Past exam of the options I course of the University of Oxford / 2011 / A / 1

In field characteristic $p$, the binomial theorem gives $(X-b{)}^p=X^p-b^p$, since the intermediate binomial coefficients are divisible by the prime number $p$. If $b\in F$ is a root of $X^p-a$, then $b^p=a$, and hence

Let $\alpha$ be a root in a splitting field. The preceding characteristic-$p$ identity gives $X^p-a=(X-\alpha {)}^p$, so there is only one distinct root. If the polynomial were reducible over $F$, choose a monic proper factor $g\in F[X]$ of degree $d$, with $1\le d<p$. In the splitting field it must be $g=(X-\alpha {)}^d$. Its coefficient of $X^{d-1}$ is $-d\alpha$, which belongs to $F$. Since $d$ is nonzero and thus invertible in the field $F$, this forces $\alpha \in F$, contradicting the absence of a root there. Therefore $X^p-a$ is irreducible over $F$.

Suppose a rational function $P(t)/Q(t)\in {\mathbb{F}}_p(t)$, with nonzero polynomials $P,Q$, had $p$th power $t$. Then $P(t{)}^p=tQ(t{)}^p$, so comparison of degrees would give $p\mathrm{deg}P=1+p\mathrm{deg}Q$, impossible modulo $p$. Thus $X^p-t$ has no root in $K$, and part (b) proves it is irreducible.

In a splitting field containing $\alpha$ with ${\alpha }^p=t$,It has one distinct root with multiplicity $p$, so it is inseparable. This is the purely inseparable polynomial over a rational function field.