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Certainly every purely inseparable simple extension must be normal, since the minimal polynomial of the generating element $a$ must look like $X^{p^r}-a$, which splits in the extension, so the example can't be simple.

Other than that I'm pretty out of ideas though.

oxeimon
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  • It seems that there is a profusion of nonsimple purely inseparable extensions, and I think that to answer this question, one would need to be armored with a big pack of examples. – Lubin Jul 26 '16 at 20:00

1 Answers1

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As you can see from my remark above, I have never thought of this question in any depth. I’m troubled by what I write below, but let’s look:

A purely inseparable extension $K\supset k$ has separable degree $1$, in other words, there is precisely one embedding $K\hookrightarrow k^{\text{ac}}$ that is identity on $k$. Since there is only one such map, it necessarily carries $K$ into itself, so that $K\supset k$ will be normal.

Lubin
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  • How do you see that there is only one embedding $K\hookrightarrow k^{ac}$ which is the identity on $k$? – oxeimon Jul 28 '16 at 02:24
  • Okay nvm this seems to be http://stacks.math.columbia.edu/tag/09HJ

    Thanks, this seems to make sense.

     – oxeimon Jul 28 '16 at 02:56
  • The total degree is the product of the separable and the inseparable degrees, and the hypothesis of pure inseparability says that the inseparable degree equals the total degree. And the separable degree counts the distinct embeddings into an algebraic closure of the base. – Lubin Jul 28 '16 at 12:49