Identify the ring $\mathbb{Z}[X]/(X^2 + X + 1)$. My guess was that 
$$
 \mathbb{Z}[X]/(X^2 + X + 1) \cong \mathbb{Z}[\alpha]
$$
with $\alpha$ being a root of $X^2 + X + 1$. But since $\alpha$ has a real part aswell, I wouldn't know if $\mathbb{Z}[\alpha]$ would be well defined.
The only definition I ever got was
$$\mathbb{Z}[\sqrt{m}] = \{a + b\sqrt{m} \mid a,b \in \mathbb{Z}\} $$
with $m$ being an integer that does not have a rational square root. Now, $\alpha$ has a real rational part, and that collides with the definition I was given. Am I going the wrong path, or should I expand my definition of $\mathbb{Z}[\sqrt{m}]$, where $\sqrt{m}$ is rational? 
 A: The roots of $X^2+X+1$ are the nontrivial third roots of unity, so let $\omega=exp(\frac{2\pi i}{3})$ then the roots are $\omega$ and $\omega^{2}$.
In fact $X^{2}+X+1$ is the minimal polynomial for $\omega$ over $\mathbb{Z}$
Consider the map $ev: \mathbb{Z}[X]\rightarrow\mathbb{Z}[\omega], f(x)\mapsto f(\omega)$, where $\mathbb{Z}[\omega]=\{a+b\omega|\ a,b\in\mathbb{Z}\}$. This evaluation map is a surjective homomorphism. It's kernel consists of all integer polynomials vanishing at $\omega$, certainly $X^{2}+X+1$ is there and any multiple of it so $(X^{2}+X+1)\subset \ker(ev)$ To show the other inclusion we can use the division algorithm. Let $g\in\mathbb{Z}[X]$ be such that $g(\omega)=0$ we can write $g(X)=q(X)(X^{2}+X+1)+r(X)$, with $deg(r)<2.$ Evaluation at $\omega$ yields $g(\omega)=q(\omega)(\omega^{2}+\omega+1)+r(\omega)=0$ which implies that $r(\omega)=0$. $r(X)$ must be a constant since if it were of degree 1 it would contradict the fact that $X^{2}+X+1$ is the minimal polynomial for $\omega$. That constant must then be $0$. So we see that $g(X)=q(X)(X^{2}+X+1)$ and hence $ker(ev)=(X^{2}+X+1)$
With this we can now apply the first isomorphism theorem to see that:
$\mathbb{Z}[X]/(X^{2}+X+1)\cong\mathbb{Z}[\omega]$
