Question from Cox's book: Show that $x^4-t^4$ is irreducible over $\mathbb{C}(t^4)$.
Q1. Is there an elegant way to show this?
If $x^4-t^4$ has a root in $\mathbb{C}(t^4)$, then it factors as $(x-a)(x^3+ax^2+a^2x+a^3)$, where $a \in \mathbb{C}(t^4)$. Then we have $a^4 = t^4$. $a$ is a rational function in $t^4$, so write $a = \frac{p(t^4)}{q(t^4)}$ where $p(x), q(x) \in \mathbb{C}[x]$. Then $p(t^4)^4=q(t^4)^4t^4$, and I think one can show that this is a contradiction.
The other case is if $x^4-t^4$ has a quadratic factor $(x^2+ax+b)$. Then I show that $x^4-t^4 = (x^2+b)(x^2-b)$, form the equation $-b^2 = t^4$, and get a contradiction in a similar way to above.
Q2. Can I check that this way is on the right track?
We write $u = t^4$. Then the question is to show that $x^4-u$ is irreducible over $\mathbb{C}(u)$. By Gauss' Lemma, it suffices to prove it is irreducible over $\mathbb{C}[u]$.
We first show that $u$ is prime in $\mathbb{C}[u]$. i.e. $(u)$ is a prime ideal in $\mathbb{C}[u]$. So suppose $f(u)g(u)= h(u) \cdot u$ for $f(x), g(x) \in \mathbb{C}[x]$. Then either $f(u)$ or $g(u)$ has constant term equal zero. So either $f$ or $g$ is divisible by $u$, and hence in $(u)$. Therefore, $u$ is prime.
By Eisenstein's criterion, $x^4-u$ is irreducible in $\mathbb{C}[u]$.
Many thanks to Jyrki!
Edit: Saw this elsewhere: To show $(u)$ is a prime ideal in $\mathbb{C}[u]$, we can use $\mathbb{C}[u] / (u) \cong \mathbb{C}$ is a field. And so $(u)$ is maximal ideal and hence prime.