# tower of separable extension

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.

• 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? – Nekojiru Dec 5 '20 at 14:54
• Yes, I do assume that. But that is part of all the definitions of seperability I know – CPCH Dec 5 '20 at 14:56
• @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. – ΑΘΩ Dec 5 '20 at 15:34
• @bellow To answer your question regarding potential counter-examples, do take a look at my comment above (to CPCH). – ΑΘΩ Dec 5 '20 at 15:35
• @ΑΘΩ And thanks for the recall that this holds true in the more general case! – 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.

• 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 – CPCH Dec 5 '20 at 17:28