Are formal derivatives studied in finite fields? If so, what is wrong or what would be the justification for the following: Let $\partial$ denote the formal derivative of a polynomial over $\mathbb{F}_p$ for $p$ prime. Then, thanks to little Fermat, we have that $X^p \equiv_p X$, allowing us to conclude $\partial X^p \equiv_p \partial X \equiv_p 1$. At the same time, we have $\partial X^p \equiv_p p X^{p-1} \equiv_p 0$.
2026-05-06 10:57:10.1778065030
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Formal derivative vs Little Fermat
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In addition to river's answer:
If
- $\mathbb F$ is a field and
- $(x_i)_{i \in \mathbb N}$ is a sequence of pairwise different elementss of $\mathbb F$ and
- $p$ is a polynomial over $\mathbb F$ and
- $p(x_i)=0, \forall i \in \mathbb N$
then $p = 0$ (the null polynomial)
From this follows that if $\mathbb F$ is a field that is not finite, like $\mathbb{ Q,R,C}$ and $p(x)=0, \forall x \in \mathbb F$ then $p=0$. But this cannot be concluded for finite fields $\mathbb F$.
In a polynomial ring $R[X]$, $X$ is not a number. So $X^n = X$ does not hold. In general, equations from the base ring $R$ will not hold on $X$. Just the basic axioms of rings like $X^a X^b = X^{a+b}$.