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If $E/K$ is a field extension we use the notation $\def\Aut{\operatorname{Aut}}\Aut(E/K)$ for the set of field automorphisms of $E$ that are the identity over $K$. It's immediate that the set $\Aut(E/K)$ is a group with the composition operation. If $H$ is a subgroup of $\Aut(E/K)$, we define $E^H=\{x \in E : \sigma (x) = x\,$ for all $\sigma \in H \}$. It can be easily verified that $E^H$ is an intermediate field of $E/K$ and we call it the field fixed by $H$.

It's known that if $E/K$ is a Galois extension with galois group $G$ then $E^G=K$. I'm interested in the converse of this claim. That is:

Let $E/K$ be an algebraic field extension and $G=\Aut(E/K)$. Is it true that $E^G=K$ implies $E/K$ Galois? What if $E/K$ is separable? What if $K$ is $\mathbb{Q}$? What if $E/K$ is finite?

I have been thinking for a while but I didn't find a counterexample nor a proof. I've tried examples with $K=\mathbb{Q}$ and $K=\mathbb{F}_2$. The problem is I've found $K \subsetneq E^G$ every time that $E/K$ was not Galois.

Thanks!

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  • $\begingroup$ What do you mean by "What if $E/K$ is $\mathbb{Q}$"? $\endgroup$
    – SMG
    Dec 16, 2014 at 2:31
  • $\begingroup$ A typo... Edited :) $\endgroup$
    – Pipicito
    Dec 16, 2014 at 2:34

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I am pretty sure Galois includes separability as a criteria. I will assume you mean by Galois "normal and separable". This is an equivalent condition to your question:

If $E/K$ is not galois, then there is some element $a \in E, a \notin K$ such that one of its conjugate roots over $K$ is not in $E$. Then let $x_i = g_i(a)$ where $g_i()$ are the automorphisms in $G$. The polynomial with coefficients the elementary polynomials in $x_i$ has roots in $E$ and coefficients in the fixed field of $G$. If this fixed field were K, then all the conjugate roots of a would be contained in E. This is a contradiction.

I am new to writing proofs in latex, so if you don't understand or want to make some edits, please feel free to ask me/edit it yourself.

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  • $\begingroup$ Why is it that some $\alpha \in E\setminus K$ must have one of its conjugate roots not in $E$? Doesn't this just violate normality and not separability? What it $E/K$ is a normal, but inseparable extension? $\endgroup$
    – Mmhmm
    Aug 24, 2015 at 10:58
  • $\begingroup$ I was wondering whether $K=\overline{\mathbb{F}_p}(t)$ and $E$ the normal closure of $\overline{\mathbb{F}_p}(t^{\frac{1}{p}})$ would work. $\endgroup$
    – Mmhmm
    Aug 24, 2015 at 11:05

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