Exercise 5.3.9

Proof.

(a)

As the characteristic is $p$, $f=x^p-a={(x-\alpha )}^p$ has only one root $\alpha$. The splitting field of $f$ over $F$ is so $F(\alpha )$, and $f$ being the minimal polynomial of $\alpha$ over $F$, $[F(\alpha ):F]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=p$.

(b)

Let $\beta \in F(\alpha )\setminus F$. As $\alpha$ is algebraic over $F$, $F(\alpha )=F[\alpha ]$ : there exists so a polynomial $p={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^d{a}_i{x}^i\in F[x]$ such that $$\beta =p(\alpha )=\underset{i=0}{\overset{d}{\sum }}{a}_i{\alpha }^i.$$

Then (by Lemma 5.3.10),

$${\beta }^p=\underset{i=0}{\overset{d}{\sum }}{a}_i^p{\alpha }^{\mathit{ip}}=\underset{i=0}{\overset{d}{\sum }}{a}_i^p{a}^i\in F.$$

(c)

Write $b={\beta }^p\in F$. Then $\beta$ is a root of $x^p-b\in F[x]$.

As $x^p-b={(x-\beta )}^p$, with $\beta \notin F$, $x^p-b$ has no root in $F$ : by Proposition 4.2.6 $x^p-b$ is irreducible over $F$. Thus $x^p-b=x^p-{\beta }^p$ is the minimal polynomial of $\beta$ over $F$.

(d)

So every element $\beta \in F(\alpha )\setminus F$ has an inseparable minimal polynomial, thus every $\beta \in F(\alpha )\setminus F$ is inseparable. By definition, the extension $F\subset F(\alpha )$ is purely inseparable.