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).


1 Answer 1


$L/K(a_1,\ldots,a_n)$ 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.

  • 1
    $\begingroup$ $\overline{\Bbb{Q}}\cap \Bbb{Q}(x,\sqrt{x^3+x})= \Bbb{Q}$ $\endgroup$
    – reuns
    Feb 19, 2021 at 19:34
  • $\begingroup$ about your example: $\Bbb{Q}(x,\sqrt{x^3+x})/\Bbb{Q}$. As far as I can follow you this explains why decomposition A2 as $K \subset K_{alg} \subset L$ fails to exist because $\Bbb{Q}_{alg}$ in $\Bbb{Q}(x,\sqrt{x^3+x})/\Bbb{Q}$ equals $\Bbb{Q}$, but $\Bbb{Q}(x,\sqrt{x^3+x})/\Bbb{Q}$ is not pure transcendental, that's the point,right? $\endgroup$
    – user267839
    Feb 19, 2021 at 19:40
  • $\begingroup$ ...sorry, previously I wrote a nonsense comment $\endgroup$
    – user267839
    Feb 19, 2021 at 19:41
  • $\begingroup$ About case $L/K$ algebraic: in several algebra textbooks it is shown that decomposition A1 always exist, but I conjecture that A2 should fail if $L/K$ is not normal. Do you know a conterexample? $\endgroup$
    – user267839
    Feb 19, 2021 at 19:47
  • 1
    $\begingroup$ Ah, the non-separable algebraic extensions decompose in the two ways: $L/K(L^{p^\infty})$ is purely separable 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. $\endgroup$
    – reuns
    Feb 19, 2021 at 20:08

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