Prove $\mathbb{Z}[\omega]/\left\langle p\right \rangle \cong \mathbb{F}_p[x]/\left \langle x^2+x+1 \right \rangle$

Question

Let $\omega=e^{2\pi\over 3}$ and let $R=\mathbb{Z}[\omega]$. Let $p>3$ be a prime number. Prove $\mathbb{Z}[\omega]/\left\langle p\right \rangle$ is isomorphic to $\mathbb{F}_p[x]/\left \langle x^2+x+1 \right \rangle$.

I have observed that $R/\left\langle p \right\rangle=\{a+b\omega: 0\leq a<p, b\in\mathbb{Z}\}.$ But I don't know how to continue.

Answer 1

First, we observe that since $\omega^3=1$, $\omega$ will be a root of $x^3-1=(x-1)(x^2+x+1)$. Further, since the second factor is irreducible and $\omega \neq 1$, it satisfies $\omega^2+\omega+1=0$. Therefore, $\mathbb{Z}[\omega]=\mathbb{Z}[x]/\langle x^2+x+1\rangle$.

Now, we just observe that $$(\mathbb{Z}[x]/\langle x^2+x+1\rangle)/\langle p \rangle> \cong \mathbb{Z}[x]/\langle x^2+x+1,p\rangle\cong (\mathbb{Z}[x]/\langle p \rangle)/ \langle x^2+x+1 \rangle$$

Answer 2

This follows from the third isomorphism theorem.

First, note that

$$\mathbb{Z}[\omega]\cong \mathbb{Z}[x]/\langle x^2 + x + 1\rangle$$

and thus

$$\mathbb{Z}[\omega]/\langle p\rangle \cong \mathbb{Z}[x]/\langle p, x^2 + x + 1\rangle$$

which is isomorphic to

$$(\mathbb{Z}[x]/\langle p \rangle)/\langle x^2 + x + 1\rangle \cong \mathbb{F}_p[x]/\langle x^2 + x + 1\rangle.$$