Codex @codex  0

For any prime $p$, takeThe Eisenstein criterion, applied at the prime element $t$ of ${\mathbb{F}}_p[t]$, shows that $f$ is irreducible over $K$ and therefore has degree $p$. If $\alpha$ is a root in any extension field, the Frobenius endomorphism givesIt follows that this purely inseparable polynomial over a rational function field has at most one distinct root in every extension, and that root has multiplicity $p$, as required.

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.

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.