 Solution

An element $x\in L$ is separable over $K$ when its minimal polynomial $m_x(T)\in K[T]$ has distinct roots in a splitting field.

Because $m_x$ is irreducible, it is separable exactly whenIf $m_x^{\prime }\ne 0$, then $\mathrm{deg}{m}_x^{\prime }<\mathrm{deg}{m}_x$, so irreducibility makes this greatest common divisor equal to one. If $m_x^{\prime }=0$, the zero-derivative criterion givesfor some $g\in K[T]$, and $m_x$ is inseparable. Therefore