Is a field perfect iff the primitive element theorem holds for all extensions, and what about function fields

Let $L/K$ be a finite separable extension of fields. Then we have the primitive element theorem, i.e., there exists an $x$ in $L$ such that $L=K(x)$.

In particular, the primitive element theorem holds for all finite extensions of a perfect field.

Question 1. Is a field $K$ perfect if and only if the primitive element theorem holds for all finite extensions of $K$?

Question 2. Suppose that $K$ is a field extension of $\mathbf{F}_p$ of transcendence degree 1, i.e., a function field over a finite field. Does the primitive element theorem hold for any finite extension of $K$?

In Question 2, I am actually only interested in the case $K=\mathbf{F}_q(t)$.

Solution 1:

Let $K$ be a field of characteristic $p\neq 0$ and assume that the degree $[K:K^p]$ is finite. Then it is equal to $p^m$ for some $m\in\mathbb{N}$. A result of Becker and MacLane (THE MINIMUM NUMBER OF GENERATORS FOR INSEPARABLE ALGEBRAIC EXTENSIONS, Bull. AMS 46 (2), 1940) says that any finite extension $L/K$ can be generated by at most $\max (1,m)$ elements.

Now $\mathbb{F}_p(t)^p =\mathbb{F}_p(t^p)$ hence $[\mathbb{F}_p(t):\mathbb{F}_p(t)^p]=p$. Consequently every finite extension of $\mathbb{F}_p(t)$ has a primitive element. Since this field is not perfect the answer to question 1 is "no". Note that one can replace $\mathbb{F}_p$ by any perfect field. Also, it easily follows that the answer to question 2 is "yes".