This question is related to this: Congruence modulo p(x) and this Congruence in F[x], proof involving infinity
I'll just restate the theorem in question before my question:
Let $F$ be a field and $p(x)$ a polynomial of degree $n$ in $F[x]$, and consider congruence modulo $p(x)$.
1) If $f(x) \in F[x]$ and $r(x)$ is the remainder when $f(x)$ is divided by $p(x)$, then $[f(x)]=[r(x)]$.
2) Let $S$ be the set consisting of the zero polynomial and all the polynomials of degree less than $n$ in $F[x]$. Then every congruence class modulo $p(x)$ is the class of some polynomial in $S$, and the congruence classes of different polynomials in $S$ are distinct.
Here is the proof for 2):
Since $r(x)=0_{F}$ or $\text{ deg }r(x) < n$, we see that $r(x) \in S$. Hence, every congruence class is equal to the congruence class of a polynomial in $S$. Two different polynomials in $S$ cannot be congruent modulo $p(x)$ because their difference has degree less than $n$, and hence is not divisible by $p(x)$. Therefore, different polynomials in $S$ must be in distinct congruence classes since $f(x) \equiv g(x) (\text{mod} p(x))$ if and only if $[f(x)]=[g(x)].$
I am trying to understand exactly what the implications of 2) is. In the specific example of Congruence in F[x], proof involving infinity I try to show that there are infinitely many distinct congruence classes modulo $x^2 -2$ in $\mathbb{Q}[x]$. Does it follow from 2) that there is a distinct congruence class for every polynomial in $S$? Or are there polynomials in $S$ which does not belong to some congruence class somehow?
I understand why every congruence class is equal to the congruence class of a polynomial in $S$, but do all congruence classes exhaust all the polynomials in $S$? I hope the question makes sense.
I am sorry if this is a stupid question, but the more I think about this, the more I seem to be confusing myself. Any response is appreciated. Thanks in advance.
Edit: As an example to be more specific, could I argue using 2) that since every congruence class modulo $x^2-2$ is on the form $ax + b$ where $a,b \in \mathbb{Q}$ then since there are infinitely many combinations of $a$ and $b$ there are infinitely many congruence classes? In my current understanding this would only work if it is the case that by 2) all polynomials in $S$ correspond to a different congruence class.