Separably closed field

Remarks

The following are equivalent for a field $K$:

1. $K$ is separably closed;
2. every separable polynomial in $K[x]$ splits into linear factors over $K$;
3. $K$ has no separable extension of degree greater than $1$.

Every algebraically closed field is separably closed. A field is algebraically closed if and only if it is separably closed and perfect.

Example: the separable closure $K^{\mathrm{sep}}$ of $K = \F_p(t)$ is separably closed but not algebraically closed. Check: the roots of a separable polynomial over $K^{\mathrm{sep}}$ are separable over $K^{\mathrm{sep}}$, hence over $K$, so they lie in $K^{\mathrm{sep}}$. But the root $t^{1/p}$ of $x^p - t$ is not separable over $K$.

Non-example: a finite field $\F_q$ is not separably closed. The polynomial $x^q - x - 1$ is separable, since its derivative is $-1$, and it has no root in $\F_q$, since $a^q - a - 1 = -1$ for every $a \in \F_q$.