$X^n - t$ is irreducible over $k(t)$

Question

How can I prove (if it's true) that $X^n - t$ is irreducible over $k(t)$, the field of fractions of $k$ ?

Answer 1

You can use Eisenstein's criterion. By Gauss' lemma, $X^n - t$ is irreducible over $k(t)$ iff it's irreducible over $k[t]$. Now, consider $\mathfrak{p} = (t)$. It is prime, and $t \not \in \mathfrak{p}^2$, so $X^n - t$ is irreducible over $k[t]$ by Eisenstein's criterion.

Answer 2

Viewing $X^n-t$ as a polynomial over $k[t]$, this is a primitive polynomial, because the $\gcd$ of the (non-zero) coefficients $1,-t$ is $1$. Therefore by a lemma of Gauss, it will be irreducible over $k(t)$ if and only if it is irreducible over $k[t]$. But over $k[t]$ the polynomial satisfies the Schönemann-Eisenstein criterion for the irreducible element $t$, hence it is irreducible.