When you partition $Z$ into cosets of Equivalence Classes using say $5Z$, then each element of the group $Z/5Z$ is an equivalence class which is defined by the relation $p \equiv q \bmod 5$ where $p$ & $q$ are from that equivalence class.
If I try something similar with a Polynomial Ring, say $Z[x]$.
For e.g. I construct $Z[x]/\langle x \rangle$ – now this again has elements which are actually an equivalence class of polynomials. How exactly do I write down the equivalence relation for each element in each coset of $Z[x]/\langle x \rangle$ the way we write $p \equiv q \bmod 5$ for $Z/5Z$?
If I want to generate all elements of the equivalence class $[\bar 2]$, the $p \equiv q \bmod 5$ relation tells me how to do it. For e.g. I can do that as {2, 2+5, 2+2*5, 2+ 3*5, 2+4*5 ….]. How would I generate all the elements of one equivalence class in $Z[x]/\langle x \rangle$ or $Z[x]/\langle x^2 + 1 \rangle$?
Best Answer
By definition, the elements of the quotient ring $R/I$ are the equivalence classes of the following equivalence relation $\sim$ on $R$: $$ \forall a,b\in R\,\ \ \ \ \ \ a\sim b\iff a-b\in I.$$ For $a\sim b$ we also write $a\equiv b\ \ \text{mod}\ I$.
For $R/I=\mathbb{Z}/5\mathbb{Z}$ we get the familiar relation \begin{align*} a\sim b &\iff a-b\in 5\mathbb{Z}\iff a-b=5k\ \text{ for some } k\in\mathbb{Z} \\ &\iff 5\mid a-b \iff a\equiv b\mod 5. \end{align*} For $R/I=\mathbb{Z}[x]/\langle x\rangle$ we get \begin{align*}f\sim g &\iff f-g\in \langle x\rangle \\ &\iff f-g=x\cdot h\ \ \text{ for some } h\in\mathbb{Z}[X] \\ &\iff x\mid f-g \ \ \text{ in }\,\mathbb{Z}[x]. \end{align*} So the analogue notation $``f\equiv g\mod \langle x\rangle"$ for $f\sim g$ makes sense.