# A finite extension is simple iff the purely inseparable closure is simple?

Question. Is it true that a finite extension $$K:F$$ is simple iff the purely inseparable 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.

EDIT: In the proof of the lemma I write that if the purely inseparable degree of $$K:I$$ is more than $$1$$, there exists $$\alpha\in K$$ which is purely inseparable over $$I$$. I am, on a careful revision, not able to justify this statement.