Let $k$ be a field and let $S\subseteq \mathbb{N}$ be an infinite subset. Let $f \in k[t]$ be a polynomial. Show that there exists a polynomial $g \in k[t]$ such that $f(t)g(t) = \sum_{i \geq 0}a_it^i$, where $a_i = 0$ for every $i \notin S$.
Can we consider the quotient $\frac{k[t] }{(f)}$? Then its dimension is equal to $deg(f)=d$(say). Then the set $\{x^k\mid k\in S\}$ is linearly dependent and hence we will get some finite subset $S'\subseteq S$ such that $$ \sum_{i \in S'}\overline{a_it^i}=0.$$
I would send the polynomial $f$ of degree $n$ to its companion matrix $C_f\in M_n(k)$.
Its minimal polynomial is $f$.
What we want is some $s_j\in S,a_j\in k$ such that $\sum_{j=1}^J a_j C_f^{s_j} = 0$,
which will give that $f$ divides $\sum_{j=1}^J a_j t^{s_j}$.
To construct some $s_j,a_j$ it suffices to use that $M_n(k)$ is a $n^2$-dimensional $k$-vector space, so for any $s_1,\ldots,s_{n^2+1}\in S$ the set $C_f^{s_1},\ldots,C_f^{s_{n^2+1}}$ is $k$-linearly dependent, ie. there are some $a_j\in k$ not all zero such that $\sum_{j=1}^{n^2+1} a_j C_f^{s_j} = 0$.