Is $ \langle x,5 \rangle $ a maximal ideal of $ \mathbb{Z}[x] $? Here, $ \langle x,5 \rangle $ is the ideal generated by $ x $ and $ 5 $ in $ \mathbb{Z}[x] $, which is the polynomial ring over $ \mathbb{Z} $. How should I approach this question?
 A: $<5>$ is a prime ideal of $\mathbb{Z}$ and therefore is a maximal ideal of $\mathbb{Z}$. (Because $\mathbb{Z}$ is a PID).
$\mathbb{Z}/<5>$ is a field. (Result well known)
Now
$$\begin{align}\mathbb{Z}[X]/<X,5>&=\mathbb{Z}[X]/<X>/<5>\end{align}$$
and $\mathbb{Z}[X]/<X>=\mathbb{Z}$ so
$$\mathbb{Z}[X]/<X,5>=\mathbb{Z}/<5>$$
A: We can the 3rd isomorphism theorem:
$$\mathbf Z[x]/(5, x)\simeq \mathbf Z[x]/5\mathbf Z[x]\color{red}/(5,x)\mathbf Z[x]/5\mathbf Z[x]\simeq \mathbf F_5[x]/x\mathbf F_5[x]\simeq \mathbf F_5. $$
A: One can in fact proceed as you mention in your comment. Using the Division Algorithm 
$$ f \,=\, x\ g + n\ \ \text{ for }\ n = f(0)\in \Bbb Z\qquad $$ 
Thus $\ (x,5,f) = (x,5,n) = (x,5)\,$ if $\ 5\mid n,\,$ else $\ \color{#c00}{(5,n)}=(1)\,\Rightarrow\,(x,\color{#c00}{5,n}) = (1)$
Since $\,(x,5,f) = (x,5)\,$ or $\,(1)\,$ it follows that $\,(x,5)\,$ is maximal in $\,\Bbb Z[x],\,$ being $\neq (1)\ $ (else $\, 1\in (x,5)\,\Rightarrow\, 1 = xg + 5h\,\Rightarrow\ 1 = 0g(0)+5h(0)\,\Rightarrow\, 5\mid 1\,$ in $\,\Bbb Z).$
Remark $\ $ It is instructive to compare this element-wise approach with the structural approach employing isomorphism theorems (e.g. in Bernard's answer), since understanding how the general results work in concrete cases helps to build better intuition for the general results.
