Let $K⊂L⊂M$ be a tower of fields. Let $L/K$ and $M/L$ be separable, is it true that $M/K$ is separable?

I guess there are counterexamples, but I cannot point out them. Thank you for your kind help.[I know if we substitute separable to normal, there are easy example.]


Yes, it is! Let us first show that it is true for finite extensions and then reduce to the finite case later:

Assume that $M/L$ and $L/K$ are finite separable extensions. Recall that a finite extension is separable if and only if its degree of separability is equal to the degree of the extension. This leaves us with: $$[M:K]_s=[M:L]_s\cdot [L:K]_s=[M:L]\cdot [L:K]=[M:K]$$ So we are done with the finite case.

Let's assume that $M/L$ and $L/K$ are (not necessarily finite) separable extensions. Let $\alpha\in M$. Assume that $a_0,...,a_n\in L$ are the coefficients of the minimal polynomial of $\alpha$ over $L$. We know that the extension $K(\alpha,a_0,\dots,a_n)/K$ is finite. Define $$F:=K(\alpha,a_0,\dots,a_n)\cap L$$ Since $F\subset L$, we know that $F/K$ is separable. Also the minimal polynomial of $\alpha$ over $F$ is the same as that over $L$ and hence separable. Thus the extension $K(\alpha,a_0,\dots,a_n)/K$ is seperable by the first paragraph since all extensions are finite.

  • $\begingroup$ Thank you ! In finite case, your proof seem to assume M is algebraic over L. Is it right? If so, there are counterexamples M/K is transcendental extension? $\endgroup$
    – dandelion
    Dec 5 '20 at 14:54
  • $\begingroup$ Yes, I do assume that. But that is part of all the definitions of seperability I know $\endgroup$
    – CPCH
    Dec 5 '20 at 14:56
  • $\begingroup$ @CPCH There is a more general notion of separability of a (associative, unitary) commutative algebra over its field of scalars, and with this more general notion the feature of "transitivity" of separability remains true: if $K$ is a commutative field, $L$ a (commutative) extension of $K$ and $A$ a separable $L$-algebra, then the $K$-algebra $A$ obtained by restricting scalars is also separable. $\endgroup$
    – ΑΘΩ
    Dec 5 '20 at 15:34
  • $\begingroup$ @bellow To answer your question regarding potential counter-examples, do take a look at my comment above (to CPCH). $\endgroup$
    – ΑΘΩ
    Dec 5 '20 at 15:35
  • 1
    $\begingroup$ @ΑΘΩ And thanks for the recall that this holds true in the more general case! $\endgroup$
    – CPCH
    Dec 5 '20 at 17:40

By the primitive element theorem, $L=K(\alpha),M=L(\beta)=K(\alpha)(\beta)$ for some $\alpha\in L,\beta \in M$. Then the minimal polynomial $m_{\beta, L} \mid m_{\beta, K}$, whence $m_{\beta, L}$ is separable. Suppose $m_{\beta, K}$ is inseparable, i.e. the characteristic of $K$ is $p\gt 0$ and $m_{\beta, K}=f(x^p)$ for some irreducible polynomial $f(x)$. Then $m_{\beta, L}=g(x^p)$ for some irreducible polynomial $g(x)$ and the characteristic of $L=K(\alpha)$ is also p, which implies that $m_{\beta, L}$ is inseparable, a contradiction. So $\beta$ is separable over $K$ and thus $M/K$ is separable.

  • $\begingroup$ But the primitive element theorem only works if we have FINITE extensions. Which doesn't have to be the case. But of course we can get around that, similarly to what I had written in my answer. So your proof is an alternative approach to the first paragraph of my proof below $\endgroup$
    – CPCH
    Dec 5 '20 at 17:28

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