Why don't we discuss (ir)reducibility over $\Bbb Z[x]$? The books (Fraliegh's and my teacher's handout) I'm reading say that we don't discuss irreducibility of a polynomial over $R[x]$ when $R$ is not a field. That is to say, we don't discuss whether a polynomial $f(x)$ in $\Bbb Z[x]$ is irreducible over $\Bbb Z[x]$. However, the book didn't say why. So why? Does it make some harm?
Ref:

 A: No -- you can certainly have an irreducible polynomial in $\mathbb{Z}[x]$. In fact, the irreducible polynomials in $\mathbb{Z}[x]$ over $\mathbb{Z}[x]$ are precisely those polynomials in $\mathbb{Z}[x]$ which are irreducible over $\mathbb{Q}[x]$, with one complication -- sometimes you can reduce a polynomial by a non-unit in $\mathbb{Z}$ (which of course is never an issue when dealing with a field). This complication renders the definition you posted insufficient.
E.g., consider the polynomial $2x^2 +2$. It is irreducible over $\mathbb{Q}[x]$ according to the definition you posted. However, over $\mathbb{Z}[x]$, we can factor it as $(2)(x^2+1)$. The factor $2 \notin \mathbb{Z}[x]\setminus\mathbb{Z}$ (so this would not be a reduction according to the definition you posted),but it is a non-unit in $\mathbb{Z}[x]$ (because it is a non-unit in $\mathbb{Z}$).
So: we can talk about irreducible polynomials over non-fields, but we might want to be more careful about the definition than the one you posted.
A: The key word you're missing is "here":

we do not talk about irreducible polynomials in $\Bbb Z[x]$ here.

Fields only, for now. Later you will revisit the topic in greater generality.  This is a common pattern in all pedagogy.
Thinking on it a little more, I think the whole point of the sentence is to suggest that if you were wondering why fields and not rings more generally, you should not worry, because rings are coming eventually. 
A: How divisibility works is extremely simple and well behaved in rings like $\mathbb{Z}$ or $F[x]$, which are in some sense one dimensional. Moving onto higher "dimension" by considering rings like $\mathbb{Z}[x]$ or $F[x,y]$ adds extra complexity to the theory.
The point of the remark is to emphasize that this text is only considering the one-dimensional case. The author is trying to help you avoid making the mistake of thinking that the theorems and methods you are about to learn work as advertised in rings like $\mathbb{Z}[x]$.
A: Eisenstein Criterion can be applied to $\mathbb{Z}[X]$: A polynomial $P(X)=a_0+a_1X+...a_nX^n$ which integers coefficients is irreducible over $\mathbb{Q}[X]$ if there exists a prime number $p$ such that $p$ divides $a_i, i<n$, $p$ does not divides $a_n$ and $p^2$ does not divide $a_0$. If moreover $gcd(a_0,...,a_n)=1$ then $P$ is irreducible over $\mathbb{Z}[X]$.
