Need help in deducing 2 arguments related to separable extension

Question

I am self studying Field Theory from class notes of a senior ( which are based on Algebra by Thomas Hungerford) and I unable to deduce the following two results related to each other:

Let $F$ be an algebraic extension field of a field $K$ of characteristic $p$ (non-zero). Prove that:
(a) if $u$ belongs to $F$, then $u^{p^n}$ belongs to $K$ for some $n\geq 0$
implies that
(b) $F$ is purely inseparable over $K$.

Can someone please tell how to do it? I have no clue about it. Also, for

(c) $F$ is generated over $K$ by a set of purely inseparable elements,
prove that (c) implies (a).

I am sorry but I am struck badly on these questions. Please help. I can't think how to proceed.

My attempt:

For the first I am unable to think how it can be done. I have no idea.

For (c) $\Rightarrow$ (a), if I assume that $u$ is purely inseparable over $K$, then I prove (a). But I am unable to get an idea on how to do it when $u$ belonging to $F$ is arbitrary.

Answer 1

Following your given definition and notation. Let $u \in F$ and $u^{p^n}=:a \in K$. Then $u$ is a root of $f:=(x-u)^{p^n}=x^{p^n}-u^{p^n}= x^{p^n}-a= \in K[x]$ (the first equality holds since we are in characteristic $p$). Now the minimal polynomial of $u$ divides $f$ so it is of the form $(x-u)^m$ for some $m\leq p^n$ but this means that $u$ is purely inseparable over $K$. Since $u$ was arbitrary this means that $F$ is purely inseparable over $K$.

Edit: Here is a more detailed sketch for part (c):

1. If $u \in F$ is purely inseparable over $K$ then the field extension $K(u)$ is purely inseparable over $K$. This is because $u^{p^n} \in K$ for some $n$ and any element $u'$ in $K(u)$ can be written as $g(u)$ for some $g \in K[x]$ but then $u'^{p^n}=g(u)^{p^n}=g(u^{p^n}) \in K$.
2. If $u \in F$ is purely inseparable over an intermediate field $L$ and $L$ is purely inseparable over $K$ then $u$ is already purely inseparable over $K$. This is because $u^{p^n} \in L$ for some $n$ and so since $L$ is purely inseparable there is some $m$ such that $(u^{p^n})^{p^m}=u^{p^{m+n}} \in K$
3. Finally, let $F$ be generated by some set $U$ of (over $K$) purely inseparable elements and $u \in F$. Then there is a finite subset $U'=\{u_1,...,u_k\}$ such that $u$ lies already in the field extension generated by $U'$. By (1.) the field extension $K[u_1,...,u_l]$ ($l \leq k$) is purely inseparable over $K[u_1,...,u_{l-1}]$ so by (2.) and a trivial induction it is actually already purely inseparable over $K$. Since $u \in K[u_1,...,u_k]$ this means $u$ is purely inseparable over $K$ which proves the assertion.