I know that the ideal $\langle x^2+x+1, 2\rangle$ is maximal ideal in $\mathbb{Z}[x]$. (I have proved this by assuming an strictly bigger ideal and showing this to be the whole ring..!!). Therefore, $F=\mathbb{Z}[x]/\langle x^2+x+1, 2\rangle$ is a field.
$\rangle$ But, now I'm suspecting that $F$ is a field of $4$ elements.
The reason behind this seems so intuitive but even I don't know whether I am doing any mistake...I include the my intuition behind this:
Any element, say, $~f(x)+ \langle x^2+x+1, 2\rangle~\in F$ can be written in the form $(ax+b)+\langle x^2+x+1, 2\rangle$, where $a,b \in \mathbb{Z}$ (using Division Algorithm). Now reducing $a,b$ by $\mod 2$ we see that there are $4$ choices of $a,b$.
In another direction, I am suspecting the following isomorphism:
$F=\mathbb{Z}[x]/\langle x^2+x+1, 2\rangle~ \cong ~\mathbb{Z}_2[x]/\langle x^2+x+1\rangle~\cong~\mathbb{F}_4 \dots\dots\dots(1)$
The last isomorphism of (1) is easy. I try to prove the 1st isomorphism appears in (1) by defining the epimorphism $\phi:\mathbb{Z}[x] \to\mathbb{Z}_2[x]/\langle x^2+x+1\rangle$ by $$f(x)\mapsto \overline{f}(x)+\langle x^2+x+1\rangle$$
Where the $~\overline{f}$ denotes the image of $~f$ under canonical epimorphism from $\mathbb{Z}[x] \to \mathbb{Z}_2[x]$.
$\rangle$ But I am unable to proof that, $\ker\phi=\langle x^2+x+1, 2\rangle$
Please help me to show the last equality...!!If I made any mistake, please let me know. Thank you.
Hint $\phi(x^2+x+1)=\phi(2)=0$ and hence $\supset$ holds.
For the other implication Let $\phi(f)=0$.
By long division $$f(x)=q(x)(x^2+x+1)+ax+b$$ and hence $$0=\phi(f)=\overline{ax+b}$$
Deduce from here that $ax+b \in <2>$.