What is the intuition behind mapping of elements from $GF(2^8)$ to $GF(((2^2)^2)^2)$? I'm finding it very difficult to understand the concept of mapping elements from the extension field $GF(2^8)$, to $((GF(2)^2)^2)^2 $. I realize that the elements of the field, $GF(2^8)$, that I want to split into simpler fields, can be represented as polynomials of maximum degree, 7, and addition is simply an XOR, multiplication and inversion is done modulo the irreducible polynomial $x^8 + x^4 + x^3 + x + 1$,  but being a beginner I have no idea whatsoever to map the elements to the composite fields. I understand that, the mapping is possible, because of field isomorphisms. In essence, I'm having trouble visualizing how the math holds together. My main motivation to understand about this, is because, this technique is used for implementing a compact version of the AES sbox.
Almost, every source, I can find on the internet, speaks complicated language!
A good source, and every help and advice will be greatly appreciated!
 A: This is not an answer just a very long comment.
I don't think you quite understand what is going on. (Forgive me if I am misunderstanding you.) We are NOT going to split the big field into smaller fields, that isn't possible, but we are going to split the additive structure of the big field into copies of the additive structure of the smaller field: the multiplication still mixes everything up.
I think it's best to start off thinking about the complex numbers $\mathbb{C}$ and the real numbers $\mathbb{R}$. When we first construct $\mathbb{C}$ we take $\mathbb{R}$ and a "new" element $i$, and we look at all the $a+bi$ with $a,b\in\mathbb{R}$: we even draw a picture with all the $a$ along the real $x$-axis, and all the $bi$ along the $y$-axis. We add these in the "obvious" way, and so at least additively we have $\mathbb{C}$ just looks like $\mathbb{R}^2$. But multiplication is different: we say "oh, let's have $i^2=-1$"; and so the multiplication tangles together the copies of $\mathbb{R}$. More formally we can manufacture $\mathbb{C}$ in this way: we let it be the set of all polynomials with real coefficients, where after we add and multiply them we reduce them all modulo the (irreducible) $X^2+1$.
We can play the same trick whenever we have a field $\mathbb{k}$ and an irreducible polynomial $\phi(X)$  of degree $d$. We can make a big field $\mathbb{K}$ by taking the set of all polynomials with coefficients in $\mathbb{k}$, using the usual addition and multiplication except that we reduce everything modulo $\phi(X)$. In this way we'll see that the additive structure of $\mathbb{K}$ is just like the additive structure of $\mathbb{k}^d$; but the multiplication tangles it all together. [It is a reasonably big theorem that this process gives a field.]
The simplest example is to start with $GF(2)=\{0,1\}$. The only irreducible quadratic is $X^2+X+1$. So we can get a field with $4$ elements by taking $GF(4)=\{0,1,\omega,1+\omega\}$ and using addition modulo $2$, and for multiplication remembering that $\omega^2+\omega+1=0$. [Note that just as we used $i$ as the new element for the complex numbers to remind ourselves to work modulo $i^2+1$, here I have used $\omega$ to remind myself to reduce modulo the irreducible.]
Final comment. There is a theorem that for each $p^n$ ($p$ prime) there is, up to field isomorphism exactly one field of order $p^n$.
