Why is $x^n+c_{n-1}x^{n-1}+\dots+c_0$ where $c_i\in\mathbb{Z}$ general? I have been studying some problems on the (ir)reducibilty and Galois groups of certain polynomials in $\mathbb{Z}[x]$. However, the papers I have examined seem to use 
$$x^n+c_{n-1}x^{n-1}+\dots+c_0,\hspace{10mm}c_i\in\Bbb{Z}$$
as the fully general form of a polynomial, instead of
$$c_nx^n+c_{n-1}x^{n-1}+\dots+c_0,\hspace{10mm}c_i\in\mathbb{Z}$$

What allows us to drop the leading coefficient?

If we were working in $\mathbb{Q}[x]$ this would make complete sense because normalization doesn't affect the properties of reducibility or the Galois group, but $\mathbb{Z}$ isn't a field, so it doesn't quite make sense in $\mathbb{Z}[x]$.
For reference, here a couple of the papers I've seen it in (though I don't think they will help answer this question - the convention is used without much preamble).
Probabilistic Galois Theory for Quartic Polynomials, Rainer Dietmann.
Hibert's Irreducibility Theorem and the Larger Sieve, David Zywina.  The relevant statement is at the top of page 3.
My suspicion is that it has to do with Hilbert's irreducibility theorem, but I don't fully understand the theorem myself yet.
 A: I think the leading coefficient can be dropped because in that case, division with remainder works. Division with remainder says the following:
Let $R$ be a ring and $f,g \in R[X]$. Assume that $g \neq 0$ and that the leading coefficient of $g$ is a unit of $R$. Then there exist unique $q,r \in R[X]$ such that
\begin{equation}
f = qg + r, \hspace{1mm} \text{ and } r = 0 \hspace{1mm} \text{ or } \hspace{1mm} \text{deg}(r) < \text{deg}(g).
\end{equation}
The condition that the leading coefficient of $g$ is a unit in $R$ cannot be dropped here, as the next example shows.
Let $R = \mathbb{Z}$, $f = X^{2}$ and let $g = 2X$. Suppose we would like to perform division of $f$ by $g$. If the remainder would be zero, then there should be a $q \in \mathbb{Z}[X]$ such that 
\begin{equation}
X^{2} = q\cdot 2X,
\end{equation}
which is nonsense. Suppose the remainder $r$ is nonzero. Then, because $\text{deg}(r) < \text{deg}(2X) = 1$, we have that $r \in \mathbb{Z} \setminus \{0\}$. Furthermore, there should exist a $q \in \mathbb{Z}[X]$ such that
\begin{equation}
X^2 = q\cdot 2X + r.
\end{equation}
However, by comparing coefficients on either side we see that there cannot exist such $r$ and $q$. 
