Construct an isomorphism between fields The first part of my question asked : State all the irreducible Polynomials in $\mathbb{Z}_2[x]$ of order 3.  
I was able to do this and get the following polynomials :   
$x^3 + x^2 + x + 1 \Rightarrow$ reducible
$x^3 + x^2 + 1 \Rightarrow$ irreducible
$x^3 + x + 1 \Rightarrow$ irreducible
$x^3 + 1 \Rightarrow$ reducible  
and the other polynomials in $\mathbb{Z}_2[x]$ are trivially reducible.
I set $f=x^3 + x^2 + 1$, $g=x^3 + x + 1$.
I then have to take $\mathbb{Z}_2[x]/f$ and $\mathbb{Z}_2[x]/g$ and construct an isomorphism between the two.  
My real problem is understanding what these two fields $\mathbb{Z}_2[x]/f$ and $\mathbb{Z}_2[x]/g$ really are- and how I can build an isomorphism between them.    
If anyone could help I'd be very grateful :) 
 A: First of all, you can treat $\mathbb{F}_2[x]/(f)$ just as a ring. As you know, a quotient of a ring is a set of cosets with the obvious operations of addition and multiplication inherited from the big ring. It just so happens that quotients of integral domains by maximal ideals are also fields, but you needn't be too worried about that at the moment. Now, to write down an isomorphism between quotient rings, the most naive way of doing it would be to write down an isomorphism between the big rings (i.e. in this case between $\mathbb{F}_2[x]$ and itself) that takes the one ideal bijectively to the other, i.e. $(f(x))$ to $(g(x))$. Again, the most naive way of accomplishing that would be to just send one generator to the other. See if you can do that and convince yourself that this procedure really does always give you an isomorphism of quotient rings. 
Another Exercise: if two isomorphic rings turn out to be fields, why is an isomorphism of rings also an isomorphism of fields?
A: The first field looks like this:
\begin{align*}
\mathbb{Z}_{2}[x]/\langle f(x)\rangle &=
\{ p(x) + \langle f(x)\rangle\mid p(x)\in\mathbb{Z}_2[x],\ \deg(p)\lt\deg(f)\}\\
&= \left\{
\begin{array}{l}
0 + \langle f(x)\rangle,\quad 1 + \langle f(x)\rangle,\quad x + \langle f(x)\rangle, \quad x^2 + \langle f(x)\rangle,\quad (1+x) + \langle f(x)\rangle,\\
(1+x^2) + \langle f(x)\rangle,\quad (x + x^2) + \langle f(x)\rangle, \quad (1+x+x^2) + \langle f(x)\rangle
\end{array}\right\}
\end{align*}
You know that you have found all elements by counting them as follows:
You have $|\mathbb{Z}_{2}| = |{ 0, 1}| = 2$ choices for any coefficient.
In any polynomial, you can have terms of degree $0$, $1$ or $2$, each such term can have any of the two possible coefficients.
So your resulting field has $2 \times 2 \times 2 = 2^3 = 8$ elements.
Regarding notation:
$\langle f(x)\rangle $ in the above means the ideal generated by $f$. This means, the set of all elements $p(x) \in \mathbb{Z}_{2}[x]$ that are multiples of $f$.
