We call a field $F$ called separably closed if the only separable algebraic extension $F\subset E$ is the trivial extension, that is $E=F$. A separable closure of a field $K$ is a separable algebraic extension $K \subset K_s$, with $K_s$ separably closed.
Q: Show that every field $K$ has a separable extension; furthermore, show that any two separable closures of $K$ are $K$-isomorphic.
Now I'm having a bit of a problem with this exercise I found in one of my old books about field theory. To begin with, this definition to me seems a bit odd, since it's not clear at all to me that such a separable closure $K_s$ should even exist for any field $K$. Maybe I'm missing something crucial here?
Anyhow, for the first part of this exercise I was so far only able to show that $K_s$ (if it exists) is indeed a field. I don't think that's what the question is asking tho'. But the part I mostly need help with is the last part of the question; I haven't been able to make much progress at all.
I appreciate any help :)
Note that for arbitrary extension $K \subset L$, $K_{s} = \{x \in K \mid \text{ x is separable and algebraic over $K$}\}$ is a subfield of $L$. Now apply this to $K \subset \overline{K}$ to show that $K_{s}$ is a separable closure of $K$.
Let $K_{1}$ and $K_{2}$ be separable closures of $K$. Let $\overline{K}$ be the algebraic closure of $K$. Then we can find subfields $L_{1}, L_{2}$ of $\overline{K}$ along with the isomorphisms $\phi_{1} \colon K_{1} \xrightarrow{\sim} L_{1}$ and $\phi_{2}\colon K_{2} \xrightarrow{\sim} L_{2}$. Then $L_{1}$ and $L_{2}$ are separable closures of $K$ in $\overline{K}$. Hence $L_{1} = L_{2}$ (since they are both maximal separable extensions of $K$ in $\overline{K}$). Thus $K_{1} \xrightarrow{\phi_{1}} L_{1} = L_{2} \xrightarrow{\phi_{2}^{-1}} K_{2}$ and any two separable closures are isomorphic.