Quotient map in polynomial ring I have come across the terminology $\mathbb{Z}[X] \stackrel{\mathrm{quot.}}{\to} \mathbb{Z}$ where $\mathbb{Z}[X]$ is the ring of integer polynomials in 
$\mathbb{Z}$
I am wondering what this quotient map is. From wikipedia:
"..given a (not necessarily commutative) ring $A$ containing $K$ and an element $a$ of $A$ that commutes with all elements of $K$, there is a unique ring homomorphism from the polynomial ring $K[X] \to A$ that maps $X$ to $a$:
$\phi: K[X]\to A, \quad \phi(X)=a.$ 
Edit: So perhaps there is no 'cannoical' map for the question? For what it is worth, I am trying to calculate homology of $0\to R \stackrel{x}{\to} R \stackrel{\mathrm{quot.}}{\to} \mathbb{Z} \to 0$ where $R = \mathbb{Z}[x]$
 A: The map isn't unique, so I don't know what this notation means without further context. A homomorphism $\mathbb{Z}[x] \to \mathbb{Z}$ is uniquely determined by where $x$ is sent, and $x$ can be chosen freely. In other words, evaluation $p \mapsto p(a)$ is a homomorphism for any $a \in \mathbb{Z}$ and every homomorphism has this form. 
A: Your edit makes clear that the quotient map you want is $\mathbb{Z}[x] \rightarrow \mathbb{Z}[x]/(x)$; the latter is indeed isomorphic to $\mathbb{Z}$ in a natural way: 
each equivalence class has a unique representative which is a polynomial of degree $0$.
In general, to give yourself a quotient map you need to give yourself an ideal $I$ to 
mod out by.  When -- as here -- you have an exact sequence -- looking at the image of 
the previous term will tell you what the kernel of the quotient map is.
A: The map $\mathbb Z[x] \to \mathbb Z$ sending a polynomial $p$ to $p(0)$ is a surjective ring homorphism.  The kernel of this is the ideal generated by $x$ so we have the projection $\mathbb Z[x] \to \mathbb Z[x]/(x) \simeq \mathbb Z$.
