Let $\overline{K}$ be an algebraic closure of a perfect field $K$. I have read that the absolute Galois group of a perfect field is simply $\text{Gal}(\overline{K}/K)$. Yet if Galois groups are only defined for Galois extensions, then presumably $\text{Gal}(\overline{K}/K)$ only makes sense if there is some a priori guarantee that $\overline{K}$ is a Galois extension of $K$.
Clearly it is an algebraic extension, but I cannot see how it would be guaranteed that the fixed field would be $K$ so that it is Galois.
Is $\overline{K}/K$ always Galois for perfect fields or am I missing something?
Best Answer
$$ \textbf{Galois = separable + normal} $$ Note that $\overline{K}$ is always a normal extension of $K$. Thus, $\overline{K}$ is Galois over $K$ if and only if it is separable over $K$.
Hence, if $K$ is a perfect field, then indeed the algebraic closure of $K$ is a Galois extension of $K$.
$\textbf{Remark:}$ The notion of perfect field has many equivalent definitions. One of them says that a field is perfect if and only if any algebraic extension of it is separable.