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Problem. Let $E:k$ be a finite extension, and let $p^r=[E:k]_i$. Assume that the characteristic of $k$ is $p>0$. Assume that there is no exponent $p^s$ with $s<r$ such that $\alpha^{p^s}$ is separable over $k$ for each $\alpha\in E$. Show that $E:k$ is simple extension.

It is easy to show this assuming $E:k$ is purely inseparable. For in such a case we can find $\beta\in E$ such that $\beta^{p^s}\notin K$ for all $s<r$. Then we are forced to have $k(\beta)=E$.

So assume that $E:k$ is not purely inseparable and let $S$ be the collection of all the elements of $E$ which are separable over $k$. Then $E:S$ is purely inseparable and $[E:S]=p^r$. Further, by hypothesis, for each $s<r$, there is $\alpha\in E$ such that $\alpha^{p^s}$ is not separable over $S$. This is because if $\alpha^{p^S}$ is separable over $S$, then it is separable over $k$ too. So by the case we considered above, we see that $E:S$ is simple extension. Say $E=S(\beta)$ for some $\beta\in E$. Also, $S:k$ is finite separable extension, and hence it is simple. So we have $S=k(\alpha)$ for some $\alpha\in S$. So we have $E=k(\alpha, \beta)$.

Here I am stuck. If by chance we can show that $\beta$ is in fact purely inseparable over $k$, then it will follow that $E:k$ is simple. But I am not able to do this. I had posted this in order to solve this problem but it seems my inference there is not correct.

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  • $\begingroup$ @GerryMyerson Perhaps I have got the terminology wrong. By $S$ I means the collection of all the elements of $E$ which are separable over $k$. $\endgroup$ – caffeinemachine Sep 20 '16 at 13:02
  • $\begingroup$ Related: math.stackexchange.com/q/431756/279515 $\endgroup$ – Brahadeesh Sep 26 '18 at 12:37
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In the first para you have shown if $E:K$ is purely inseparable then $E=K(\beta)$.

Now for general case let F be the maximal separable extension of $K$ contained in $E$. Then $E$ over $F$ is p.i. So $E=F(\beta)$. Now $F$ over $K$ is finite separable and so is simple. Hence $F=K(\alpha)$ for some $\alpha$ separable over $K$. So $E=K(\alpha,\beta)$.

Now there is a famous exercise of Kaplansky,which says if $\alpha$ is separable and $\beta$ is algebraic over $K$ then $K(\alpha,\beta)$ over $K$ is simple.

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