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Here is a proof I came up with in the exam I just took. But I suspect there may be some issues since I think it seems too simple.

Proof
Let $p_n(x)$ denote a complex polynomial of order $n$ (coefficients from $\mathbb C$), and suppose $p_n(x)$ has $s$ distinct roots ($s>n$). And let $$q_s(x)=\prod_{i=1}^{s}(x-x_i)$$ where $x_i$ ($i=1,2,\cdots,s$) is one root for $p_n(x)$. Therefore, $q_s(x)$ must divide $p_n(x)$ (is it sufficient to say this?), i.e., $$q_s(x)\mid p_n(x)$$ which requires $$\deg(q_s(x))\le\deg(p_n(x))$$ or $$s\le n$$ which contradicts the hypothesis. So the original statement is right.

Since this question appeared at the end of my exam, I think it wouldn't be this easy to do. Maybe there are indeed some issues in my proof? Can you point it out, or verify that my proof is just ok? Thanks in advance.

Arpit Kansal
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Vim
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    Do you mean distinct roots? – Bill Dubuque Apr 10 '15 at 15:10
  • @BillDubuque oh yes.. apologies for my poor english. – Vim Apr 10 '15 at 15:11
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    You must specify the coefficient ring, e.g. $,x^2-1,$ has $4$ roots $,x\equiv \pm1,\pm3$ over $,\Bbb Z_8 = $ integers mod $8$. – Bill Dubuque Apr 10 '15 at 15:16
  • @BillDubuque the coefficients are picked from $\mathbb C$, I'll add it to my post. – Vim Apr 10 '15 at 15:18
  • Yes, then a simple induction using the Factor Theorem can prove what you claim. Take note where the proof cancels differences of distinct roots $,x_i - x_j$ in order for the induction to work. That's why the above case fails, because $,\Bbb Z_8$ has nonzero elements that are not cancellable, i.e. it is not an integral domain, e.g. $,2\cdot 4 = 0\ \ $ Post a proof as an answer and we can give you feedback on it. – Bill Dubuque Apr 10 '15 at 15:21
  • @BillDubuque ... I don't quite understand how things in that $\mathbb Z_8$ work. It's not even a field after all. – Vim Apr 10 '15 at 15:23
  • That's ok, in your case you only need to understand how they work in the field $,\Bbb C,,$ where $, a\neq 0,\ ab = 0,\Rightarrow, b = 0,,$ since you can multiply by $,a^{-1},$ to cancel $,a.\ $ Do you see how to do the inductive proof that $\ q\mid p ?\ \ $ – Bill Dubuque Apr 10 '15 at 15:25
  • @BillDubuque thank u ! Then I don't have to worry over that and can put my mind at rest:) – Vim Apr 10 '15 at 15:27
  • But you cannot simply claim that $,q\mid p.,$ That requires proof. As the example I gave shows, it can fail in other rings. – Bill Dubuque Apr 10 '15 at 15:28
  • @BillDubuque well I deemed it to be just too trivial to prove.. And to be honest, I don't know how ( for the exam I would have done it if I could) – Vim Apr 10 '15 at 15:30
  • The argument is either incomplete or incorrect without a proof of that claim. See this answer for one way to do the proof. – Bill Dubuque Apr 10 '15 at 15:33
  • @BillDubuque This answer is for $\mathbb Z_p$? – Vim Apr 10 '15 at 15:35
  • The question is about the field $,\Bbb Z_p,$ but the answer works in any field, or any integral domain, i.e. any ring where $, ab = 0,\Rightarrow, a=0,$ or $,b = 0.,$ In fact a ring $R$ is an integral domain $\iff$ this property holds true, i.e. nonzero polynomials $f$ with coefficients in $R$ have no more than $,\deg f,$ roots. – Bill Dubuque Apr 10 '15 at 15:38

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