Seperability degree and roots of $f=X^{p^2}-TX^p-T\in\mathbb{F}_p(T)[X]$ Consider the field $K=\mathbb{F}_p(T)$ for a prime number $p$ and the polynomial
$f=X^{p^2}-TX^p-T\in K[X]$.
I think that $f$ is irreducible in $K[X]$ but not seperable. This means, that there are less than $p^2$ distinct roots of $f$ in $\overline{\mathbb{F}_p(T)}$. My question now is:
$\textbf{How can I decide exactly how many distinct roots $f$ has in $\overline{\mathbb{F}_p(T)}$?}$
I tried substituting $Y=X^p$ but this gives me two roots
$X_{1,2}=0$ which cannot be true...

$f$ is irreducible by Eisenstein's criterion. If $X_0$ is a root, then $T=\frac{X_0^{p^2}}{X_0^p+1}=(\frac{x_0^{p}}{X_0+1})^p$, that is the splitting field of $f$ contains a $p$-th root of $T$. And in any extension field with $\sqrt[p]{T}$, we have the factorization $f(X)=(X^p-\sqrt[p]{T}X-\sqrt[p]{T})^p = g(X)^p$. Note that the polynomial $g(X)$ satisfies $g'(X)=-\sqrt[p]{X}\not=0$ with no zero hence no common zero with $g$, thus $g$ and hence $f$ has $p$ distinct roots.
Note that $F(T)\simeq F(\sqrt[p]{T})$ where $F:=\mathbb F_p$, and we shall use $S=\sqrt[p] T$ as an independent variable. Let $X_1, X_2$ be two distnct roots of $g(X)$, then we have $(X_1-X_2)^p=S(X_1-X_2)$, hence $(X_1-X_2)^{p-1}=S$. That is the splitting field of $g(X)$ contains a $(p-1)$-th root of $S$, but $[F(S)(X_1):F(S)]=p$ has no nontrivial intermediate field. Therefore the splitting field of $g(X)$ over $F(X)$ is not just $F(S)$ joining one root.
Let $\alpha$ be one $(p-1)$-th root of $S$, then $a\alpha$ for $a\in\mathbb F_p\setminus\{0\}$ are all the $(p-1)$-th root. And it's easy to check that $X_1+a\alpha$ forms all the roots of $g(X)$. As $\gcd(p, p-1)=1$, we must have the splitting field of $g(X)$ has degree $p(p-1)$ over $F(S)$, hence $p^2(p-1)$ over $F(T)$.
Knowing that the splitting field contains $(\sqrt[p-1] S)^p = \sqrt[p-1]{T}$, it's not hard to show that $F(\sqrt[p-1] T)$ is the separable closure of $F(T)$.
