Here is the problem:
R=GF(2)[x] mod x^3 + 1 = 0
Now, I know these are the 8 elements in it.
0, 1, x, x+1, x2, x2 + 1, x2 + x, x2 + x + 1
According to the question I asked here: Constructing add/multi tables for GF(2)
The 8 elements come from any element in GF(2)[x] divided by x^3 + 1. Then there are 8 possible remainders, and those remainders are the 8 elements.
I don't understand this though. If 0 is an element, and I divide that by x^3 + 1, I get 0 as an answer, with no remainder. Where are these 8 remainders coming from? Does the amount of remainders change if instead of x^3 + 1, it was x^4 + 1, or x^3 + x + 1?
‘No remainder’ means a remainder equal to $0$.
The remainders are all the polynomials of degree at most $2$ in $\mathbf F_2[x]$: $$\{0,1,x, x^2, 1+x, 1+x^2, x+x^2,1+x+x^2\}.$$
For $x^3+x+1$ it would be the same, but the Cayley tables for multiplication would be different.
For $x^4+1$ as well as for $x^3+1$, you don't obtain a field, because the polynomials are not irreducible, as pointed @fkraiem 4: $$x^4+1=(x^2+1)^2,\quad x^3+1=(x+1)(x^2+x+1).$$