Let $K \subset L$ be arbitrary extension of fields and assume (for the first time) it has a finite
transcendence degree $n:= \operatorname{Trdeg}_K(L)$.
Let $\{a_1,…,a_n\}$ it's transcendence basis. I'm interested in
canonical ways how the extension $K \subset L$ can be decomposed und where
I should be more careful.
At first, it seems that it is always possible to decompose it in two
ways since I not see what could fail there:
T1) as $K \subset K(a_1,…,a_n) \subset L$ where
$K \subset K(a_1,…,a_n)$ is pure transcendental and
$K(a_1,…,a_n) \subset L$ algebraic
T2) as $K \subset L_{alg} \subset L$ where
$K_{alg} := \{a \in L \vert a \text{ algebraic } \}$ is the algebraic
closure of $K$ in $L$ such that
$K \subset L_{alg}$ is tautologically algebraic and
$L_{alg} \subset L$ pure transcendental
First of all du these two decompositions always exist and work or
should one be more carefully at this point? Do we obtain always
same two decompositions if we assume $\operatorname{Trdeg}_K(L)
= \infty$ if we replace $\{a_1,…,a_n\}$ by infinite trans basis
$\{a_1, a_2, …\}$?
If we have now always such two compositions we can how study the algebraic
part separately, therefore assume from now that $K \subset L$ is an
algebraic extension. Can we always decompose it as:
A1) as $K \subset K_s \subset L$ where $K_s$ is the separable closure
of $K$ in $L$. It's a standard exercise in algebra that such
$K_s \subset L$ always exist and
$K \subset K_s$ is again tautologically a separable extension
and $K_s \subset L$ pure inseparable
A2) about this hypothetical one I'm not sure: does there exist a decomposition
$K \subset K_i \subset L$ where $K_i$ might be the 'inseparable
closure' of $K$ in $L$ and $K \subset K_i$ is pure inseparable ext and
$K_i \subset L$ separable.
About A2) I not sure if such decomposition exist because I haven't found
any textbook refering to 'inseparable closure'. On the other hand it
seems to be always possible to define $K_i$ as the subfield
$K_i := \{a \in L \vert a \text{ inseparable } \}$ where we recall that
an element $a \in L$ is inseparable over $K$ if it's minimal polynomial
$f_a \in K[t]$ has the form $t^{p^m}-c$ where $p >0$ is the characteristic
of $K$.
i know that if $L/K$ is normal and $G= \operatorname{Aut}(L)$ then
one can choose $K_{G}= L^{\text{Fix(G)}}$ as the the subfield of elements
fixed by $G$ and check that $K_{G}$ 'plays the role' of my
'inseparable closure'. What is going wrong with $K_i$ I defined
if $L/K$ is not normal?
Can the $K_i$ from my definition replaced in case of not normal ext by
by another itermediate field $K \subset M \subset L$ having the property
$K \subset M$ pure inseparable and $M \subset L$ separable.
In summary: If $L/K$ is arbitrary field extension do decompositions
$T1$ and $T2$ always exist and if $L/K$ algebraically what is going wrong with
$A2$ and my $K_i$? (since I assume that construction $A1$ always exist).
Best Answer
$L/K(a_1,\ldots,a_n)$ is algebraic and $K(a_1,\ldots,a_n)/K$ is purely transcendental.
In general there is no other canonical (*) decomposition as shown with
$\Bbb{Q}(x,\sqrt{x^3+x})/\Bbb{Q}$ (an elliptic curve) whose field of constants is $\Bbb{Q}$.
(*) the decomposition depends on a choice of transcendental basis so it is not really canonical, it is interesting to check if there are some natural conditions giving a particular one. Minimizing $[L:K(a_1,\ldots,a_n)]$ is natural.
For your second question, the non-separable normal algebraic extensions decompose in the two ways: $K(L^{p^\infty})$ means $\bigcap_{n\ge 1} K(L^{p^n})$ where $p$ is the finite characteristic, then $L/K(L^{p^\infty})$ is purely inseparable and $K(L^{p^\infty})/K$ is separable, while with $H$ the $K$-embeddings $L\to \overline{L}$ and $L^H$ the subfield fixed by all the embeddings then $L/L^H$ is separable and $L^H/K$ is purely inseparable.
When $L/K$ is not normal then $L/L^H$ doesn't have to be separable as shown there but I didn't know it before so I don't have much to say about it.