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It is a fairly well known fact that sparse polynomials $f(x)$ cannot have large order zeros other than at $x=0$. If $f(x)=a_1x^{r_1}+\cdots+a_kx^{r_k}$ then at $c \neq 0$, $f$ has a zero of order at most $k$. The proof is by induction on $k$. First if $k=1$ then the only possible root of $f$ is $0$. Now for $k+1$, by writing $f(x)=x^jg(x)$ with $g(0)\neq 0$, we know $g'$ has a zero of order at most $k$ at $c$ since it has at most $k$ terms in its expansion. This means $g$ has a zero of order at most $k+1$.

This argument runs into trouble mod $p$ once the degree of $f$ exceeds $p$, since for example $(x-c)^p$ has only two terms once expanded, but has a zero of order $p$ at $c$. My question is, can this be reconciled with some additional hypotheses? Can one deduce anything about the nature of zeros of sparse polynomials?

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    $\begingroup$ Base $p$ expansion. If $f$ over a field of characteristic $p$ has a root $\alpha \neq 0$ of exact order $m = \sum_i c_i p^i$, $0 \leq c_i < p$, then $f$ has at least $\prod_i (c_i + 1)$ non-zero coefficients. This is sharp for every $m$. To prove it, reduce first to the case $m < p$, to which you can apply the same argument as in characteristic zero. $\endgroup$
    – Vesselin Dimitrov
    Sep 29, 2015 at 4:46

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