Consider an infinite field $K$. For $f, g \in K[t]$, show that $f \neq g$ implies $f_K \neq g_K$, where $f_K, g_K: K \rightarrow K$ denote the usual polynomial functions.
My attempt: By Euclidean Algorithm, there exist unique $q, r \in K[t]$ such that $$f=qg+r$$ where we have assumed $deg(f) \geq deg(g)$ and $deg(r)<deg(g)$. Then for any $x \in K$, $$f_K(x)=q_K(x)g_K(x)+r_K(x)$$ and now I am stuck. I don't see the significance of 'infinite' in the proof. I also think of using the roots of $f$ and $g$, but the existence of roots is not guaranteed. Any help will be appreciated.
As Derek Holt hinted, you should consider the polynomial $h=f-g$. By hypothesis $h\ne 0$. Then consider the associated polynomial function of $h$, denoted here by $h_K$. Suppose the contrary that $h_K=0$, that is, $h_K(a)=0$ for all $a\in K$. This means that all elements of $K$ are roots of $h$. But over a field a polynomial can't have more (distinct) roots than its degree, a contradiction. (Here is where you use that $K$ is infinite.)