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 $\Bbb Z_2$ I've been working through Rotman's Galois Theory and am stumped by exercise 60:


*Use Kronecker's theorem to construct a field with four elements by adjoining a suitable root of $x^4 - x$ to $\mathbb{Z}_2$.


I can split $x^4 - x$ in $\mathbb{C}[x]$ (which is not a field but which is contained in the field $\operatorname{Frac}(\mathbb{C[x]})$). 
The roots are $F = \{ 0, 1, \frac{-1 - i \sqrt{3}} {2}, \frac{-1 + i \sqrt{3}} {2} \}$.
So what does it mean to adjoin a suitable root to $\mathbb{Z}_2$? I have been unable to construct a four element field using either complex root.
So I went and tried to use the proof of Thm 33 (Galois) which gives a construction. In this case, F is as above and has the required four elements with $p = 2$, $n = 2$, and $q = 4$, and indeed $F \setminus \{0\}$ behaves as desired for multiplication. 
The problem is addition. How is addition defined? Normal addition is not closed, nor do I see how to define it to make it work. Further, given Kronecker's and Galois's theorems, I would assume that addition must be defined as it would be in the containing field.
 A: 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.
A: 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.
A: 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
$$\begin{matrix}
   + & 0 & 1 & \zeta & 1+\zeta\\
   0 & 0 & 1 & \zeta & 1+\zeta\\
   1 & 1 & 0 & 1+\zeta & \zeta\\
   \zeta & \zeta & 1+\zeta & 0 & 1\\
   1+\zeta & 1+\zeta & \zeta & 1 & 0\\
\end{matrix}$$
A: I don't understand why you are talking about the field $\mathbb C$. It has nothing to do with this exercise.
Anyway, $x^4-x=x(x-1)(x^2+x+1)$ and $x^2+x+1$ has no roots in $\mathbb{Z}_2$. So, consider the field $\mathbb{Z}_2[x]/\langle x^2+x+1\rangle$, which has exactly $4$ elements.
A: The field is $\mathbb Z_2(\xi)$, where $\xi^2+\xi+1=0$.  It is a $\mathbb Z_2$ -vector space of dimension $2$, with basis $\{1,\xi\}$. Thus has $4$ elements.
It can also be written as $\frac{\mathbb Z_2[x]}{(x^2+x+1)}$.
A: The kicker, here, is that $0,1$ are already roots of $f,$ since $x$ and $x-1$ are factors of $f(x).$ In particular, $f(x)=x(x-1)(x^2+x+1),$ but neither $0$ nor $1$ are roots of $x^2+x+1,$ so we need to adjoin elements that are roots to this.
To accomplish this, we define $$\Bbb Z_2[\alpha]:=\Bbb Z_2[t]/(t^2+t+1),$$ where $(t^2+t+1)$ indicates the ideal of $\Bbb Z_2[t]$ generated by $t^2+t+1.$ Letting $\alpha=\overline t,$ we find that $\alpha$ and $\alpha+1$ are elements of $\Bbb Z_2[\alpha]$ that are roots of $x^2+x+1.$ At that point, we simply need to verify that $\Bbb Z_2[\alpha]$ is a field of four elements, which is straightforward.
