1
$\begingroup$

Lifting the Exponent Lemma

I am referring to page 3 of the linked paper. In the paper, the author proves that $v_p(x^p-y^p) = v_p(x-y) +1$. To prove this, he first proves that p divides $\frac{x^p-y^p}{x-p}$. The proof for this makes sense to me although the reason behind his next step is quite unclear to me. For some reason, he proves that $p^2$ does not divide $\frac{x^p-y^p}{x-p}$. What is the purpose of doing this? Isn't it obvious that $p^2$ can't divide $px^{p-1}$ if $p$ can't divide $x$?

$\endgroup$
2
$\begingroup$

The reason why he proves that $p^2$ doesn't divide $\frac{x^p-y^p}{x-y}$ is to prove that $p$ is the highest power of $p$ dividing the integer.

I feel that the confusion stems from one of the first line in the proof where we have that $x^{p-1}y + \cdots xy^{p-1} \equiv px^{p-1} \pmod p$. Note that here we have $p$ as a modulo and not $p^2$. Thus this isn't enough to conclude that the number on the left is equal to $px^{p-1}$ modulo $p^2$. Therefore we need the extra work.

$\endgroup$
  • $\begingroup$ aaaah i see! Thanks! $\endgroup$ – Dude156 Jul 31 '18 at 1:59

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.