Perfect field

Remarks

A field $K$ is perfect if and only if every irreducible polynomial in $K[x]$ is separable, if and only if every algebraic extension of $K$ is separable.

Example: a finite field $\F_q$ of characteristic $p$ is perfect. Check: the Frobenius map is an injective ring homomorphism, and an injective map from a finite set to itself is surjective.

Non-example: $K = \F_p(t)$ is not perfect. The element $t$ is not a $p$-th power. If $(f/g)^p = t$ with polynomials $f, g$, then $f^p = t g^p$. The degree of the left side is divisible by $p$, while the degree of the right side is not. Accordingly, $x^p - t$ is irreducible over $K$ but not separable.