What are the advantages of separable extensions?

Question

I understand the definition of separable extensions very well. But I want to understand whether it holds importance as an individual concept, or does it only make sense when it's paired along with the extension being normal and thus Galois. I mean, what advantage does it give you when your polynomial doesn't have repeated roots? (I understand when it's Galois, we'll have our Galois group permuting all the roots but if it's not Galois, could we still have some benefits?)

Answer 1

You can do Galois theory with separable extensions, you don't actually need them to be Galois, altough the situation is a bit better for Galois extensions. Let $K$ be a base field, and let us fix $\bar{K}$ a separable closure of $K$; let $\Gamma = Gal(\bar{K}/K)$ the absolute Galois group of $K$ (wich technically depends on the choice of $\bar{K}$, but not really).

For any extension $L/K$, you can associate the set $X(L)$ of $K$-algebra morphisms $L\to \bar{K}$. There is a canonical action of $\Gamma$ on $X(L)$, simply by acting naturally on $\bar{K}$. If we can write $L=K(\theta)$, and $P\in K[X]$ is the minimal polynomial of $\theta$, then we can naturally identify $X(L)$ with the set of roots of $P$ in $\bar{K}$; this is what is usually done in Galois theory, but $X(L)$ is more fundamental, and does not depend on the choice of some $\theta$, and furthermore also works when $L/K$ is infinite.

Then when $L/K$ is (finite) separable, you have a nice behaviour: $|X(L)|=[L:K]$, and the action of $\Gamma$ on $X(L)$ is transitive and continuous (if $X(L)$ is seen as discrete, and $\Gamma$ has its usual Krull topology). If $L/K$ is actually Galois, then all points in $X(L)$ have the same stabilizer $G\subset \Gamma$ which is an open normal subgroup (usually all points have conjugate stabilizers), and $Gal(L/K)=\Gamma/G$, so the action of $\Gamma$ factors through that of $Gal(L/K)$ and you recover the usual action of the Galois group on roots of a polynomial.

Furthermore, you can recover $L$ from $X(L)$: if $X$ is any finite discrete transitive $\Gamma$-set, then we can define $L(X)$ as the $K$-algebra of morphisms $X\to \bar{K}$ of $\Gamma$-sets. This is a separable extension of $K$, and the processes $L\mapsto X(L)$ and $X\mapsto L(X)$ are inverse to each other. We actually get a anti-equivalence between finite separable extensions of $K$ and finite discrete transitive $\Gamma$-sets (this is a stronger statement of the usual fundamental theorem of Galois theory).

In the end, you see that Galois theory is really about separable extensions, the case of Galois extensions is just the nicer case where the $\Gamma$-action factorizes through a normal subgroup. On the other hand, none of this works for non-separable extensions: there is of course no chance to recover $L$ from $X(L)$ when $L/K$ is inseparable, since $X(L)$ may very well be a singleton even though $L/K$ is non-trivial (this happens when $L=K(\sqrt[p]{a})$ in characteristic $p$).

Answer 2

When a finite extension $L/K$ is separable then we have the following equality which is very useful:

$$\bigl|\mathrm{Hom}_K(L,K^{\mathrm{alg}})\bigr|=[L:K]$$

You also have the primitive element theorem and more!