Galois Theory by Rotman, Exercise 60, a field of four elements by using Kronecker's theorem and adjoining a root of $x^4-x$ to $\Bbb Z_2$

You are not required to adjoin a complex root to $\mathbb{Z}_2$. You can't do that even if you try because $\mathbb{C}$ and $\mathbb{Z}_2$ have different characteristic.

Instead, the hint is to use Kronecker's theorem, so let's do that.

First, what are the obvious roots of $f(x) = x^4 - x$ in $\mathbb{Z}_2$? Clearly, $\bar{0}$ and $\bar{1}$ are both roots of $f$. So, we can factor out $x$ and $x - 1$ to get $f(x) = x(x-1)(x^2+x+1)$. Note that $x^2 + x + 1$ is irreducible over $\mathbb{Z}_2$. By Kronecker's theorem, there exists an extension field of $\mathbb{Z}_2$ that contains a root of $x^2+x+1$, which is also a root of $f(x)$. The degree of the extension is equal to the degree of the polynomial $x^2 + x + 1$, since it is irreducible over $\mathbb{Z}_2$. A degree two extension of $\mathbb{Z}_2$ has to contain $4$ elements (do you see why?) and so we are done.

Hint:

Over any commutative ring, you have the factorisation $$x^4-x=x(x^3-1)=x(x-1)(x^2+x+1).$$ Now over the field with two elements $\mathbf F_2$, it is easy to check $x^2+x+1$ is irreducible, so the quotient $\;\mathbf F_2[x]/(x^2+x+1)$ is a field. Denoting $\xi=x\bmod x^2+x+1$, you have by construction $\xi^2+\xi+1=0$, hence $\;\xi^4-\xi=0$.

This field is a $\mathbf F_2$-vector space with dimension $2$, so it has $2^2=4$ elements.

In the exercise, you're asked to work on the field $\mathbb Z_2$ which is of characteristic $2$. So you can't split $p(x) = x^4-x$ on $\mathbb C$.

Now you can write on $\mathbb Z_2$: $$p(x) = x(x^3-1)=x(x+1)(x^2+x+1)$$

If $\zeta$ is a root of $q(x) = x^2+x+1$ (which is irreducible on $\mathbb Z_2$) in a splitting field, the other one is $1+\zeta$ as $(1+\zeta)^2+(1+\zeta) + 1=1+\zeta^2+ 1 +\zeta +1 = 1+1=0$.

Regarding your problem which is addition

So the four elements of a splitting field of $p$ over $\mathbb Z_2$ are $z_1=0, z_2=1, z_3=\zeta, z_4=1+ \zeta$. And regarding addition, you have

$$\begin{matrix} + & 0 & 1 & \zeta & 1+\zeta\\ 0 & 0 & 1 & \zeta & 1+\zeta\\ 1 & 1 & 0 & 1+\zeta & \zeta\\ \zeta & \zeta & 1+\zeta & 0 & 1\\ 1+\zeta & 1+\zeta & \zeta & 1 & 0\\ \end{matrix}$$