Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
@Thomas my question is more general. – user42912 Oct 23 '12 at 12:50
Here is a similar question in the case $F=\mathbb C$:… – Martin Sleziak Nov 1 '15 at 16:21
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 $(x-a_1)....(x-a_{n+1})$..

share|cite|improve this answer
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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.