I have just proved the first part $p\mid n$ in the following manner.
Let $f(x)=a_0+a_1 x+\cdots+a_n x^n$ where $a_n\ne0$
Now, $f'(x)=a_1+2a_2 x+\cdots+na_n x^{n-1}=0\implies ia_i=0\ \forall i=1, 2,\ldots,n$
$\implies $either $p\mid i$ or $a_i=0\ \forall i=1, 2,\ldots,n$
In particular since $a_n\ne 0$, $p\mid n$(Done)
But I can't prove the second part of this problem that is $f(x)$ has at most $n/p$ distinct roots.
My thought is- if I can prove any root of $f(x)$ has multiplicity at least $p$, then we are done. But I can't prove it. I even don't know whether my intuition is correct or not.
Can anybody solve the second part? Thanks for assistance in advance.


Under these conditions, your $f(x)$ has the form $$f(x)=a_0+a_px^p+a_{2p}x^{2p}+\cdots+a_n x^n.$$ Let $b_j$ be a $p$-th root of $a_{pj}$ in some extension field of $F$. Then $f(x)=g(x)^p$ where $$g(x)=b_0+b_1x+b_2x^2+\cdots+b_{n/p}x^{n/p}.$$ This is now defined over an extension field of $F$, not necessarily $F$ itself, but no matter, each zero of $f$ is a zero of $g$, and $g$ has $\le n/p$ zeroes.

  • $\begingroup$ Ok, but at the last line I think Every Zero of $g$ is a zero of $f$, since if $g(\alpha)=0\implies g(\alpha)^p=0\implies f(\alpha)=0$ $\endgroup$ – Biswarup Saha Oct 26 '18 at 5:38
  • $\begingroup$ Both $\implies$s are $\iff$s. @BiswarupSaha $\endgroup$ – Lord Shark the Unknown Oct 26 '18 at 6:27
  • $\begingroup$ Lord Shark the Unknown, one moment why such $b_j$ will always exist. If $F$ is finite or perfect then okay, but else? $\endgroup$ – Biswarup Saha Dec 5 '18 at 14:38
  • $\begingroup$ @BiswarupSaha "...in some extension field of $F$." $\endgroup$ – Lord Shark the Unknown Dec 5 '18 at 15:04

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