$F/K$ finite extension, then $\exists !$ intermediate field $K \subset L \subset F$ such that $L/K$ separable and $F/L$ purely inseparable

Question

I've completed an introductory course in Galois Theory, but feel my understanding of separability is poor. I think my confusions boil down to the following question:

What is the relationship between a separable extension $L/K$ and the space $\mathrm{Hom}_K(L,E)$ for some field $E$ in which $K$ can be embedded?

Any explanations or references would be greatly appreciated.

Thanks

EDIT: Perhaps this is a bad questions, since it is too vague. I'm trying to solve the following question:

Let $F/K$ be a finite extension. Show that there is a unique intermediate field $K \subset L \subset F$ such that $L/K$ is separable and $F/L$ is purely inseparable (i.e. $|\mathrm{Hom}_L(F,E)| \leq 1$ for every extension $E$ of $L$).

Firstly, if $\mathrm{char}K = 0$ then we must have $L = F$ and the conditions are satisfied. So now suppose $\mathrm{char}K = p > 0$.

I took a guess and said $L$ is probably the smallest field containing all the elements of $F$ which are separable over $K$. Then if $\alpha \in F$ with minimal polynomial $f_L$ over $L$, and $\theta: F \to E$ is a homomorphism that fixes $L$, then $\alpha$ must be mapped to a root of $\theta(f_L)$. So if each element of $F\backslash L$ has a minimal polynomial over $L$ with only one distinct root, then I can see that the $|\mathrm{Hom}_L(F,E)| \leq 1$ will be satisfied. We also know that $\alpha$ has minimal polynomial $f_K$ over $K$ which is not separable, and so we have $f_L | f_K$ and $f_K \in K[X^p]$.

I'm stuck where to go from here, though. Is my approach correct?

Answer 1

Suppose that $L/K$ is a finite separable extension, and $E/K$ is an arbitrary extension. Then $\mathrm{Hom}_K(L,E)$ consists of at most $[L:K]$ elements, because a field homomorphisms $\sigma :L\rightarrow E$ that fixes $K$ is uniquely determined by the image $\sigma (x)$ of a primitive element of $L/K$, which in turn must be a root of the minimal polynomial $f$ of $x$ over $K$.

To characterize separability you could say that the finite extension $L/K$ is separable if and only if there exists a field extension $E/K$ such that $|\mathrm{Hom}_K(L,E)|=[L:K]$. (For $E$ you can then take the normal hull of $L/K$, or any normal extension $N/K$ containing $L$.)

Answer 2

Hagen answers your original question pretty well.

As for your edit, you're on the right track. You've said "$L$ is the smallest field containing all the elements of $F$ which are separable over $K$". This is a slightly strange thing to say, since if this $L$ has any elements other than those separable over $K$ then you're in trouble. So you need to check if the set of elements of $F$ which are separable over $K$ does indeed form a field (it turns out it does, so you don't need to refine your guess).

Having done this, we now need to check that $F/L$ is purely inseparable. What elements of $F$ are separable over $L$? Well, if $x \in F$ is separable over $L$, then $L(x)/L$ is separable; but $L/K$ is separable then implies $L(x)/K$ is separable, forcing $x \in L$. So we've shown that everything in $F\backslash L$ is inseparable over $L$. This is often the definition of a purely inseparable extension (it's the definition Wikipedia gives). So we need to check that this definition is equivalent to the one in your question. This is probably the trickiest bit of the question, which is a bit annoying.

Let $f$ be the minimal polynomial of $x$ over $L$. $f$ is irreducible and inseparable, so must be of the form $g(X^p)$ for $g \in L[X]$ irreducible. But if $g$ is not separable, we can apply this again, and continue to do so until we have $f = h(X^{p^n})$ for some $n$ and $h \in L[X]$ irreducible and separable. But the only separable polynomials over $L$ are linear, so $f = X^{p^n} - a $ for some $a \in L$. But then $f$ has only one root (Frobenius map is an automorphism of $L$). So you can construct an inductive argument on the generators of $F/L$ to show that there is only one $L$-embedding of $F$ into $E$.