The statement to be proven is: "Show that there are infinitely many distinct congruence classes modulo $x^2 -2$ in $\mathbb{Q}[x]$." My question is if the following solution suffices. My apologies if this is unclear.
Denote the congruence class of $f(x)$ modulo $p(x)$ by $[f(x)]$ and define $[f(x)] = \{g(x) \vert g(x) \in F[x] \textrm{ and } g(x) \equiv f(x)(\textrm{mod } p(x))\}$
We can rewrite this as $[f(x)]= \{f(x) + k(x)p(x) \vert k(x) \in F[x]\}$. In this form, one can see from the division algorithm for polynomials that the congruence class of $[f(x)]$ are all the polynomials in $F[x]$ that leaves remainder $f(x)$ when divided by $p(x)$. Since, by the division algorithm for polynomials, the degree of $f(x)$ is strictly smaller than the degree of $p(x)$, we get that the different congruence classes modulo $x^2-2$ in $\mathbb{Q}[x]$ are on the form $ax+b$, where $a,b \in \mathbb{Q}$. Since there are infinitely many elements in $\mathbb{Q}$, there are infinitely many distinct polynomials on the form $ax+b$ in $\mathbb{Q}[x]$, and since two congruence classes modulo $p(x)$ are either disjoint or identical, there are infinitely many congruence classes modulo $x^2 -2$ in $\mathbb{Q}[x]$.
I feel like there is something missing in this solution, but I cannot quite put my finger on it. Any thoughts/hints on the subject would be much appreciated. Thanks in advance.
Your idea is good, but not well explained.
Any polynomial is congruent modulo $x^2-2$ to a polynomial of the form $ax+b$, by taking the remainder of its division by $x^2-2$.
Now distinct polynomials of the form $ax+b$ and $cx+d$ are not congruent modulo $x^2-2$, because their difference cannot be divisible by $x^2-2$, unless they're equal.
Thus there are infinitely many equivalence classes with respect to congruence modulo $x^2-2$, because there are infinitely many degree one polynomials.