Is algebraic multiplicity of a root of a polynomial always smaller than the characteristic of a field? I'm trying to prove a criteria for algebraic multiplicity of a root of a polynomial.
Let $F$ be a field and $f\in F[X]\setminus\{0\}$.
Let $r$ be the algebraic multiplicity of $c$ as a root of $f$.
Then, there exists $g\in F[X]$ such that $f=(X-c)^r g$ and $g(c)\neq 0$.
Then, $(D^rf)(c)=g(c)r!$. However, does this gurantee thay $(D^rf)(c)$ is nonzero even though the characteristic of $F$ is not zero?
(I'm not really familiar with abstract algebra language, so i hope you explain this in relatively simple words.. Thank you in advance)
 A: I'm not sure what you are asking. But it seems that you want to know under what conditions the following holds: If $c$ is a root of a polynomial $f$ with (algebraic) multiplicity $r$, then $(D^r f)(c) \neq 0$. Well, using your formula $(D^r f)(c) = g(c) r!$ and $g(c) \in F^*$, this holds if and only if $r! \in F^*$, which holds if and only if the characteristic of $F$ is either zero or a prime number larger than $r$.
A: No, if $r\ge p$ in your field of characteristic $p$, then $r!=0$ and the right side is zero regardless of $f(c)$. For example, if $p = 2$, then every factorial is even after the first, so all higher derivatives vanish.
A: I'm not sure that this answers your question either. Consider characteristic $p$.
The polynomial
$$
(x-1)^{p^2}=x^{p^2}-1
$$
has $x=1$ as a zero of multiplicity $p^2$, which is certainly larger than the characteristic. Note that already the first (formal) derivative of this polynomial vanishes. Thus this also shows that vanishing of derivatives is not as closely related to the multiplicity of the root as it is in characteristic zero.
