A few days ago I found this question here on math.stackexchange, which gave a sufficient criterion for a separable, algebraic extension $E/K$ to be an algebraic closure of $K$. However it was claimed by KCd, in a comment below the question I'm referring to, that we can drop the separability condition on the extension and still get the same result
My question is: how does the proof work in the non-separable case?
Put precisely: Let $E/K$ be an algebraic extension such that every non-constant polynomial in $K[X]$ has a root in $E$, then $E$ is the (up to isomorphism) algebraic closure of $K$.
It is pretty clear to me that Makotos proof in the separable case (which can be found on the page the link above is leading to) won't work for the case of a non-separable extension (e.g. because the primitive element theorem may fail). I had some ideas of working with the separable closure but didn't try much, because I didn't see a real perspective in my approach. In other words, I'm stuck.
Lastly an apology: I refrained from asking KCd this question directly because it might be of common interest.