Showing that $f=X^p-X+T$ is irreducible over $\mathbb{F}_p(T)[X]$

Question

Let $K=\mathbb{F}_p(T)$ be the field of rational functions on one variable T over $\mathbb{F}_p$, and $f=X^p-X+T \in K[X]$. I want to show that $f$ is an irreducible polynomial.

I know that $T$ is a prime element in the field $K$, so I’m trying to apply Eisenstein criterion for that element. The only point where I was having some doubts is verifying that $T$ divides $-1$, but I think that is true since $-1/T \in K$. Is that right?

Edit: As discussed in the comments, Eisenstein’s criterion can’t be applied here. I don’t really see how to apply this solution here though, since $T \notin \mathbb{F}_p$, so $T^p \neq T$ as far as I know…

Answer 1

Since $\mathbb{F}_p[T]$ is a UFD, $f$ is irreducible in $K[X]$ $\iff$ $f$ is prime in $\mathbb{F}_p[T][X]=\mathbb{F}_p[X][T]$. On the other hand, $\mathbb{F}_p[X][T]/(f)=\mathbb{F}_p[X]$, an integral domain.

Answer 2

This is a special case of an additive analogue in characteristic $p$ of the following multiplicative theorem about binomial polynomials.

Theorem. For a field $K$ and $a \in K^\times$, $x^p - a$ is reducible in $K[x]$ if and only if this polynomial has a root in $K$.

Proof: See Theorem 3.1 here.

Theorem. For a field $K$ of characteristic $p$ and $a \in K$, $x^p - x - a$ is reducible in $K[x]$ if and only if this polynomial has a root in $K$.

Proof: Exercise. Mimic the proof of the previous theorem using additive ideas in place of multiplicative ones (e.g., the roots of $x^p - x - a$ in terms of one root $r$ are $r + c$ for $c \in \mathbf F_p$).

Example. Consider $x^p - x + T$ in $K[x]$ where $K = \mathbf F_p(T)$. To show it is irreducible, by the second theorem with $a = -T$ it suffices to show this polynomial has no root in $K$. Do this as follows.

Step 1: Show that an element of $\mathbf F_p(T)$ that is a root of $x^p-x+T$ has to be in $\mathbf F_p[T]$ and can't be constant.

Step 2: By counting degrees, show a nonconstant polynomial in $\mathbf F_p[T]$ is not a root of $x^p - x + T$.