How to prove that $\mathbb{Z}[X]/g\mathbb{Z}[X]$ is not a field? 
How to prove that $\mathbb{Z}[X]/g\mathbb{Z}[X]$ is not a field, where $g$ is a non constant polynomial in $\mathbb{Z}[X]$?

(The key is to show that $f : \mathbb{Z}[X]/g\mathbb{Z}[X] \to \mathbb{Z}/p\mathbb{Z}$ is not injective.)
edit 1:
Now I'm trying to find the characteristic of the given ring.
Consider the map: $g: \mathbb{Z} \to \mathbb{Z}[X]/g\mathbb{Z}[X]$, where $m \mapsto 1 + 1 + \cdots + 1 $ (m times). Since the $\ker f$ is an ideal of $\mathbb{Z}$, it has to be of the form $n\mathbb{Z}$, where $n$ is the characteristic. So I guess the characteristic is $0$...no?
edit 2: Thank you for all the replies below! I looked through the answer but it's a little advanced for me... I figured it out eventually using some more basic ideas and I'll put it here when I get some time...Thanks again!
edit 3:
My solution:
Step 1: Prove there is a $a \in \mathbb{Z}$ such that $g(a) \neq 0, \pm 1$ and let $p$ be a prime number that divides $g(a)$.
Suppose $g(X) = \sum_{i = 1}^na_iX^i$, then $g(X) = 0$ has at most $n$ integer solutions, $p(X) = g(X) -1 = 0$ has at most $n$ integer solutions, and so is $q(X) = g(X)+1 = 0$. There are at most $3n$ integers that satisfy $g(a) = 0,\pm1$, but $\mathbb{Z}$ is infinite, so there has to be a $a \in \mathbb{Z}$ such that $g(a) \neq 0, \pm 1$.
Step 2: Prove that there is a unique well-defined surjective morphism of rings $f: \mathbb{Z}[X] \to \mathbb{Z}/p\mathbb{Z}$ sending $X$ to $a$ mod $p$ and that it yields a homomorphism of rings $\phi: \mathbb{Z}[X]/g\mathbb{Z}[X] \to \mathbb{Z}/p\mathbb{Z}$.
Observe that $g\mathbb{Z}[X]$ is in the kernel, so the map factor through $\mathbb{Z}[X]/g\mathbb{Z}[X]$.
Step 3: Show that the resulting map is not injective.
Note that the map $\phi$ is injective if and only if $\ker f = g\mathbb{Z}[X]$, but actually $g\mathbb{Z}[X] \subset \ker f$. Alternatively, according to my prof, if the map was injective, $\mathbb{Z}[X]/g\mathbb{Z}[X]$ would identify with a subring of $\mathbb{Z}/p\mathbb{Z}$ so it would have characteristic $p$, contradiction!
Step 4: We use that if $g: K \to A$ where $K$ is a field, then $g$ is either injective or the zero map.
Since the above map is not a zero map and we proved that it's not injective, $\mathbb{Z}[X]/g\mathbb{Z}[X]$ can't be a field.
 A: My answer expands on JonHales comment. Suppose your polynomial is $g(x)=a_nx^n + \cdots + a_0$, with $a_i\in\Bbb{Z}$, $a_n\ne 0$, and $n > 0$. Choose a prime $p$ with $p\nmid a_n$. Then mod $p$, $g(x)$ still has degree $n$.
Thus $\Bbb{Z}[x]/(g,p)\simeq \Bbb{F}_p[x]/(g)\ne 0$, so $(g,p)$ is a proper ideal of $\Bbb{Z}[x]$. Moreover, it properly contains $(g)$, since $(g)\cap \Bbb{Z}=(0)$. Thus $(g)$ is not maximal, and $\Bbb{Z}[x]/(g)$ is not a field.
Edit 
In response to the OP's edit, saying that they found this solution a bit advanced, I thought I'd translate the solution into more basic ideas for future readers.
Again, if our polynomial is $g(x)=a_nx^n+\cdots + a_0$, then if $p$ is a prime integer with $p\nmid a_n$, $p$ is nonzero, but not a unit in $\Bbb{Z}[x]/(g)$. Therefore $\Bbb{Z}[x]/(g)$ is not a field.
Proof.
By contradiction. Assume $p \cdot c(x) \equiv 1 \pmod{g(x)}$ for some polynomial $c(x)$. Then $pc(x)-1 = g(x)h(x)$ for some polynomial $h(x)$, so $pc(x) = g(x)h(x)-1$. Write $h(x)=pq(x)+r(x)$, where the coefficients of $r$ are either between $0$ and $p-1$ (inclusive).
Then $$pc(x)=g(x)(pq(x)+r(x))-1=pg(x)q(x) + g(x)r(x)-1,$$
and 
$$p(c(x)-g(x)q(x))=g(x)r(x)-1.$$
Now if $r(x)=0$, we have that the left hand side is divisible by $p$, and the right hand side is $-1$, so this is impossible. 
On the other hand, if $r(x)\ne 0$, $g(x)r(x)$ has positive degree, and it's leading term is $a_nb_m$, where $b_m$ is the leading term of $r(x)$. However $p$ must divide $g(x)r(x)-1$, so it must divide the leading term, so $p\mid a_nb_m$. However $p\nmid a_n$, and $1\le b_m\le p-1$, so $p\nmid b_m$ either. Thus $p\nmid a_nb_m$. Contradiction.
