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I am trying to show that if $E$ is an extension of a field $F$ with char$(F)$=p, prime, then the extension is separable if and only if $E=F(E^p).$ I have proven that if $\{e_1,\ldots ,e_n\}$ is a basis for $E$ over $F$, then $\{e_1^p,\ldots,e_n^p\}$ is a basis for $F(E^p).$

My approach has been to form the minimal polynomial, $m(X)$, of $e_k$ then show that if it's a polynomial in $p$ then this leads to a contradiction with the irreducibility of min$(e_k^p,F)$. But I am having trouble even getting started.

user460693
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  • For future visitors, the claim about the basis is proved https://math.stackexchange.com/questions/4184233/if-the-elements-y-1-y-2-dots-y-r-in-e-are-linearly-independent-over-f-sho/4188099#4188099 – D.R. Mar 27 '25 at 03:20

1 Answers1

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  • $E/F(E^p)=F(e_1,\ldots,e_n)/F(e_1^p,\ldots,e_n^p)$

    If $E\supsetneq F(E^p)$ then $E/F(E^p)$ is purely inseparable so $E/F$ is not separable.

  • Conversely if $E/F$ is not separable then let $k\ge 1$ be the least integer such that $\forall a\in E, F(a^{p^k})/F$ is separable.

    $\forall b\in E^p, F(b^{p^{k-1}})/F$ is separable, therefore $\forall b\in F(E^p), F(b^{p^{k-1}})/F$ is separable, so it can't be that $E=F(E^p)$.

reuns
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  • Trying fill in details: $(\Rightarrow)$ We have $F\leq F(E^p) \leq E$ and $E/F$ separable implies $E/F(E^p)$ is separable. If $F(E^p)\subsetneq E,$ choose basis element $e_k \in E \setminus F(E^p)$. Then $X^p-e_k^p$ does not factor over $F(E^p)$ and has $e_k$ as a root but is equal to $(X-e_k)^p$ over $E;$ a condradiction. Hence $F(E^p) = E.$ – user460693 Feb 10 '23 at 18:55
  • Filling in details: ($\Leftarrow$) Such a least integer $k\geq 1$ exists, because $E$ is a finite extension. – user460693 Feb 10 '23 at 19:29