$X^p-t\in\mathbb F_p(t)[X]$ is irreducible but has one root.

Question

I don't really understand the incredible thing with this polynomial.

i) In what it is a special polynomial ?

ii) Do you agree that $\mathbb F_p(t)=\{1,\sqrt[p]t,\sqrt[p]{t^2},...,\sqrt[p]{t^{p-1}}\}$ ?

Answer 1

It's irreducible because over $\mathbb{F}_p(t)$, relative to the $t$-adic valuation, it is an Eisenstein polynomial (The proof of Eisenstein's criterion extends to this case too! Alternatively, you can read about the theory of the Newton Polygon). (There's probably an easier way to see that it's irreducible).

Either way, it certainly has no roots in $\mathbb{F}_p(t)$. Obviously $\sqrt[p]{t}\notin\mathbb{F}_p(t)$.

It's special because it's an example of an irreducible polynomial with only one root (ie, an irreducible inseparable polynomial). Note that in a splitting field, it factors as $(X-\sqrt[p]{t})^p$. If you replace $\mathbb{F}_p(t)$ with a field of characteristic 0 or prime to $p$, then it would factor as $\prod_{i=0}^p(X-\zeta_p^i\sqrt[p]{t})$, but over $\mathbb{F}_p(t)$, there are no $p$th roots of unity!

Exercise: Can you find an irreducible polynomial with only one root over $\mathbb{F}_p$?