Why does Cryptography use Polynomial Modular Arithmetic in Finite Fields?
Polynomial Modular Arithmetic done in Finite Fields is used in a lot of places in Cryptography  for e.g. in AES/Rijndael, GHASH, POLYVAL etc. What is the reason for this?
First, why Finite Fields at all?
Let’s start with what is a Finite Field? For that, we will start with a Group & from there go all the way up to Finite Fields.
A Group is a set G with a binary operation on G such that for all \(a, b, c \in G\)

\(a * (b * c) = (a * b) * c\)  This property is called associativity

There exists an element e \(\in G\) (called the identity element) such that for all \(a \in G, a * e = a = e * a\)

For all \(a \in G\) there exists \(a' \in G\) (called the inverse of a) such that \(a * a' = e = a' * a\)

Any operation done on the group using the operator will always produce a result which is also a member of the group. This property is called Closure
An Abelian Group is a Group which is also Commutative  i.e. for all \(a, b \in G, a * b = b * a\)
An Abelian Group under Addition which also has the following two properties for Multiplication is called as a Ring

Associative multiplication
\[a * ( b * c ) = ( a * b ) * c\] 
Left & right distributive over multiplication
Left: \(a*(b+c) = a*b + a*c\)
Right: \((a+b)*c = a*c + b*c\)
An Integral Domain is a Ring which is

Commutative under multiplication  \(a*b = b*a\)

Cancellation under multiplication  \(ac = bc\), means \(a = b\)

an identity element under multiplication

one which has no zero divisors  i.e. \(a*b = 0\) only if either \(a = 0\) or \(b = 0\)
A particular type of Integral Domain is called a field. A field is a commutative ring with an identity element under multiplication where every nonzero element has a multiplicative inverse. Every finite Integral Domain is a Finite Field.
Finite Fields are also called Galois Fields
So, from all of the above (going all the way from a Group to a Finite Field)  these are the different properties which a Finite Field must possess
Under Addition

Closure

Associativity

Identity

Inverse

Commutativity
Under Multiplication

Closure

Associativity

Distributivity

Commutativity

Identity

Inverse

No zero divisors
Many of the different properties which Finite Fields possess are very useful in cryptography.
Let’s look at a few of them

When you are multiplying 2 numbers in a cryptographic operation, it’s very useful to have the result also in the same closed set. For e.g. in AES, when you multiply 2 single bytes, you want the result also to be a byte  the multiplication shouldn’t result in something which is bigger than a byte. Finite Fields have Closure, so the result will always be a byte

Many operations in Cryptography need to be invertible. For e.g. if while encrypting, you do some mathematical operation, you want it to invertible so that you can reverse the operation while decrypting. All elements in a Finite Field have an Inverse under addition & all nonzero elements have an Inverse under multiplication. In AES, in the MixColumns operation, we multiply by a matrix & during decryption we multiply by the inverse of the matrix.

In AES’s MixColumns, we need the associativity property  \(a*(b+c) = a*b + a*c\)
So, these are some of the reasons many cryptographic operations are done in Finite Fields.
Why not regular modular arithmetic?
We want to do arithmetic with bytes. A byte can be represented by using the bits as the coefficients of a polynomial.
For e.g. if we want to represent a byte with value 51. In binary, 51 is 00110011.
This bit pattern can be represented as the coefficients of the polynomial \(x^{5} + x^{4} + x + 1\)
Likewise 249 = 11111001 = \(x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + 1\)
So, if we are multiplying 2 bytes, then the 2 bytes can be represented as 2 polynomials \(f(x)\) & \(g(x)\)  the degree of each of the polynomial will be 7 or lower.
But when we multiply 2 polynomials of degree 7 or lower, we can get a result which can be of a degree higher than 7, so it can no longer be represented in a byte. So we need closure. We can achieve closure by reducing by modulo in a finite field.
A finite field of order q exists if and only if q is a prime power  i.e. \(q = p^{n}\)
A finite field of order \(p^{n}\) is written as \(GF(p^{n})\)
There are 2 cases here
1) n = 1. This is written as \(GF(p)\)
2) n > 1. This is written as \(GF(p^{n})\)
Both these kinds of finite fields have different structures. \(GF(p^{n})\) is called an extension field of \(GF(p)\), because it can embed \(GF(p)\). For e.g. \(GF(2^{8})\) is an extension of \(GF(2)\), \(GF(2^{2})\), \(GF(2^{4})\)
We can do the reduction by modulo 8, which would ensure that the degree of the polynomial cannot be greater than 7. However, $\mathbb{Z_8}$ is not a finite field under regular modular arithmetic because some of the elements in $\mathbb{Z_8}$ do not have an inverse.
But, 2 is a prime & 8 is an integer, so it should be possible to form a finite field \(GF(2^{8})\) even if not through regular modular arithmetic. \(2^8\) is 256 which is the number of bits in one byte, so we can use \(GF(2^{8})\) as the finite field.
Why Modular Polynomial Arithmetic?
Both \(f(x)\) and \(g(x)\) are polynomials of degree 7 or less. So \(f(x) * g(x)\) can result in a polynomial of degree greater than 7.
So, we reduce \(f(x) * g(x)\) by modulo an irreducible polynomial of degree 8, so that the result \(r(x)\) is a polynomial of degree 7 or less.
So in AES, if 2 bytes have to be multiplied, they are each represented as a polynomial (the bits of the bytes form the coefficients of the polynomial) of degree 7 or less. After multiplying the 2 polynomials, they are reduced by modulo an irreducible polynomial of degree 8, which results in a polynomial of degree 7 or lesser which will again fit in a byte.
Hence Modular Polynomial Arithmetic done in a Finite Field is used in Cryptography.
AES uses \(x^{8} + x^{4} + x^{3} + x + 1\) as the irreducible polynomial