I need to prove the following:

Let $L/K$ be a normal field extension. Denote by $H=\operatorname{Aut}(L/K)$ the Galois group of the extension, and by $L^H$ the fixed field of $H$ in $L$. Prove that $L/L^H$ is separable, and that $L^H/K$ is purely inseparable.

For now, we tried to use the fact that an extension $M/F$ is separable (purely inseparable) if the number of homomorphisms $\phi:M\to \overline{M}$ that preserve $k$ is $[M:F]$ ($1$). We did not manage to conclude any results, and moreover this works only for finite extensions. How should we proceed?


For the first part take any $\alpha \in L$. Then let $\{\alpha_1,\dots \alpha_n\}$ be the set of distinct elements obtained by $\text{Aut}(L/K)$ acting on $\alpha$. Note that this set is finite, as the extension is algebraic. Now consider:

$$h(x) = \Pi_{i=1}^{n} (x-\alpha_i)$$

Now it's not hard to see that $h(x)$ is fixed by $\text{Aut}(L/K)$, as it only permutes the factors on the right and so we have that $h(x) \in L^H[x]$. Moreover it's irreducible, as if $g = \min(\alpha,L^H)$ then by the transitivity of the Galois group on the set of distinct elements we have that $(x-\alpha_i)$ is a factor of $g$ for any $i$. Hence we conclude that $h = \min(\alpha,L^H)$ and as it's separable $\alpha$ is separable over $L^H$ and so we conclude that $L^H \subseteq L$ is a separable extension.

However I'm not able to prove the second part for infinite extensions. Anyway here's proof for finite extensions.

We first prove that the extension $\text{Aut}(L/K) = \text{Aut}(L/L^H)$. As $K \subseteq L^H \subseteq L$ we have that every automorphism of $L$ fixing $L^H$, also fixes $K$ and so $\text{Aut}(L/L^H) \subseteq \text{Aut}(L/K)$. However from the condition we have that any automorphism on $L$ fixing $K$, also fixes $L^H$ and so we must have $\text{Aut}(L/K) \subseteq \text{Aut}(L/L^H)$. From here we conclude that $\text{Aut}(L/K) = \text{Aut}(L/L^H)$

This will give us that $K \subseteq L^H$ is also a normal extension and by Galois correspondence we have that $|\text{Aut}(L/K)| = 1$. (Here's the part where I need finiteness).

Now let $\beta \in L^H$ and consider $f = \min(\beta,K)$. Let $L_f$ be the splitting field of $f$ over $K$. As $K \subseteq L^H$ is normal we must have $L_f \subseteq L^H$. But then $|\text{Aut}(L_f/K)| = \frac{|\text{Aut}(L/K)|}{|\text{Aut}(L/L_f)|} = 1$, as $\text{Aut}(L/L_f)$ is normal in $\text{Aut}(L/K)$. But now $\text{Aut}(L_f/K)$ acts transitively on the roots of $f$ and so we must have that the only root of $f$ is $\beta$. So if $\beta \not \in K$, then $\deg f \ge 2$ and as it's only root is $\beta$ we have that $f$ isn't separable and hence $\beta$ isn't separable. From here we conclude that $K \subset L^H$ is purely inseparable extension.

  • $\begingroup$ Hi, I believe you have some typos and mistakes in the proof of the second part. (1) I think wherever you talk about the field $L$, you actually mean $L^H$, in the last two paragraphs. For instance, the Galois correspondence gives you $|\mathrm{Aut}(L^H/K)| = 1$, not $\mathrm{Aut}(L/K) = 1$. Similarly, $|\mathrm{Aut}(L_f/K)| = \frac{|\mathrm{Aut}(L^H/K)|}{|\mathrm{Aut}(L^H/L_f)|}$, as $\mathrm{Aut}(L^H/L_f)$ is normal in $\mathrm{Aut}(L/K)$. $\endgroup$ – Brahadeesh Jun 4 at 6:53
  • $\begingroup$ (2) There is a difference between $\beta$ being "not separable" and $\beta$ being "purely inseparable". What you have shown at the end is that $\min(\beta,K)$ has a unique distinct root, namely $\beta$ itself, and hence $\beta$ is purely inseparable; but you concluded something weaker, namely that $\beta$ is merely not separable. The reason I'm emphasising this is that one cannot say that $K \subset L^H$ is purely inseparable if one has only shown that each $\beta \in L^H$ is not separable over $K$. $\endgroup$ – Brahadeesh Jun 4 at 6:57
  • $\begingroup$ (3) Your proof can be streamlined further: once you have that $|\mathrm{Aut}(L^H/K)| = 1$, you need not go through the Galois theory in the last paragraph. Since $K \subset L^H$ is a normal extension, every embedding of $L^H$ over $K$ into an algebraic closure of $L$ is actually an automorphism of $L^H$, hence $|\mathrm{Aut}(L^H/K)| = 1$ implies that there is only one embedding of $L^H$ over $K$ into an algebraic closure of $L$. Hence, the separable degree of $L^H/K$ is $1$, and so $L^H/K$ is purely inseparable. $\endgroup$ – Brahadeesh Jun 4 at 7:01
  • $\begingroup$ (4) Lastly, with the above simplification, the same proof works for infinite extensions as well, because there is no more counting to be done. The part where we concluded that $|\mathrm{Aut}(L^H/K)| = 1$ is true for infinite extensions as well, because what is more generally true under the given conditions is that $\mathrm{Aut}(L^H/K)$ is isomorphic to $\mathrm{Aut}(L/K)/\mathrm{Aut}(L/L^H)$. $\endgroup$ – Brahadeesh Jun 4 at 7:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.