation 1. "Definition of the elliptic curves" definies elliptic curves over any field, it doesn't have to the real numbers. Notably, the definition also works for finite fields, leading to elliptic curve over a finite fields, which are the ones used in Elliptic-curve Diffie-Hellman cyprotgraphy.

The elliptic curve group of an elliptic curve is a group in which the elements of the group are points on an elliptic curve.

The group operation is called elliptic curve point addition.

Bibliography:

Elliptic curve point addition is the group operation of an elliptic curve group, i.e. it is a function that takes two points of an elliptic curve as input, and returns a third point of the elliptic curve as its output, while obeying the group axioms.

The operation is defined e.g. at en.wikipedia.org/w/index.php?title=Elliptic_curve_point_multiplication&oldid=1168754060#Point_operations. For example, consider the most common case for two different points different. If the two points are given in coordinates:then the addition is defined in the general case as:with some slightly different definitions for point doubling $P+P$ and the identity point.

This definition relies only on operations that we know how to do on arbitrary fields:and it therefore works for elliptic curves defined over any field.

Just remember that:means:and that $y^{-1}$ always exists because it is the inverse element, which is guaranteed to exist for multiplication due to the group axioms it obeys.

The group function is usually called elliptic curve point addition, and repeated addition as done for DHKE is called elliptic curve point multiplication.

One major difference between the elliptic curve over a finite field or the elliptic curve over the rational numbers the elliptic curve over the real numbers is that not every possible $x$ generates a member of the curve.

This is because on the Equation "Definition of the elliptic curves" we see that given an $x$, we calculate $x^3+ax+b$, which always produces an element $y^2$.

But then we are not necessarily able to find an $y$ for the $y^2$, because not all fields are not quadratically closed fields.

For example: with $a=1$ and $b=1$, taking $x=1$ gives:and therefore there is no $y\in \mathbb{Q}$ that satisfies the equation. So $x=1$ is not on the curve if we consider this elliptic curve over the rational numbers.

That $x$ would also not belong to Elliptic curve over the finite field ${\mathbb{F}}_4$, because doing everything $\mathrm{mod}4$ we have:Therefore, there is no element $y\in {\mathbb{F}}_4$ such that $y\times y=2$ or $y\times y=3$, i.e. $2$ and $3$ don't have a multiplicative inverse.

For the real numbers, it would work however, because the real numbers are a quadratically closed field, and $\sqrt{3}\in \mathbb{R}$.

For this reason, it is not necessarily trivial to determine the number of elements of an elliptic curve.