Show that if $K \subset L$, then the separable closure of $K$ in $L$ is a field

Question

Let $K \subset L$ be a field extension. Consider the separable closure $K_s$ of $K$ in $L$ defined as $$ K_s = \left\{ {x \in L \mid x \text{ is algebraic and separable over } K} \right\} $$ Prove that $K_s$ is a field.

I know how to prove that the algebraic elements are closed under operations. If also the separable elements were closed under addition and multiplication, then I'm done (I think that this happens) but I don't know how to prove it.

As far as I know, there is no simple proof of this fact. – Oct 02 '12 at 23:43 — , Oct 02 '12 at 23:43
@user18119 still once can even give the idea of the proof. – Our Apr 03 '19 at 13:38 — Our, Apr 03 '19 at 13:38

score 2 · Answer 1 · answered Nov 06 '13 at 01:32

Let $\alpha,\beta\in K_{s}$, $\beta\neq0$. Then by theorem, $K(\alpha,\beta)\subset L$ is a separable extension (adjoining a finite number separable elements gives a separable extension). Then in particular, $\alpha-\beta$, $\alpha\cdot\beta^{-1}$ are separable so that both are in $K_{s}$. Then you have it, closed under subtraction and multiplication by inverse, so $K_{s}\subset L$ is a subfield.

I believe that does it.

score 1 · Answer 2 · answered May 31 '19 at 17:23

One way to show this could be to first show that if $\alpha_1,\dots,\alpha_n \in L$ are algebraic and separable over $K$, then $K(\alpha_1,\dots,\alpha_n)$ is a separable extension of $K$ (that is, every element of $K(\alpha_1,\dots,\alpha_n)$ is separable over $K$). From this, it would follow that $$K_s = \bigcup K(\alpha_1,\dots,\alpha_n),$$ where the union is taken over all finite subsets $\{\alpha_1,\dots,\alpha_n\}$ of $L$ that consist of algebraic and separable elements over $F$. Then, one can show that $K_s$ is a field by a similar argument as used in showing that algebraic elements are closed under addition and multiplication.

One can use Galois theory to show that if $\alpha_1,\dots,\alpha_n \in L$ are algebraic and separable over $K$, then $K(\alpha_1,\dots,\alpha_n)$ is a separable extension of $K$. You can take a look at this answer of mine for a proof that uses Galois theory.

score 1 · Answer 3 · edited Mar 06 '24 at 09:45

If $A/B$ and $B/C$ are separable finite extensions then $A/C$ is a separable finite extension. This follows essentially from that $B(c)/B$ is separable iff $[B(c):B]=\deg\mathrm{minpoly}_B(c)=|\mathrm{Hom}_B(B(c),\overline{B(c)})|$.
If $a,b$ are separable over $K$ then the $K(a)$-minimal polynomial of $b$ (which divides its separable $K$-minimal polynomial) is separable so $K(a,b)/K(a)$ is separable and hence $K(a,b)/K$ is separable.
This proves that if $a,b\in K_s$ then $ab,a+b,-a,a^{-1}\in K_s$, i.e., $K_s$ is a field.

Show that if $K \subset L$, then the separable closure of $K$ in $L$ is a field

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