1
$\begingroup$

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 $\partial X^p \equiv_p p X^{p-1} \equiv_p 0$.

$\endgroup$
4
$\begingroup$

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 $X^a X^b = X^{a+b}$.

$\endgroup$
2
$\begingroup$

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 $\mathbb F$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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