[Math] Finite field, I don’t quite understand the concept

cryptographyself-learning

What is the concept behind AES's finite field? I understand most of you will laugh at my lack of understanding of the concept. I'm trying to learn a greater amount of higher math concepts and I am implementing AES to do so. I understand that in a basic term, polynomial is an equation of only positive integer expodetials and variables but what does it mean by treated as a polynomial over GF(2^8) and is 2^8, 2 to the power of 8 or 2 XOR 8?

Corrected: from Wikipedia AES

In more general sense, each column is treated as a polynomial over $GF(2^8)$
and is then multiplied modulo $x^4 + 1$ with a fixed polynomial $c(x) = 0x03*x^3 + x^2 + x + 0x02$. The coefficients are displayed in their hexadecimal equivalent of the binary representation of bit polynomials from $GF(2)[x]$.

Best Answer

I will try to give a layman's terms description; hoping this helps clarify some of your confusions. But you should be warned that this is not a precise definition.

$\mathbb{Z}$ is the set of all integers. $\mathbb{Z}_2$ is the set of all integers modulo 2. That is, ${0,1}.$ Arithmetic operations in $\mathbb{Z}_2$ are performed modulo $2.$ Consider a vector of length 8, whose entries are elements in $\mathbb{Z}_2$. In other words, it's a bit vector of length 8. A given vector $(b_1,b_2,\dots,b_8) \in (\mathbb{Z}_2)^8$ can be represented as a polynomial: $b_1 x^7 + b_2 x^6 + \dots + b_7 x + b_8.$ That is, a polynomial in $x$ with coeffiecients from $\mathbb{Z}_2.$ $\mathrm{GF}(2^8)$ is defined as a set of such vectors (equivalently such polynomials. This is what the article refers to as each column is treated as a polynomial..). In order to construct $\mathbb{Z}_2$ we need the element $2,$ and performed all arithmetic modulo $2.$ If you want to construct $\mathbb{Z}_5,$ will perform operations mod $5$ etc. Similarly, In order to construct $\mathrm{GF}(2^8)$, you need a specific (irreducible) polynomial $p$ of degree $8.$ All arithmetic operations in $\mathrm{GF}(2^8)$ are then performed modulo $p.$

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