Vtyjkyuk

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:

$$g(t) = fracf(t)h(t) quad textfor some f,h in k[t]$$

Note that $f,h$ above are polynomials in the ring of polynomials of $k$.

The plugging it in the equation we get:

$$left(fracf(t)h(t)right)^p - t = 0 iff f(t)^p = th(t)^p$$

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.

I wouldn’t bother with the Lemma you quoted.

The field $F$ is the fraction field of the Unique Factorization Domain $k[t]$, in which $t$ is an indecomposable element. Your polynomial is $k[t]$-irreducible by Eisenstein. So it’s $k(t)$-irreducible, by Gauss.