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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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@Thomas my question is more general. –  user42912 Oct 23 '12 at 12:50

3 Answers 3

up vote 4 down vote accepted

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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4  
Beware: this hint glosses over a crucial point. Namely, this property holds true precisely because $\rm\:ab = 0\:\Rightarrow\: a=0\:$ or $\rm\:b = 0\:$ in a field (i.e. a field is an integral domain). Otherwise a polynomial can have more roots than its degree, e.g. $\rm\:x^2\!-1 = 0\:$ has $4$ roots $\rm\,x = \pm1,\pm3\in\Bbb Z/8 = $ integers mod $8.\ \ $ –  Bill Dubuque Oct 23 '12 at 2:11
    
@BillDubuque thank you both, it helps a lot –  user42912 Oct 23 '12 at 15:33

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 $\rm\ n > deg\ f\:.$

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