Show that the principal ideal $(x-1)$ in $\mathbb{Z}[x]$ is prime but not maximal. 
Show that the principal ideal $(x-1)$ in $\mathbb{Z}[x]$ is prime but not maximal.

I have tried defining an isomorphism for the problem but am extremely confused on how this helps me in any way. 
 A: Hint: consider the map $\varphi\colon\mathbb{Z}[x]\to\mathbb{Z}$ defined by
$$
\varphi(f)=f(1)
$$
Is it a ring homomorphism? What are its kernel and image?
A: Hint: An ideal is maximal iff the quotient ring is a field, and is prime iff the quotient is a integral domain.
 Can you find $\mathbb{Z}[x]/(x-1)$?
A: Hint $\,\ \Bbb Z[x]/(x\!-\!1)\cong \Bbb Z$ is a $\rm\color{#c00}{nonfield}\ \color{#0a0}{domain} \Rightarrow\,(x\!-\!1)\,$ is a $\rm\color{#c00}{nonmaximal}\ \color{#0a0}{prime}$ ideal.
If quotient rings are unfamiliar one may prove it as follows. Note that $\,(x\!-\!1)\subsetneq (2,x\!-\!1)\subsetneq (1),\,$ since $\ x\!-\!1\nmid 2\,$ and $\,1\not\in (2,x\!-\!1),\ $ else $\,1 = 2f + (x\!-\!1)g\,$ so evaluating at $\,x=1\,$ yields the contradiction that $\, 1 = 2f(1).\,$ Thus $\,(x\!-\!1)\,$ is not maximal, but it is prime since
$\quad x\!-\!1\mid fg=h \iff h(1)=0\iff f(1)g(1)=0\iff\!\! \begin{array}{r} f(1)=0\\{\rm or}\,\ \ g(1)=0\end{array}\iff\!\! \begin{array}{r}x\!-\!1\mid f\\ {\rm or}\,\ \ x\!-\!1\mid g\end{array}$
