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Let $K$ be a finite extension of $F$ with char $F = p > 0$ and $K^p\subseteq F$. Thus, $K/F$ is purely inseparable. A set $\{a_1,\dots, a_n\}\subseteq K$ is said to be a $p$-basis for $K/F$ provided that there is a chain of proper extensions

$$F\subset F(a_1) \subset F(a_1, a_2) \subset \dots \subset F(a_1, \dots , a_n) = K.$$ Show that if $\{a_1, \dots ,a_n\}$ is a $p$-basis for $K/F$, then $[K: F] = p^n$, and conclude that the number of elements in a $p$-basis is uniquely determined by $K/F$. The number $n$ is called the $p$-dimension of $K/F$. Also, show that any finite purely inseparable extension has a $p$-basis.

Note: This has already been posted in the following link: Show that any finite purely inseparable extensions has a $p-$basis., but in an incorrect and incomplete way. They have not given an answer, so I ask the question again in a corrected way.

My attempt: \begin{align} [K:F]&=[F(a_1,\dots, a_n):F]\\ &=[F(a_1,\dots, a_n):F(a_1,\dots, a_{n-1})][F(a_1,\dots, a_{n-1}):F(a_1,\dots, a_{n-2})] \cdots [F(a_1,a_2):F(a_1)][F(a_1):F]. \end{align} So, we would have to show that the minimal polynomial of each $a_i$ is $x^p-a_i^p=(x-a_i)^p$. I am not sure of this; with this will the first part of this exercise be solved?

For the second part, if $K$ is a purely inseparable finite extension then $K=F(a_1,\dots, a_n)$ where each $a_i$ is purely inseparable over $F$ and so $F\subset F(a_1) \subset F(a_1, a_2) \subset \dots \subset F(a_1, \dots , a_n) = K$.

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Since $a_i$ is a root of $x^p-a_i^{p}=(x-a_i)^p$, the minimal polynomial of $a_i$ is $(x-a_i)^m$ for $m\leq p$. Notice that $p=mk+r$ with $r<m$, this means that

$(x-a_i)^p=(x-a_i)^{mk}(x-a_i)^r$

and this implies that $(x-a_i)^{r}$ is a polynomial with coefficients in the base field with degree smaller than $m$. Since $r<m$ we have that $r=0$ and that $m=p$.

Your observation is enough to prove the first part of the exercise because the minimal polynomial of an element that is not in the base field, will always have degree $p$ since $K^p\subseteq F$.

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  • $\begingroup$ In general we have a finite extension $K/E$, we find the maximal separable subextension $F/E$, so that $K/F$ has no separable element $\not \in F$. How to show $K^{p^l} \subset F$ from this ? $\endgroup$ – reuns Nov 10 '17 at 11:51
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    $\begingroup$ From $f \in F[x]$ the non-separable minimal polynomial of $\alpha$, I should be able to say $f = h^m$ and show $m< p \implies h \in F[x]$. Thus $f' = 0$ and hence $f(x) = g(x^p)$ and we can look at $K/F(\alpha)/F(\alpha^p)/F$ and apply it again to $K/F(\alpha)$ and $F(\alpha^p)/F$. $\endgroup$ – reuns Nov 10 '17 at 14:55
  • $\begingroup$ Thanks Reuns, that should do the trick $\endgroup$ – JoseCruz Nov 10 '17 at 15:04

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