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Let $k\subset K$ be a normal extension of fields of characteristic $p>0$ with $G=\text{Aut}_k(K)$ (the automorphisms $\varphi\colon K\to K $ with $\varphi|_k = \text{id}\colon k\to k$). Show that the extension $k\subset K^G$ is purely inseparable.

So we have $k\subset K^G\subset K$ with $k\subset K$ normal, which implies that $K^G\subset K$ is normal as well. How are purely inseparable extensions and normal extensions related?

user346096
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1 Answers1

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Let $a\in K^G$ have minimal polynomial $f\in k[x]$. Since $K/k$ is normal and $f$ has one root in $K$, we know that $f$ factors completely in $K[x]$, and that $G$ acts transitively on the roots of $f$. But we know that $a$ is one root of $f$ and that $\sigma(a)=a$ for any $\sigma\in G$, so therefore $a$ is the only root of $f$, or in other words, $f=(x-a)^n$ for some $n$. This is the definition of $K^G/k$ being purely inseparable.

(Note that actually, characteristic $p$ is not necessary for this argument at all. It just happens to be the case that in characteristic $0$ every extension is separable, and if $L/F$ is both separable and purely inseparable, then $L=F$.)

Zev Chonoles
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  • To conclude that $k\subset K^G$ is purely inseparable, I think we would need that $n = p^e$ for some $e\geq 0.$ But I don't see how it would follow from this argument. – user346096 Aug 03 '16 at 15:41
  • Personally, I would not consider that a requirement of being purely inseparable, but rather a theorem about such extensions. Here is the argument: for $k$ a field of characteristic $p$, a polynomial $f\in k[x]$ is separable (has distinct roots) if and only if either $f$ is linear or $f'= 0$ (the latter of which is the case if and only if $f(x)=g(x^p)$ for some other polynomial $g\in k[x]$). – Zev Chonoles Aug 03 '16 at 18:40
  • Therefore if a polynomial $f\in k[x]$ has only one root, then either $f$ is linear or $f(x)=g(x^p)$ for some $g\in k[x]$, if the latter then $g$ certainly can't have multiple roots since then $f$ would too, so $g$ is either linear or $g(x)=h(x^p)$ for some $h\in k[x]$, etc., leading to the conclusion that $f(x)=x^{p^e}-b$ for some $e$ and some $b$. – Zev Chonoles Aug 03 '16 at 18:40