Explicit description of quotient ring of $\mathbb{Z}[x]$ I am studying for a prelim and I stumbled on this problem:
Describe explicitly the elements in the quotient ring $\dfrac{\mathbb{Z}[x]}{(3,x^3-x+1)}$. First of all I don't see why the ideal $(3,x^3-x+1)$ is a maximal ideal in $\mathbb{Z}[x]$. If there is anyone who can help me with this will be greatly appreciated.

Let $m$ be a maximal ideal of $\mathbb Z[x]$ with $(3,x^3-x+1)\subseteq m$. Assume tha $(3,x^3-x+1)\not=m$ and let $f\in m\setminus(3,x^3-x+1)\subseteq m$. Since $x^3-x+1$ is monic we have $f=g(x^3-x+1)+h$, where $g,h\in \mathbb Z[x]$ and $1\leqslant deg(h)\leqslant 2$. Now since $x^3-x+1\in m$, we have $h\in m$. We consider the following: 
Case 1) $deg(h)=1$: Let $h=ax+b$. Since $3\in m$,  we assume that $a, b\in\{1, 2\}$. Thus we have the following subcasis:
Subcase 1a) $a=b=1$: In this case $x+1\in m$ and since $x^3-x+1\in m$, we have $x(x^2-2)=x^3-2x\in m$. Thus, $x\in m$ or $x^2-2\in m$. Hence, $1\in m$ or $x(x+2)=x^2+2=x^2-2+2(x+1)\in m$.Thus, $1\in m$ or $x\in m$ or $x+2\in m$ and so $1\in m$, a contradiction.
Other subcases are simillar.
Case 2) $deg(h)=2$: Let $h=ax^2+bx+c$. Since $3\in m$,  we assume that $a, b, c\in\{1, 2\}$. Thus we have the following subcases:
Subcase 2a) $a=b=c=1$: In this case $x^2+x+1\in m$ and since $x^3-x+1\in m$, we have $x(x^2-x-2)=x^3-x^2-2x\in m$. Thus, $x\in m$ or $x^2-x-2\in m$. Hence, $1\in m$ or $(x^2-x-2)-(x^2+x+1)\in m$, if $1\in m$, a contradiction. Now let $(x^2-x-2)-(x^2+x+1)\in m$, hence a polynominal of degree one is in $m$ and by case 1 we have a contradiction.
Other subcases are simillar.
