# 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 $R = \mathbb{Z}[x]$

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no it isn't, the wikipedia page gives a formula for the map. –  quanta Mar 14 '11 at 23:56
@quanta - thanks I will update the question –  Juan S Mar 14 '11 at 23:58
The map that takes every element to 1 is not a ring homomorphism. –  lhf Mar 14 '11 at 23:59

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.

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Thanks, I am accepting this as the answer (the others were good as well!). How do you know this is an exact sequence? (Of course, once I know that the map is $\mathbb{Z}[x] \to \mathbb{Z}[x]/(x)$ then I can see that it is exact) –  Juan S Mar 15 '11 at 10:20

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.

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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 $\mathbb Z[x] \to \mathbb Z[x]/(x) \simeq \mathbb Z$.

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