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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.

user26857
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am_11235...
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  • Let $n > p_i$ for all given $p_i$ and $f=x^n$, then there is no non-zero polynomial $g$ such that the coefficient for $x^{p_i}\neq 0$ in $fg$ and $fg$ can't be $0$. Maybe there's some additional constraint on $f$ or you mean to say "given $f$ you can find primes ${p_i}$ and polynomial $g$ such that ...". – Derek Elkins left SE Apr 13 '17 at 16:10
  • No, there are no additional constraints on $f$. And all the primes are previously given, we need not to find them. – am_11235... Apr 13 '17 at 18:02
  • Okay, the issue with my putative counter-example is that if you are given $n+1$ distinct primes (in order, for convenience) then $p_{n+1} > n$. You can write $f = \sum_{i=0}^n a_i x^i$ and $g = \sum_{i=0}^{p_{n+1}-n} b_i x^i$, multiply them, and then you have a system of linear equations for the $c_i$s. The goal would then be to show that that system of equations has at least one solution. That is to say, the resulting matrix doesn't send every vector to $0$. – Derek Elkins left SE Apr 13 '17 at 18:37
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    To be clear, you get a matrix corresponding to multiplying a polynomial of degree $p_{n+1}-n$ by $f$. You split that into two matrices by extracting the rows corresponding to terms with powers in ${p_i}$ into one matrix $M$ and the remainder into another $N$. You then find the kernel of $N$, call the matrix corresponding to the inclusion $K$, then you need to show that $MK$ doesn't send every input vector to $0$. I have no idea if this is a tractable approach. It's just the direct approach. – Derek Elkins left SE Apr 13 '17 at 18:56
  • OK. I'll try this and let you know if that works. – am_11235... Apr 13 '17 at 19:17

1 Answers1

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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

user26857
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Suman Kundu
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