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Multiplicative inverse by

Ciro Santilli 37 Updated 2025-07-16

Distributive property by

Ciro Santilli 37 Updated 2025-07-16

One of the defining properties of algebraic structure with two operations such as ring and field:

a (b + c) = ab + a c

This property shows how the two operations interact.

Finite field by

Ciro Santilli 37 Updated 2025-07-16

A convenient notation for the elements of

GF (n)

of prime order is to use integers, e.g. for

GF (7)

we could write:

GR (7) = {- 3, - 2, - 1, 0, 1, 2, 3}

which makes it clear what is the additive inverse of each element, although sometimes a notation starting from 0 is also used:

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

(2)

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:

0 \times 2 = 01 \times 2 = 22 \times 2 = 43 \times 2 = 04 \times 2 = 25 \times 2 = 4

(3)

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.

Video 1.

Finite fields made easy by Randell Heyman (2015)

Source. Good introduction with examples

Classification of finite fields by

Ciro Santilli 37 Updated 2025-07-16

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^{or d er} = 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!

GF(4) by

Ciro Santilli 37 Updated 2025-07-16

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.:

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

(2)

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:

(1 X + 0) \times (1 X + 1) = (1 \times 1) X^{2} + (1 \times 1) X + (0 \times 1) X + (0 \times 1) = 1 X^{2} + 1 X + 0

(3)

Without modulo, that would not be one of the elements of the field anymore due to the

1 X^{2}

So we take the modulo, we note that:

1 X^{2} + 1 X + 0 = 1 (X^{2} + X + 1) + (0 X + 1)

(4)

and by the definition of modulo:

(1 X^{2} + 1 X + 0) mod (X^{2} + X + 1) = (0 X + 1)

(5)

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.

Quadratically closed field by

Ciro Santilli 37 Updated 2025-07-16

Vector field by

Ciro Santilli 37 Updated 2025-07-16

A function:

R^{m} \to R^{n}

Algebra over a field by

Ciro Santilli 37 Updated 2025-07-16

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

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.

complex numbers, i.e. $R^{2}$ with complex number multiplication
$R^{3}$ with the cross product
quaternions, i.e. $R^{4}$ with the quaternion multiplication

Division algebra by

Ciro Santilli 37 Updated 2025-07-16

An algebra over a field where division exists.

Frobenius theorem (real division algebras) by

Ciro Santilli 37 Updated 2025-07-16

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

Notably, the octonions are not associative.

Associative property by

Ciro Santilli 37 Updated 2025-07-16

Algebraic geometry by

Ciro Santilli 37 Updated 2025-07-16

Algebraic curve by

Ciro Santilli 37 Updated 2025-07-16

Elliptic curve by

Ciro Santilli 37 Updated 2025-07-16

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"

y^{2} = x^{3} + a x + b

Equation 1.

Definition of the elliptic curves

Figure 1.
Plots of real elliptic curves for various values of $a$ and $b$
. Source.

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.

Elliptic curve group by

Ciro Santilli 37 Updated 2025-07-16

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:

mathoverflow.net/questions/6870/why-is-an-elliptic-curve-a-group

Elliptic curve point multiplication by

Ciro Santilli 37 Updated 2025-07-16

Domain of an elliptic curve by

Ciro Santilli 37 Updated 2025-07-16

belongs to the elliptic curve over a non quadratically closed field by

Not every

x

Ciro Santilli 37 Updated 2025-07-16

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} + a x + 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:

y^{2} = 1^{3} + 1 \times 1 + 1 = 3

and therefore there is no

y \in 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

F_{4}

, because doing everything

mod 4

we have:

0 * 0 1 * 1 2 * 2 3 * 3 = 0 = 1 = 4 = 9 = 0 = 1 mod 4 mod 4 mod 4 mod 4

(2)

Therefore, there is no element

y \in F_{4}

such that

y \times y = 2

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

3 \in R

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

Number of elements of an elliptic curve by

Ciro Santilli 37 Updated 2025-07-16