Residue of $x$ in polynomial ring $\mathbb{Z}[x]/(f)$ Someone asked this question here, but I don't understand the answer. When we say that $\alpha$ is the residue of $x$ in $\mathbb{Z}[x]/(f)$, and $f=x^4+x^3+x^2+x$, wouldn't $\alpha$ just be $x$? Because if we divide with the remainder, we would get $x=0{f}+x$?
Thank you!
 A: To expand Randall's comment into an answer:
The elements of a quotient ring $R/I$ (where $I$ is a two-sided ideal of the ring $R$) are the cosets of $I$, i.e. the sets of the form $a + I := \{a + i : i \in I\}$. In particular, the elements of $R/I$ are subsets of $R$!
When we talk about the residue of an element $x \in R$ in the quotient $R/I$, we mean the coset $x + I$.
So, the residue of $x$ in $\mathbb{Z}[x]/(f)$ is (by definition) the coset
$$x + (f) = \{x + i : i \in (f)\} = \{x + gf : g \in \mathbb{Z}[x]\}.$$
Be warned: sometimes it gets annoying to keep writing "$+~I$" everywhere, so people will sometimes just use "$x$" to refer to the residue of $x$ in some quotient. E.g. someone might say "$x^2 = 0$ in $R/I$" to mean "$x^2 + I = 0 + I$". It'll be up to you to keep track of the context and interpret elements as cosets when necessary! This can be confusing at first but gets easy once you've had enough experience :)
A: Still to add on the previous answer (by diracdeltafunk):
Another approach, at least for an irreducible polynomial $f$, is to consider a hypothetical element $\alpha$ which satisfies $f(\alpha)=0$, and extend the original ring $R$ by $\alpha$ (and whatever formal elements that makes a ring), analogously as one extends $\Bbb R$ by an element $i$ which satisfies $i^2+1=0$ to obtain the field of complex numbers $\Bbb C$.
Actually the quotient ring construction (as the set of cosets of the ideal $I$ generated by $f$) shows that the above hypothetical construction is feasible, namely we can set $\alpha:=x+I$.
