Exercise 4.2.9

Proof. If $f$ has a root in $k(t)$, then there exists a rational function $u∕v,u,v\in k[t],u\wedge v=1$ such that

which is equivalent to the equality in $k[t]$:

As $u\wedge v=1$, then $u\wedge v^p=1$, and $u$ divides $t{v}^p$, thus $u$ divides $t$.

Since $t$ is irreducible (as every polynomial of degree 1), $u(t)=\lambda$, or $u(t)=\mathit{\lambda t}$, $\lambda \in k^{\text{∗}}$.

The case $u(t)=\lambda$ implies $t\mid 1$, which is false.

The case $u(t)=\mathit{\lambda t}$ gives ${\lambda }^p{t}^p=\mathit{tv}{(t)}^p$, thus ${\lambda }^p{t}^{p-1}=v{(t)}^p$, and as $p>1$, $t$ divides also $v$, which contradicts $u\wedge v=1$.

Conclusion: If $p$ is prime, $f=x^p-t$ is irreducible over $F=k(t)$. □

Proof.

For a shortest proof (see [Leintner], ex. 3.1.13, with the hint "Consider degrees of polynomials"):

If $t$ has a $p$-th root in $k(t)$, then there exists $u∕v\in k(t)$, where $u,v\in k[t],v\ne 0$, such that

which is equivalent to the following equality in $k[t]$:

Write $n=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u),m=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)$. The comparison of degrees gives

Therefore $p(n-m)=1$, thus $p\mid 1$, where $p$ is a prime number: this is a contradiction. □