showing that some quotient ring is a field. To show that $$\mathbb{Z}[x] / \langle 5, x^3+x+1\rangle$$ is a field; I tried to show that $\langle 5, x^3+x+1\rangle$ is a maximal ideal of $\mathbb{Z}[x]$, but I failed. 
Because I have not seen the maxmal ideal generated by 2 generateors, the problem is hard.
Please inform me how to show that.
 A: HINT: Note that $\mathbb{Z}[x]/\langle5, x^3+x^2+1\rangle\cong \mathbb{Z}_5[x]/\langle x^3+x^2+1\rangle$. Show that $\langle x^3+x^2+1\rangle$ is a maximal ideal in $\mathbb{Z}_5[x]$. Note that it is enough to show that $x^3+x^2+1$ has no root in $\mathbb{Z}_5$, this would imply that the polynomial is irreducible and hence the ideal is maximal.
A: the most general case, which basically has the same proof (try it!) as what pritam and Thomas Andrews wrote, is
Theorem: An ideal in $\;\Bbb Z[x]\;$ is maximal iff it is of the form $\;\langle\,p\,,\,f(x)\,\rangle\;$ , with $\;f(x)\in\Bbb Z[x]\;$ s.t. $\;f(x)\pmod p\in\Bbb F_p[x]\;$ is irreducible, $\;p\;$ a prime .
Since $\;x^3+x+1\in\Bbb F_5[x]\;$ is irreducible we're done.
A: Hint: Suppose you add a polynomial $P(X)$ to your ideal. By multiplying $x^3 + x + 1$ by powers of $x$ and constants, and then and subtracting, you can assume that $P(X)$ is at most cubic. If $P(X)$ is a constant not equal to $5$, it is coprime with $5$ and you get the full ring. If $P(X)$ is quadratic not in the ideal, you can again try a "division algorithm" style argument...
This is admittedly uglier than pritam's wonderful argument, but it's good to see for yourself how you can computationally produce the unit $1 \in R$ by adding any new polynomial.
