Prove that the ideal $(x-2)$ is not maximal in the ring $\mathbb{Z}[x]$ 
In the ring $\mathbb{Z}[x]$, I want to prove that the principal ideal $I=(x-2)$ is not maximal.

Surely I just need to find an ideal $J$ that contains $I$ plus some other elements of $\mathbb{Z}[x]$. My instinct is to just say $J = (x-2) + (x)$ (not confident this is correct), or any other principal ideal that is not $(x-2)$ for that matter. However, I'm not sure how I would prove $I$ is contained in $J$ without them being equal even if I am correct.
 A: Hint:
Notice that $f(x)(x-2)$ cannot be a nonzero constant polynomial for any $f(x) \in \mathbb{Z}[x]$.  Given this, can you think of an ideal that properly contains $\langle x-2 \rangle$ that is also a proper subset of $\mathbb{Z}[x]$?  It'll be easiest to construct this new ideal in terms of a set of generators (which will include $x-2$ of course), and note that there may be more than one generator as $\mathbb{Z}[x]$ is not a principal ideal domain.

In addition to explicitly constructing an ideal that properly contains $\langle x-2 \rangle$, there is another approach.  One can show that an ideal $I \subset R$ is maximal $\iff R/I$ is a field.  I claim that $\mathbb{Z}[x] / \langle x-2 \rangle \cong \mathbb{Z}$ (not a field!).  To prove this, find a surjective homomorphism $\phi:\mathbb{Z}[x] \rightarrow \mathbb{Z}$ with the appropriate kernel.  The result would follow from the isomorphism theorem.
A: The fact that $\mathbb{Z}$ is not a field is key here. For example $(x-2,2)$ is a proper ideal containing yours.
A: Applying the isomorphism theorem to the evaluation map $P(x)\mapsto P(2)\in\mathbb{Z}$ yields
$$
\mathbb{Z}[x]\longrightarrow\mathbb{Z}\simeq\frac{\mathbb{Z}[x]}{(x-2)}.
$$
This shows that the ideal $(x-2)$ is prime since the quotient is an integral domain, but not maximal since the quotient is not a field.
The maximal ideals containing $(x-2)$ are in bijection with the maximal ideal of $\mathbb{Z}$, thus are all ideals of the form $(x-2,p)$ with $p$ a prime number.
