Separability is transitive ... for infinite extensions Let $L/M/K$ be a tower of fields. The proof that $L/K$ is separable iff $L/M$ and $M/K$ are also separable is contained in a lot of notes and texts I've come across, subject to the assumption that the extension $L/K$ is finite. An equivalent condition to separability, that $L\otimes_K\overline{K}$ is reduced, is used to actually define separability in the case of infinite extensions. Here I can prove that $L/K$ is separable iff for all $a\in L$, either $a$ is transcendental or separable over $K$ (in the usual sense that its minimal polynomial is separable). I have nearly proven that separability is transitive, too, but there is one piece I am missing: I need to show that $L/M$ and $M/K$ separable imply $L/K$ is separable.
Presumably we assume $a\in L$ is inseparable and get a contradiction, but I'm unsure how to do this, although it feels like it must be something simple that I'm missing. (Here is what I have so far.)
 A: So I'm writing this answer for the case where $L$ is algebraic over $K$. I'm in a bit of a hurry so I will add the transcendental part later.
The main idea used here is that while $L/K$ is infinite, every element of $L$ is contained in a finite extension of $K$.
Let $a$ be an arbitrary element of $L$. We need to show that the minimal polynomial of $a$ over $K$ has no repeated roots i.e that $a$ is separable. Well, we know that $L/M$ is separable and hence $m_{a, M}$ has no repeated roots. Let $\alpha_{1}, \ldots, \alpha_{n} \in M$ be the coefficients of $m_{a, M}$. Additionally, as $M/K$ is separable, each $\alpha_{i}$ is separable over $K$.
Let us now define the extension 
$$L' = K[\alpha_{1}, \ldots, \alpha_{n}, a]$$
Since $\alpha_{i}$ is separable over $K$, $M' = K[\alpha_{1}, \ldots, \alpha_{n}]$ is a finite separable extension of $K$. Additionally, $m_{a, M'} = m_{a, M}$ by construction and hence $a$ is separable over $M'$, as this polynomial has no repeated roots. Thus, we have $L'/M'$ is finite separable over $M'$. Applying transitivity for finite separable extensions, we get that $L'/K$ is separable. Hence, the minimal polynomial of $a$ over $K$ has no repeated roots.
In the case where the extension is not necessarily algebraic, separability is the existence of a transcendence base over which the extension is separable algebraic. Let $\{\alpha_{i}\}_{i \in I}$ and $\{\beta_{i}\}_{i \in J}$ be separating transcendence bases for $L/M$ and $M/K$ respectively. Then, $\{\alpha_{i}, \beta_{j}\}$ is a transcendence base for $L/K.$ We need to show that $L$ is separable over $K(\alpha_{i}, \beta_{j}).$
So, we have a chain of algebraic extensions
$$L \supseteq M(\beta_{j}) \supseteq K(\alpha_{i}, \beta_{j})$$
and we also know that $L/M(\beta_{j})$ is separable (by definition of separating transcendence base) and we know that $M/K(\alpha_{i})$ is separable and hence so is $M(\beta_{j})/K(\alpha_{i}, \beta_{j})$. Thus, by the argument that was made in the algebraic case, $L$ is separable over $K(\alpha_{i}, \beta_{j})$ as desired.
