Prove that if has solutions , there is a polynomial such that .

I feel that I need to somehow show that there is a polynomial such that and that has solutions , but I don't know start from there and finish the proof using induction.

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- November 5th 2009, 08:34 PMkeityoProof using induction
Prove that if has solutions , there is a polynomial such that .

I feel that I need to somehow show that there is a polynomial such that and that has solutions , but I don't know start from there and finish the proof using induction. - November 5th 2009, 08:46 PMBruno J.
We proceed by induction on the degree of . If has degree 1, there is nothing to show because it is already in the form . So suppose it holds for all polynomials having degree . Let be of degree , and suppose is a solution. Then . So we have

where is of degree . Now use the fact that is prime to show that any other roots of must in fact be roots of , and apply the induction hypothesis to obtain the desired factorization for .

Hope that helps!