Let $f\in \mathbb{Q}[x]$ be a polynomial of degree $n>0$. Let $p_1, \dots , p_{n+1}$ be distinct prime numbers. Show that there exists a non-zero polynomial $g\in \mathbb{Q}[x]$ such that $fg=\sum_{i=1}^{n+1} c_ix^{p_i}$ with $c_i\in \mathbb{Q}$.
I tried to solve the problem using the division algorithm over $\mathbb{Q}[x]$ but could not get any satisfactory result. Also I don't understand how the distinct prime powers be particularly accounted in the expression of $fg$. Any help on this one will be appreciated.
Answer
Consider the vector space of all polynomials in $\mathbb{Q}[x]$ with degree at most $n$.
Now consider the remainders $r_i(x)=x^{p_i} - f(x)h_i(x), \forall i=1 ,2,...,n+1$ (obtained from division algorithm). There are $n+1$ $r_i$'s in the $n$ dimensional vector space $V$. So there is a non-zero linear combination of $r_i(x)$ equal to $0$, i.e. there exist $c_1,c_2,...,c_{n+1} \in \mathbb{Q} $ such that $\sum_{1}^{n+1} c_i r_i(x)=0$.
Now set $g(x)=\sum_{1}^{n+1} c_i h_i(x)$.
So $f(x)g(x)=\sum_{1}^{n+1} c_i f(x) h_i(x)=\sum_{1}^{n+1} c_ix^{p_i}-\sum_{1}^{n+1} c_i r_i(x)=\sum_{1}^{n+1} c_ix^{p_i}$ .
QED
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