Zeroes of $x^{p^n} - x$ are distinct in splitting field of $\mathbb{F}_p$

Question

I am attempting to show that the zeroes of $x^{p^n} - x$ are distinct (as a part of showing the splitting field over $\mathbb{F}_p$ has $p^n$ elements). I understand that it is sufficient to show that $g(x)=\frac{x^{p^n} - x}{x-\alpha}$ is non-zero for all $\alpha$ that are zeroes of $x^{p^n} - x$ on the splitting field. Clearly, $0$ is one such $\alpha$, and $g(0)$ is undefined. But for the purpose of my proof, being undefined still implies that $g(0) \not = 0$, is that correct? Just asking for a sanity check.

Answer 1

It is really bad idea to call concepts that are "undefined" an instance of not being $0$. You're supposed to be using legitimate calculations the whole time.

Method 1: Use derivatives of polynomials and the fact that $f(x)$ is separable if and only if $(f(x),f'(x))$ is a nonzero constant. In characteristic $p$, $x^{p^n}-x$ has an especially simple derivative.

Method 2: If $\alpha$ is a root of $x^{p^n} - x$ in some field $K$ containing $\mathbf F_p$ then $$ x^{p^n} - x = x^{p^n} - x - (\alpha^{p^n} - \alpha) = (x-\alpha)^{p^n} - (x-\alpha), $$ where I rewrote $x^{p^n} - \alpha^{p^n}$ as $(x-\alpha)^{p^n}$ by taking advantage of the fact that the $p$th power mapping (as well every iteration of it) is additive in rings of characteristic $p$ like $K[x]$. Now go ahead and take out a factor of $x-\alpha$ from each terms in $(x-\alpha)^{p^n} - (x-\alpha)$ and see what the value at $\alpha$ is of the remaining part. You should get a nonzero value, so $x-\alpha$ is not a repeated factor of $x^{p^n} - x$ in $K[x]$.