Why are the elements of $\frac{\mathbb{Z}_5[x]}{\langle x^2\rangle}$ of the form $ ax +b $. why not $(ax+b)+\langle x^2\rangle$?

Question

$$\dfrac{\mathbb{Z}_5[x]}{\langle x^2\rangle}$$

My question : Why are the elements of $\frac{\mathbb{Z}_5[x]}{\langle x^2\rangle}$ of the form $ ax +b $. why not $(ax+b)+\langle x^2\rangle?$

My thinking : I know that $R/A= \{r +A |r \in R\}$ where $R$ is a ring and $A$ is a subring of $R$

I think elements of $\frac{\mathbb{Z}_5[x]}{\langle x^2\rangle}$ are of the form $(ax+b)+\langle x^2\rangle$ where $a ,b \in \mathbb{Z}_5$

My confusion : Where $x^2$ has gone ? What is the reason for omitting $\langle x^2\rangle$ from the $(ax+b)+\langle x^2\rangle$?

Answer 1

When working with quotients, we often choose convenient representatives for the equivalence classes. For instance, we might look at $\mathbb{Z}/5\mathbb{Z}$ and think of it as $\{0,1,2,3,4\}$ with $+/\cdot$ modulo 5 instead of as the five cosets: $$ \{\{\ldots, 0, 5, 10, \ldots\},\{\ldots, 1, 6, 11, \ldots\},\{\ldots, 2, 7, 12, \ldots\},\{\ldots, 3, 8, 13, \ldots\},\{\ldots, 4, 9, 14, \ldots\}\}. $$ At the very least, it's less writing and everyone knowns what you mean.

The same thing is done in your example, where the linear polynomials are a collection of convenient representatives.

The same thing is done in, e.g. geometry/topology, where one might choose representatives $\{[x:y:1],[x:1:0], [1:0:0]\}$ for the projective plane instead of considering the collection of linear subspaces.