Exercise 5.4.4

Proof.

(a)

As $\alpha ,\beta \in L$, and as $F\subset L$, $F(\alpha ,\beta )\subset L$.

Since $F$ has characteristic $p$, $f=(x^p-t)(x^p-u)={(x-\alpha )}^p{(x-\beta )}^p$ has only the roots $\alpha ,\beta$. The splitting field of $f$ over $F$ is so $F(\alpha ,\beta )$.

$$L=F(\alpha ,\beta ).$$

The polynomial $x^p-u$ has no root in $k(t,u,\alpha )=F(\alpha )$ by Exercise 4.2.9. applied to the field $k(t,\alpha )$. Moreover, $p$ is prime, so Proposition 4.2.6 shows that $h=x^p-u$ is irreducible over $F(\alpha )$. Therefore $h$ is the minimal polynomial of $\beta$ over $F(\alpha )$. Consequently,

$$[L:F(\alpha )]=[F(\alpha ,\beta ):F(\alpha )]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(x^p-u)=p.$$

With the same argument, $x^p-t$ has no root in $F(t)$ and is irreducible. $x^p-t$ is the minimal polynomial of $\alpha$ over $F$, thus $[F(\alpha ):F]=p$.

Finally

$$[L:F]=[L:F(\alpha )][F(\alpha ):F]=p^2.$$

(b)

Let $\gamma \in L\setminus F$.

We have proved in Example 5.4.4 that the extension $F\subset L$ has no primitive element, thus $F(\gamma )\ne L$ :

$$F⊊F(\gamma )⊊L.$$

So $d=[F(\gamma ):F]$ divides $p^2=[L:F]$. Moreover $d\ne 1$, otherwise $F(\gamma )=F,\gamma \in F$, and $d\ne p^2$, otherwise $F(\gamma )=L$, thus

$$[F(\gamma ):F]=p$$

.

(c)

By part (b), the minimal polynomial $g$ of $\gamma$ over $F$ has degree $p$. Moreover $b={\gamma }^p\in F$ by Example 5.4.4, so $\gamma$ is a root of $x^p-b\in F[x]$. Thus $g\mid x^p-b$. As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(x^p-b)$, and as $g$ and $x^p-b$ are monic, $x^p-b=g$ is the minimal polynomial of $\gamma$ over $F$. Since $g={(x-\gamma )}^p$, this polynomial is not separable. Consequently every $\gamma \in L\setminus F$ is inseparable, so the extension $F\subset L$ is purely inseparable.