TITLE: Algebraic closure of a perfect field.
QUESTION [5 upvotes]: I don't know if this result is true or not, if we are in the first case, how can I prove it ?
$$k \subset \overline{k} \text{ is Galois Extension } \Leftrightarrow k \text{ is a perfect  field} $$

REPLY [5 votes]: This is true. Perfect fields have the nice property that their separable closure (the maximal separable extension of $k$ contained in $\bar{k}$) is exactly their algebraic closure, and both of these are normal extensions, so we have
$k$ perfect $\Rightarrow$ $\bar{k}/k$ Galois. Indeed, perfect fields are the only fields for which the algebraic closure is separable, so if $\bar{k}/k$ is Galois, then $k$ must be perfect.