Exercise 5.3.10

Proof. Suppose that $F$ has characteristic $p$ and $F\subset L$ is a finite extension.

(a)

$\text{∙}$ Suppose that the extension $F\subset L$ is purely inseparable. Let $\alpha$ any element in $L$. $\alpha$ is algebraic over $F$ since $[L:F]<\infty$.

If $\alpha \in F$, the minimal polynomial of $\alpha$ over $F$ is $x-\alpha =x^{p^0}-a$, where $a=\alpha \in F$.

Suppose now that $\alpha \notin F$. By definition of a purely inseparable extension, the minimal polynomial $f$ of $\alpha$ over $F$ is not separable.

By Proposition 5.3.16 (see Ex. 7),

$$f=g(x^{p^e}),g\in F[x],e\ge 1,$$

where $g$ is a separable irreducible polynomial.

If $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)>1$, then $g$ has the root $\beta ={\alpha }^{p^e}\in L$, and $\beta \notin F$, otherwise $g$ would be divisible by $x-\beta$ and so would be reducible over $F$. As $g$ is irreducible, the minimal polynomial of $\beta$ over $F$ is $g$, which is separable. Thus $\beta$ is separable, and $\beta \notin F$, in contradiction with the hypothesis " $F\subset L$ is purely inseparable". Hence $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=1$. As $f$ is monic, $g$ is also monic, thus $g=x-a,a\in F$, and $f=x^{p^e}-a$.

So the minimal polynomial over $F$ of every $\alpha \in L$ is of the form $x^{p^e}-a,a\in F,e\ge 0$.

$\text{∙}$ Conversely, suppose that the minimal polynomial over $F$ of every $\alpha \in L$ is of the form $f=x^{p^e}-a,a\in F,e\ge 0$.

If $\alpha \in L\setminus F$, then $e\ge 1$, otherwise $\alpha =a\in F$.

Consequently, $f^{\prime }=p^e{x}^{p^e-1}=0$, since $p\mid p^e,e\ge 1$. So $f\wedge f^{\prime }=f\ne 1$, thus $f$ is not separable. No element of $L\setminus F$ is separable, so the extension $F\subset L$ is purely inseparable.

Conclusion : $F\subset L$ is purely inseparable if and only if the minimal polynomial of every $\alpha \in L$ is of the form $x^{p^e}-a$ for some $e\ge 0$ and $a\in F$.

(b)

Lemma. If $F\subset L$is a finite purely inseparable extension, and if $F\subset K\subset L$, then $K\subset L$is purely inseparable.

Proof (of Lemma). Let $\beta \in L\setminus K$, and let $f\in F[x]$ be the minimal polynomial of $\beta$ over $F$, and $f_K\in K[x]$ be the minimal polynomial of $\beta$ over $K$. As $f\in F[x]\subset K[x]$ and $f(\beta )=0$, $f_K$ divides $f$.

By part (a), $f$ is of the form $f=x^{p^e}-a,a\in F,e\ge 1$. As $f=x^{p^e}-a=x^{p^e}-{\beta }^{p^e}={(x-\beta )}^{p^e}$, $x-\beta$ is the only monic irreducible factor of $f$. Since $f_K\mid f$, $f_K={(x-\beta )}^k,k\ge 1$. As $\beta \notin K$, $k\ge 2$, thus $\beta$ is not separable over $K$, and this is true for every $\beta \in L\setminus K$, so $K\subset L$ is a purely inseparable extension. $\square$.

Suppose now that $F\subset L$ is a purely inseparable extension. As $F\subset L$ is finite, there exists ${\alpha }_1,\dots ,{\alpha }_n\in L$ such that $L=F({\alpha }_1,\dots ,{\alpha }_n)$. Let $F_0=F$ and $F_i=F({\alpha }_1,\dots ,{\alpha }_i),1\le i\le n$

Reasoning by induction, suppose that $[F_i:F]$ is a power of $p$. This is true for $i=0$ since $[F_0:F]=[F:F]=1=p^0$.

By the preceding Lemma, $L$ is purely inseparable over $F_i$. By part (a) applied to $F_i$, we know that the minimal polynomial $f_{i+1}$ of ${\alpha }_{i+1}$ over $F_i$ is of the form $f_{i+1}=x^{p^e}-a,a\in F_i,e\ge 0$. Thus $[F_{i+1}:F_i]=[F_i({\alpha }_{i+1}):F_i]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f_{i+1})=p^e$. Consequently, $[F_{i+1}:F]=[F_{i+1}:F_i][F_i:F]$ is a power of $p$, which conclude the induction.

Finally, $[L:F]=[F({\alpha }_1,\dots ,{\alpha }_n):F]$ is a power of $p$.