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I have slight problems understanding a thing about algebraic closures of fields. It seems to me that any algebraic closure $C$ of a field $K$ is a Galois extension, but I read that this is not true. Following are the definitions I use:

An extension $F/K$ is Galois if the fixed field of $Aut_K F$ is $K$.

Note that this definition is equivalent to the definition of F being simultaneously normal and separable over K but I want to use this one as it is this one that I have problems with.

Following is a theorem from my book:

Let $\sigma : K \rightarrow L$ be an isomorphism of fields. If $S$ is a set of polynomials in $K[x]$, $F$ a splitting field of $S$, and $S'$ a set of corresponding polynomials in $L[x]$, $P$ a splitting field of $S'$ then $\sigma$ extends to $F \cong P$.

All right. So my argument is as follows. Take any $u \in C \setminus K$. Then $u$ is algebraic with a nontrivial minimal polynomial, and at least another root $v \in C$. It is elementary knowledge that $K(u) \cong K(v)$. C is an algebraic closure, and therefore splitting field of all irreducible polynomials over $K(u)$ and $K(v)$. Then by the theorem, I should be able to extend the isomorphism to a K-automorphism of $C$ that switches $u,v$. These were chosen arbitrarily in $C$, hence there is a $K$-monomorphism affecting nontrivially every element of $C \setminus K$, hence by the definition, extension is Galois.

Where does this reasoning fail?

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Strictly speaking in your definition of Galois you need to say algebraic extension. –  Qiaochu Yuan Jun 15 '12 at 13:22

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The reasoning fails at the last step when you assume $u \neq v$. Consider $K = \mathbb{F}_p(u^p)$.

More generally, if $K$ is not perfect it admits inseparable extensions and the algebraic closure cannot be Galois. The maximal Galois extension is the separable closure.

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