Let $A = \Bbb Z [\omega]$ be the subring of complex numbers where $\omega$ is a non-real $6$-th root of unity. In the polynomial ring $A[x],$ let $I$ be the ideal generated by $(x^2+x+1).$ Prove or disprove that $A[x]/I$ is an integral domain.
Let $\omega_1, \omega_1^2$ be the primitive cube root of unity. Let $\overline A(x) = (x - \omega_1) + \left \langle x^2 + x + 1 \right \rangle$ and $\overline B(x) = (x - \omega_1^2) + \left \langle x^2 + x + 1 \right \rangle.$ Then $\overline A(x), \overline B (x)$ are two non-zero elements of $A[x]/I$ whose product is zero. Hence $A[x]/I$ is not an integral domain. Can it be proved in a diffrent way i.e. by the help of Chinese remainder theorem for rings?
Thanks in advance.