Irreducibility of $x^n-a^n$ on $F(a^n)$ where $a$ is transcendental on $F$

Question

I'm looking for a short proof of the following : If $F$ is a field where $a$ is transcendental, $x^n-a^n$ is irreducible on $F(a^n)$. I've managed to prove it in a tedious way but I feel like there should be shorter way for this. Any help is appreciated!

Answer 1

Suppose $x^n-a^n$ reducible in $F(a^n)$. Then this gives $a$ has degree $d<n$ in $F(a^n)$, i.e., $$(x-a)\mid x^d+c_{d-1}x^{d-1}+\dots+c_0,\quad c_j\in F(a^n)$$ in $F(a)$. So clearing denominators, $$p_d(a^n)a^d+\dots+p_1(a^n)a+p_0(a^n)=0\tag{*}$$ where $p_i(x)\in F[x]$. Note that $p_j(a^n)a^j$ contributes only powers of $a$ which are $j$ mod $n$ so are not cancelled by other $p_k(a^n)a^k$. Since $p_d(a^n)\neq 0$ (as $a$ is transcendental over $F$), this gives (*) is a nonzero polynmial in $a$, i.e., $a$ is algebraic over $F$, contradiction.

Answer 2

We may enlarge $F$ and assume it contains all $n$th roots of unity, so that $F(a)$ is the splitting field of $x^n-a^n$ over $F(a^n)$. For any $n$th root of unity $\zeta$, there is then an automorphism of $F(a)$ that maps $a$ to $\zeta a$ and fixes $F(a^n)$ (namely, send a rational function $f(a)$ to $f(\zeta a)$). Thus the automorphism group of $F(a)$ over $F(a^n)$ acts transitively on the roots of $x^n-a^n$, so they all have the same minimal polynomial $p$ over $F(a^n)$. This means $x^n-a^n$ is a power $p^d$ of $p$. Looking at the constant terms in the equation $p^d=x^n-a^n$, we see the constant term of $p$ must be a $d$th root of $-a^n$. But for $d>1$, there is no $d$th root of $-a^n$ in $F(a^n)$ (since $a^{n/d}\not\in F(a^n)$), so we can only have $d=1$, and thus $x^n-a^n$ is irreducible.