Polynomial t^p-a irreducible if p odd prime, a not 0, a not p-th power

Question

Let p be an odd prime. Let F be a field of characteristic 0, and let a be an element of F, a not equal to 0. Assume that a is not a p-th power in F. Prove that t^p-a is irreducible in F[t].

Would anyone know how to approach this problem/solve it?

Answer 1

Hint: Consider the splitting field $L$ of $Q := t^p - a$, and let $b\in L$ be any $p$-th root of $a$ in $L$. Now, show that if $\sigma$ is an element of the Galois group $\text{Gal}(L/F)$ such that $\sigma(b)\neq b$, then $\sigma^i(b)\neq b$ for all $0<i<p$. This shows that $b$ has at least $p$ conjugates, so $Q$ must be irreducible. For the proof, note that $L$ contains all $p$-th roots of unity $\zeta_p^i$, and that $\sigma(b)=b\cdot\zeta_p^l$ and $\sigma(\zeta_p)=\zeta_p^k$ for some $k$ and $l$.