When is $\overline{K}/K$ a Galois extension of $K$? When is $\overline{K}/K$ a Galois extension of $K$, where $\overline{K}$ stands for the algebraic closure of $K$? I have the following three extensions:

$\overline{\mathbb{Q}}/\mathbb{Q}$,$\overline{\mathbb{F}_p}/\mathbb{F}_p$,$\overline{\mathbb{F_p} (t)}/\mathbb{F}_p (t)$

I know that algebraic closure is always normal. So I only need to check whether these algebraic closures are separable or not. But algebraic extension of characteristic zero field and finite field is always separable, so all three are Galois extension.
Is my reasoning correct? Thank you.
 A: The argument for the first two is correct. The third field is neither charateristic $0$ nor finite. 
In fact, it turns out in the last case the extension is not separable, and the extension not Galois. To see this consider the polynomial $X^p -t$. It is irreducible over $F_p(t)$ yet is not separable.
A: As others have said, your reasoning is correct for the first two examples but not the third.  Let me discuss how to answer the question for a general field.
In general, if $K$ is a field, then $\overline{K}/K$ is Galois iff either $K$ has characteristic $0$ or $K$ has characteristic $p$ and every element of $K$ has a $p$th root.  Such a field is called perfect.  Indeed, if some element $a\in K$ does not have a $p$th root, then $\sqrt[p]{a}\in \overline{K}$ is not separable over $K$.  Conversely, if every element of $K$ has a $p$th root and $a\in\overline{K}$ is not separable, let $f(x)$ be the minimal polynomial of $a$.  Since $a$ is not separable, we can write $f(x)=g(x^p)$ for some polynomial $g(x)\in K[x]$.  But since every element of $K$ has a $p$th root, we can take the $p$th roots of all the coefficients of $g$ to get $h(x)\in K[x]$ such that $f(x)=h(x)^p$.  This contradicts irreducibility of $f(x)$.
In particular, a finite field is perfect since $x\mapsto x^p$ is injective and hence surjective since any injection from a finite set to itself is surjective.  But $\mathbb{F}_p(t)$ is not perfect, since $t$ has no $p$th root.
