How to find an extension field of $x^3 + x + 1$ over $\Bbb Z_2$? Let $F = \Bbb Z_2$ and let $f (x) = x^3 + x + 1 \in F[x]$. Suppose that $a$ is a
zero of $f (x)$ in some extension of $F$. How many elements does $F(a)$
have? Express each member of $F(a)$ in terms of $a$.
I can't think of a single extension that this would have a root in.

Since the degree of the irreducible--no roots and low degree--polynomial, $f$, is $3$ we know that the degree of the extension it generates will be $3$. This means that as a vector space, the dimension of $F(a)$ over $\Bbb Z_2$ is $3$, so it is isomoprhic to $\Bbb Z_2 \times\Bbb Z_2\times\Bbb Z_2$ which has $8$ elements. By the first ring isomorphism theorems,
$$F(a)\cong \Bbb Z_2[x]/(x^3+x+1)$$
by the map $\varphi:  \Bbb Z_2[x]\to F(a)$ where $p(x)$ maps to $p(a)$, so for example $a+1, a^2+a, a^2+a+1$ et cetera are all elements of $F(a)$.
