My question is a little difficult to describe, so let me first start with an analogy
In an elliptic curve over a finite field, there are 2 groups - the first group is a finite field over which the elliptic curve is defined. The 2nd group is the group which is formed by all the points of the elliptic curve. These are the 2 different groups.
My actual question:
In AES256 we use a polynomial to represent each byte. The coefficients of the polynomial are from $\operatorname{GF}(2)$ - i.e. it's polynomial ring over $\operatorname{GF}(2)$. The polynomial addition is done mod 2. The multiplication is done in 2 steps. First the 2 polynomials are multiplied mod 2. Then they are reduced mod the irreducible polynomial
So I am confused as to where exactly the $\operatorname{GF}(2^8)$ comes into the picture?
I am guessing that each byte which is a represented by a polynomial is a member of field $\operatorname{GF}(2^8)$ - i.e. $\operatorname{GF}(2^8)$ is a field of bytes.
And just like for elliptic curves we arbitrarily define addition using the tangent & chord method, here we arbitrarily define the addition & multiplication of the field elements (the bytes) as
addition of 2 elements of the field $\operatorname{GF}(2^8)$ - add coefficients mod 2.
multiplication of 2 elements of the field $\operatorname{GF}(2^8)$ - multiply the coefficients mod 2 & then reduce it mod the irreducible polynomial.
Is my interpretation correct or am I totally missing the field abstraction & operations here?
If it is correct, then my next question is about the concept of extension fields here - $\operatorname{GF}(2^8)$ is an extension field of $\operatorname{GF}(2)$ - what exactly does does it mean here - does it just mean that each byte contains 8 bits (each bit being an element of $\operatorname{GF}(2)$). Likewise what do the sub-fields $\operatorname{GF}(2^2)$ & $\operatorname{GF}(2^4)$ represent here?