Finite Extension Simple iff Purely Inseparable Closure Simple

[caffeinemachine]

Question. Is it true that a finite extension $K:F$ is simple iff the purely inseprable closure is simple over $F$?

I think have an argument to support the above.

First we show the following:

Lemma. Let $K:F$ be a finite extension and $S$ and $I$ be the separable and purely inseparable closures. Then $K=SI$.
Proof. We note that $K$ is separable over $I$. This is because if $K$ is not separable over $I$, then the purely inseparable degree of $K$ over $I$ is greater than $1$. So there would exist an element $\alpha\in K\setminus I$ which is purely inseparable over $I$. But then $I(\alpha)$ would be purely inseparable over $F$, giving $I(\alpha)=I$, that is $\alpha\in I$, a contradiction. Similarly we have $K$ purely inseparable over $S$.
Therefore $K$ is both separable and purely inseparable over $SI$, meaning $K=SI$.

The following is a well known theorem:

A finite extension is simple iff there are only finitely many intermediate fields.

Now we show that if $K:F$ is a finite simple extension, then so is $I:F$, where $I$ is the purely inseparable closure. Since there are only finitely many intermediate fields between $K$ and $F$, therefore there are only many intermediate fields between $I$ and $F$. Therefore $I:F$ is simple.

Conversely, assume that $I:F$ is simple. We show that there are only finitely many intermediate fields between $K$ and $F$. Let $M$ be an intermediate field between $K$ and $F$.
Let $M_I$ be the purely inseparable closure of $M$ over $F$ and $M_S$ be the separable closure of $M$ over $F$. Then by the lemma above we have $M=M_IM_S$.
Also, $M_I=M\cap I$ and $M_S=M\cap S$ are subfields of $I$ and $S$ which contain $F$, and knowing that $S:F$ is simple (since finite separable extensions are simple), there are only finitely many composites $M_IM_S$.
Therefore there are only finitely many intermediate field between $K$ and $F$ and thus $K:F$ is simple.

 

[caffeinemachine]

Question. Is it true that a finite extension $K:F$ is simple iff the purely inseprable closure is simple over $F$?

I think have an argument to support the above.

First we show the following:

Lemma. Let $K:F$ be a finite extension and $S$ and $I$ be the separable and purely inseparable closures. Then $K=SI$.
Proof. We note that $K$ is separable over $I$. This is because if $K$ is not separable over $I$, then the purely inseparable degree of $K$ over $I$ is greater than $1$. So there would exist an element $\alpha\in K\setminus I$ which is purely inseparable over $I$. But then $I(\alpha)$ would be purely inseparable over $F$, giving $I(\alpha)=I$, that is $\alpha\in I$, a contradiction. Similarly we have $K$ purely inseparable over $S$.
Therefore $K$ is both separable and purely inseparable over $SI$, meaning $K=SI$.

The following is a well known theorem:

A finite extension is simple iff there are only finitely many intermediate fields.

Now we show that if $K:F$ is a finite simple extension, then so is $I:F$, where $I$ is the purely inseparable closure. Since there are only finitely many intermediate fields between $K$ and $F$, therefore there are only many intermediate fields between $I$ and $F$. Therefore $I:F$ is simple.

Conversely, assume that $I:F$ is simple. We show that there are only finitely many intermediate fields between $K$ and $F$. Let $M$ be an intermediate field between $K$ and $F$.
Let $M_I$ be the purely inseparable closure of $M$ over $F$ and $M_S$ be the separable closure of $M$ over $F$. Then by the lemma above we have $M=M_IM_S$.
Also, $M_I=M\cap I$ and $M_S=M\cap S$ are subfields of $I$ and $S$ which contain $F$, and knowing that $S:F$ is simple (since finite separable extensions are simple), there are only finitely many composites $M_IM_S$.
Therefore there are only finitely many intermediate field between $K$ and $F$ and thus $K:F$ is simple.

In the above I have claimed that $K$ is separable over the purely inseparable closure $I$ of $F$ in $K$. This seems to be in error. What is true is the following: A finite extension $K:F$ is simple if and only if $K$ is a simple extension of the separable closure of $F$ in $K$.