What is the condition for a field to make the degree of its algebraic closure over it infinite? As we all know, the algebraic closure often has an infinite degree.
Also, this shows the necessary and sufficient condition for a Galois extension to be a finite extension of fields.
However, we may want to characterize the cases of the case where the extension is not Galois, which is actually my question.
And this question is related to this question.
 A: Let me summarize the results of $\S 12.4$ of my notes.  (This involves making explicit some things which were left as "exercises", but I'm okay with that.)
Let $K$ be any field, let $K^{\operatorname{sep}}$ be any separable closure and let $\overline{K}$ be any algebraic closure containing $K^{\operatorname{sep}}$.  Then:
1) Suppose first that $K = K^{\operatorname{sep}}$, i.e., $K$ is separably closed.  Then either:
1a) $K$ is algebraically closed, or
1b) It isn't, i.e., $K$ has positive characteristic $p$ and there exists $a \in K$ such that 
the polynomial $t^p - a$ is irreducible.  Then $t^{p^n}-a$ is irreducible for all $n$, so 
$[\overline{K}:K]$ is infinite.  (For this, see Lemma 32 in $\S 6.1$.)
2) Suppose that $K$ is not separably closed, so $G = \operatorname{Aut}(K^{\operatorname{sep}}/K)$ is nontrivial.  Then by the Artin-Schreier theorem, either
2a) $\# G = 2$, in which case $K$ is formally real and $K(\sqrt{-1})$ is algebraically closed, or
2b) $\# G > 2$, in which case $G$ is infinite, which implies $[\overline{K}:K]$ is infinite.   
Note that some of the exercises outline further extensions of the Artin-Schreier Theorem, i.e., if $\overline{K}/K$ is "small" in other ways then $K$ is either real closed or algebraically closed.
A: The Artin-Schreier theorem asserts that, if the field $k$ is not algebraically closed and $[\bar{k} : k]$ is finite, then in fact it equals $2$ and $\bar{k} = k[i]$ and $k$ is real closed. So I guess the condition you're looking for is: if and only if $k$ is not algebraically closed and also not real closed. 
I am answering the title question, not the body question. The body question is not equivalent to the title question; the answer to it is that the algebraic closure is Galois if and only if it's separable, hence if and only if $k$ is perfect. Perhaps you have only seen the definition of Galois extension for finite extensions. This is not the definition which works in maximal generality. The definition which does work in maximal generality is the following:
Definition: An algebraic extension $k \to L$ is Galois if the fixed field of $\text{Aut}_k(L)$ is $k$.
The separable closure $k_s$ of $k$ is always Galois (in fact it is the maximal Galois extension of $k$), and agrees with the algebraic closure if and only if $k$ is perfect. 
