TITLE: What do the elements of the field $\mathbb{Z}_2[x]/(x^4+x+1)$ look like? What is its order?
QUESTION [5 upvotes]: Background: I'm looking at old exams in abstract algebra. The factor ring described was described in one question and I'd like to understand it better.
Question: Let $F = \mathbb{Z}_2[x]/(x^4+x+1)$. As the polynomial $x^4+x+1$ is irreducible over $\mathbb{Z}_2$, we know that $F$ is a field. But what does it look like? By that I am asking if there exists some isomorphism from $F$ into a well-known field (or where it is straightforward to represent the elements) and about the order of $F$.
In addition: is there something we can in general say about the order of fields of the type $\mathbb{Z}_2[x]/p(x)$ (with $p(x)$ being irreducible in $\mathbb{Z}_2[x]$)?

REPLY [3 votes]: You were asking for other fields that are isomorphic to $F=\Bbb{Z}_2[x]/\langle x^4+x+1\rangle=\Bbb{Z}_2[\alpha]$ with $\alpha=x+\langle x^4+x+1\rangle$.
I proffer one such. Unfortunately it doesn't easily generalize to all finite fields. Thus it is a bit ad hoc.
Consider the ring $R=\Bbb{Z}[\zeta]$, where $\zeta=e^{2\pi i/5}$ is a primitive fifth root of unity. In $\Bbb{Z}_2[x]$ we have
$$
x^{12}+x^9+x^6+x^3+1=(x^4+x+1)(x^8+x^4+x^2+x+1).
$$
Therefore the element $\beta=\alpha^3\in F$ satisfies the equation
$\beta^4+\beta^3+\beta^2+\beta+1$. As this coincides with the minimal polynomial of $\zeta$ there exists a homomorphism of rings $f:R\to F$ determined by $f(\zeta)=\beta$.
This homomorphism is surjective. One way of seeing that follows from the calculation
$$
\begin{aligned}
\alpha&=\alpha\cdot(1+0)^2=\alpha\cdot(1+1+\alpha+\alpha^4)^2\\
&=\alpha\cdot(\alpha+\alpha^4)^2=\alpha\cdot(\alpha^2+\alpha^8)\\
&=\alpha^3+\alpha^9=\beta+\beta^3.
\end{aligned}
$$
What about $\ker f$? Clearly $2R\subseteq \ker f$. Because $R$ is a free abelian group of rank four, we know that $|R/2R|=2^4=16$. This forces $2R=\ker f$.
Conclusion:
$$
F\cong R/2R=\Bbb{Z}[\zeta]/\langle 2\rangle.
$$