Counting elements of a field and more

[PsychonautQQ]

Homework Statement

Consider the field K=Z_2[X]/<x^3+x+1>. How many elements does this field have?

Then let f(x) = x^3+x+1 and b be a root of f(x) in Z_2[X], Find all other roots of f(x) in K and describe the Galois group

Homework Equations

The Attempt at a Solution

So I'm trying to understand what K looks like, perhaps something like this:
K={ax^3 + bx^2 + cx + d | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}

Then is it possible that since x^3 = x + 1 mod 2 that I could make this substitution in the description of the field? i.e.:
K={a(x+1) + bx^2 + cx + d | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}
={ax+a + bx^2 + cx + d | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}
= {bx^2 + (c+a)x + (d+a) | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}
={bx^2 + ex + f | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}

So there would be 8 elements maybe? Is this logic correct?

 

[fresh_42]

I think this gives you an upper bound for the number of elements in $\mathbb{Z}_2[x] / (x^3+x+1)$ as the reasoning:
"Since $x^3$ can be expressed by lower powers of $x$, we will at most get polynomials of degree $2$ and thus $8$ possible coefficient arrays."
would do. But it's not quite obvious (to me), whether all possibilities will actually occur, i.e. no double counting took place. (My guess is it is so, but I don't see it yet.)

Have you tried to decompose $x^3+x+1\,$?