Problem: Let $k$ be a field, and let $F = k(t)$ be the field of rational functions in $t$ with coefficients in $k$. Then consider $f(x) = x^p - t \in F[x]$ where $p$ is prime. Prove that $f$ is irreducible over $F$.
Attempt: I know there exists a Lemma (David Cox, Galois Theory, page 86) that says:
Lemma: Let $F$ be a field and let $p$ be prime. then $f = x^p - a \in F[x]$ is irreducible over $F$ if and only if $f$ has no roots in $F$.
Hence, by the above lemma, I wish to show that $f(x) = x^p - t \in F[x]$ has no roots in $F$. How can I show this? A 'root' would be a rational function in $t$ with coefficients in $k$. Let this function be denoted by $g(t)$. Then this would have to satisfy $g(t)^p = t$ in order to be a root of $f$.
I'm not sure why this is not possible. Is it because $p > 1$ is prime?
As you've noticed the supposed root has to be a rational function, so we get that:
$$g(t) = \frac{f(t)}{h(t)} \quad \text{for some } f,h \in k[t]$$
Note that $f,h$ above are polynomials in the ring of polynomials of $k$.
The plugging it in the equation we get:
$$\left(\frac{f(t)}{h(t)}\right)^p - t = 0 \iff f(t)^p = th(t)^p$$
Now just compare the degrees of the polynomials. If $\deg f = n$ and $\deg h = m$, then we get that the LHS has degree $pn$, while the RHS has degree $pm + 1$. It's obvious this two can't be equal. Thus we obtain a contradiction.
Hence the proof.