Seperability degree and roots of $f=X^{p^2}-TX^p-T\in\mathbb{F}_p(T)[X]$

Question

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...

Answer 1

As mentioned in comments, for $Y := X^p$ we have

$$f= Y^p-TY-T.$$

This is a separable polynomial (in $Y$, over $\mathbb F_p(T)$), let its (distinct!) roots be $y_1, ..., y_p \in \overline{\mathbb F_p(T)}$.

For each $y_i \in \overline{\mathbb F_p(T)}$ there is a unique $\eta_i \in \overline{\mathbb F_p(T)}$ such that $\eta_i^p =y_i$. Consequently, in $\overline{\mathbb F_p(T)}$, we have the factorization

$$f(X) = \prod_{i=1}^p (X^p-\eta_i^p) = \prod_{i=1}^p (X-\eta_i)^p . $$

In other words, the roots of $f$ are the $\eta_i$, each of them with multiplicity $p$, giving a total numebr of roots (with multiplicities) of $p^2$.

Note that in a way, here I am making a tower of fields $\mathbb F_p(T) =: K \subset K(y_1, \dots, y_p) \subset K(\eta_1, \dots, \eta_p)$ where the bottom extension is separable and the top is purely inseparable. The other answer first puts the purely inseparable $E:=K(\sqrt[p]{T})$ on top of $K$, and then constructs a separable extension of $E$ with its polynomial $g \in E[X]$.

Answer 2

$f$ 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 $F(T)$.

Answer 3

We can write in $\mathbb{F}_p(T^{\frac{1}{p}}) \supset \mathbb{F}_p(T)$

$$f(X) = (X^p - T^{\frac{1}{p}} X - T^{\frac{1}{p}})^p$$

The polynomial $g(X) \colon =X^p - T^{\frac{1}{p}} X - T^{\frac{1}{p}}$ has simple roots, since its derivative is a constant $\ne 0$ : $g'(x) = - T^{\frac{1}{p}}$. We conclude that $g(X)$ has $p$ simple roots, and $f(X)$ has $p$ distinct roots, each with multiplicity $p$.