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It's a really basic question,in these days, I've been thinking why a polynomial $p(x)\in F[x]$ ($F$ a field) with degree $n$ can have at most n roots. It seems easy to prove, but I've been trying to prove this since yesterday, maybe I forgot some important details necessary to prove, I don't know. I'm solving some questions about Field Theory and I noticed that almost every question I should use this theorem, I need help.

Thanks

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    Here is a similar question in the case $F=\mathbb C$: http://math.stackexchange.com/questions/25822/how-to-prove-that-a-polynomial-of-degree-n-has-at-most-n-roots2015-11-01

3 Answers 3

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It follows immediately from the following lemma

Lemma $P(a)=0 \Leftrightarrow x-a | P(x)$.

Now if $P(X)$ has at least $n+1$ roots, it follows that $P(X)$ is divisible by $(x-a_1)....(x-a_{n+1})$..

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    @BillDubuque thank you both, it helps a lot2012-10-23
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Use the factor theorem and induction. This obviously only works if you're working in an integral domain.

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If you know a little ring theory then you can reformulate the mentioned inductive proofs using the factor theorem in the following more conceptual form. Suppose that $\rm\,R\,$ is an integral domain.

Note $\rm\ a_i\ne a_j\ \Rightarrow\ x-a_i\ $ are nonassociate primes in $\rm\,R[x],\:$ since $\rm\: R[x]/(x-a) \cong R\:$ is a domain.

Therefore $\rm\ \ x-a_1\ |\ f(x),\ \ldots\:,\: x-a_n |\ f(x)\ \ \Rightarrow\ \ (x-a_1)\cdots (x-a_n)\ |\ f(x) $

since LCM = product for nonassociate primes. But this is contra degree if $\rm\ n > deg\ f\:.$