When are nonintersecting finite degree field extensions linearly disjoint?

Let $F$ be a field, and let $K,L$ be finite degree field extensions of $F$ inside a common algebraic closure. Consider the following two properties:

(i) $K$ and $L$ are linearly disjoint over $F$: the natural map $K \otimes_F L \hookrightarrow KL$ is an injection.

(ii) $K \cap L = F$.

It is well known that (i) $\implies$ (ii): see e.g. $\S$ 13.1 of my field theory notes. This implication ought to be (and usually is) followed up with the comment that (ii) does not imply (i) without some additional hypothesis: for instance take $F = \mathbb{Q}$, $K = \mathbb{Q}(\sqrt[3]{2})$, $L = \mathbb{Q}(e^{\frac{2 \pi i}{3}}\sqrt[3]{2})$, or more generally, any two distinct, but conjugate, extensions of prime degree. Thus some normality hypothesis is necessary. What is the weakest such hypothesis?

The following is a standard result: see e.g. $\S$ 13.3, loc. cit.

Theorem: If $K/F$ and $L/F$ are both Galois, then (ii) $\implies$ (i).

I remember this point coming up in a course I took as a graduate student, and the instructor claimed in passing that it was enough for only one of $K$, $L$ to be Galois. None of the standard field theory texts I own contains a proof of this. But by online searching I found an algebra text of P.M. Cohn which proves something stronger.

Theorem: If at least one of $K,L$ is normal and at least one is separable [it is permissible for the same field to be both normal and separable], then (ii) $\implies$ (i).

I wasn't able to freely view the proof, so if someone can pass it along to me I'd be appreciative. Still, I think I know of no counterexamples to the following stronger

Claim: If at least one of $K,L$ is normal, then (ii) $\implies$ (i).

Is this in fact true? (I believe I have seen it claimed in certain research papers, e.g. one by Piatetski-Shapiro. But because terminology and running separability hypotheses are not so standardized, I don't take this as conclusive evidence.)

There are counterexamples to the Claim. Recall that an extension of a field $F$ of characteristic $p>0$ is separable if and only if it is linearly disjoint from $F^{1/p^\infty}$ over $F$. Note that $F^{1/p^\infty}/F$ is a normal extension. There is an insparable algebraic extension $K/F$ which does not intersect $F^{1/p^\infty} \smallsetminus F$ (see Does an inseparable extension have a purely inseparable element?). This gives a counterexample.