This question has escalated from math.stackexchange. I'm doing this because it has been a while since the question has been open, receiving no satisfactory answers, even when subject to a bounty. I hope it is adequate for MO, I apologize if it is not.

Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.

Wikipedia, following Isaacs, Algebra, A Graduate Course p.301, says:

On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.

The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it?

To clarify, I'm looking both for a counterexample and for intuition regarding the impediment of the situation to be symmetrical.

ADDED: After posting the link to this answer on Math.SE, Georges Elencwajg kindly answered there also, providing further intuition on this subject.

This question has escalated from math.stackexchange. I'm doing this because it has been a while since the question has been open, receiving no satisfactory answers, even when subject to a bounty. I hope it is adequate for MO, I apologize if it is not.

Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.

Wikipedia, following Isaacs, Algebra, A Graduate Course p.301, says:

On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.

The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it?

To clarify, I'm looking both for a counterexample and for intuition regarding the impediment of the situation to be symmetrical.

ADDED: After posting the link to this answer on Math.SE, Georges Elencwajg kindly answered there also, providing further intuition on this subject.

This question has escalated from math.stackexchange. I'm doing this because it has been a while since the question has been open, receiving no satisfactory answers, even when subject to a bounty. I hope it is adequate for MO, I apologize if it is not.

Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.

Wikipedia, following Isaacs, Algebra, a graduate course p.301, says:

On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.

The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it?

To clarify, I'm looking both for a counterexample and for intuition regarding the impediment of the situation to be symmetrical.

This question has escalated from math.stackexchange. I'm doing this because it has been a while since the question has been open, receiving no satisfactory answers, even when subject to a bounty. I hope it is adequate for MO, I apologize if it is not.

Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.

Wikipedia, following Isaacs, Algebra, A Graduate Course p.301, says:

On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.

The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it?

To clarify, I'm looking both for a counterexample and for intuition regarding the impediment of the situation to be symmetrical.

ADDED: After posting the link to this answer on Math.SE, Georges Elencwajg kindly answered there also, providing further intuition on this subject.