If every irreducible polynomial in $K[x]$ is separable then every algebraic closure $\bar{K}$ of $K$ is Galois over $K$?

Question

Let $K$ be a field.

1. If every irreducible polynomial in $K[x]$ is separable then every algebraic closure $\bar{K}$ of $K$ is Galois over $K$?

2. If every algebraic closure $\bar{K}$ of $K$ is Galois over $K$ then every algebraic extension of $K$ is separable over $K$?

Here is the list of related statements that I learned in class:

1. Let $F$ be an extension field of $K$ with $\mathrm{char}K=p\neq 0$. If $u\in F$ is algebraic over $K$, then $u^{p^n}$ is separable over $K$ for some $n\geq 0$.

2. If $F$ is an extension field of $K$, $X$ is a subset of $F$ such that $F=K(X)$, and every element of $X$ is separable over $K$, then $F$ is a separable extension of $K.$

3. If $F$ is a separable extension field of $E$ and $E$ is a separable extension field of $K,$ then $F$ is separable over $K.$

Definition: Let $F$ be an extension field of $K$ such that the fixed field of the Galois group $\mathrm{Aut}_KF$ is $K$ it self.

The statements in number 1 and 2 are the equivalent definitions of perfect field. I searched through algebra books that I have but couldn't find a reference. Any reference or ideas are greatly welcome.

Answer 1

You should look in books that discuss infinite Galois theory. Most general abstract algebra books don't have a section on that, so you won't find your question answered in such books.

For your first question, every element $\alpha$ of $\overline{K}$ has a separable minimal polynomial over $K$, so by Galois theory $\alpha$ lies in a finite Galois extension: the splitting field over $K$ of its minimal polynomial over $K$. Thus each finite extension of $K$ in $\overline{K}$ is contained in a finite Galois extension of $K$, so $\overline{K}/K$ is Galois.

Your definition makes no sense. You write "Let $F$ be an extension field of $K$ such that the fixed field of the Galois group ${\rm Aut}(F/K)$ is $K$ itself." That is not a definition of anything.