Separable degree

Remarks

Let $L/K$ be finite. Then $[L:K]_s$ divides the degree $[L:K]$. If $K$ has characteristic $p > 0$, the quotient $[L:K]_i = [L:K] / [L:K]_s$ is a power of $p$, called the inseparable degree; in characteristic $0$ it is $1$. The extension $L/K$ is separable if and only if $[L:K]_s = [L:K]$. Separable degree is multiplicative in towers: $[M:K]_s = [M:L]_s [L:K]_s$. It equals the degree over $K$ of the separable closure of $K$ in $L$.

Example: $L = \Q(\sqrt{2})$ over $\Q$ with $\Omega = \C$. Check: an embedding is determined by the image of $\sqrt{2}$, which must be a root of $x^2 - 2$, so there are exactly $2$ embeddings and $[L:\Q]_s = 2 = [L:\Q]$.

Non-example of the equality $[L:K]_s = [L:K]$: let $K = \F_p(t)$ and $L = K(s)$ with $s^p = t$. Over an algebraic closure, $x^p - t = (x - s)^p$ has one root, so $[L:K]_s = 1$ while $[L:K] = p$.