Is $\mathbb{Z}[x]/\langle 4,x^2+x+1 \rangle$ a field?

Question

I know that $\mathbb{Z}[x]/\langle4, x^2+x+1\rangle$ is isomorphic to $(\mathbb{Z}/4\mathbb{Z})[X]/\langle x^2+x+1\rangle $. Since $(\Bbb Z/4\Bbb Z)[X]$ is not a PID, irreducibility/reducibility of $x^2+x+1$ is not enough to say whether or not this is a field. My idea is to prove that this ring is not a domain. For example, $(\mathbb{Z}/4\mathbb{Z})[X]/\langle x^2+x+1\rangle$ is not a domain if $2$ is not in the ideal $\langle x^2+x+1\rangle$ but I can't prove this.

Answer 1

Hint $ $ Let $\,h = x^2\!+\!x\!+\!1.\ $ If $\,2\,$ is a unit: $\,2 f = 1 + 4g + hh'\,$ in $\Bbb Z[x]\,$ $\overset{\bmod 2}\Longrightarrow\,h\mid 1\,$ in $\,\Bbb Z_2[x]\ \Rightarrow\!\Leftarrow$

And if $\,2=0\,$ then $\,2 = 4g + hh'\,$ so $\,2\mid h'\,$ hence $\,1 = 2g+ h(h'/2)\,$ $\overset{\bmod 2}\Longrightarrow\, h\mid 1\,$ in $\Bbb Z_2[x]\ \Rightarrow\!\Leftarrow$

Answer 2

Hint: $2$ is in $(x^2+x+1){\bf Z}/4{\bf Z}[x]$ if and only if for some $k\in {\mathbf Z}$ such that $k\equiv 2 \pmod 4$, $k$ is in $(x^2+x+1){\bf Z}[x]$.