Let f∈Q[x] be a polynomial of degree n>0. Let p1,…,pn+1 be distinct prime numbers. Show that there exists a non-zero polynomial g∈Q[x] such that fg=∑n+1i=1cixpi with ci∈Q.
I tried to solve the problem using the division algorithm over 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 Q[x] with degree at most n.
Now consider the remainders ri(x)=xpi−f(x)hi(x),∀i=1,2,...,n+1 (obtained from division algorithm). There are n+1 ri's in the n dimensional vector space V. So there is a non-zero linear combination of ri(x) equal to 0, i.e. there exist c1,c2,...,cn+1∈Q such that ∑n+11ciri(x)=0.
Now set g(x)=∑n+11cihi(x).
So f(x)g(x)=∑n+11cif(x)hi(x)=∑n+11cixpi−∑n+11ciri(x)=∑n+11cixpi .
QED
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