Prime degree ⇒ linearly disjoint? 
If F is a field of characteristic 0 with subfields K, L such that F is the compositum of K and L and [ L : L ∩ K ] is prime, must be K and L be linearly disjoint over L ∩ K?
In other words, must [ KL : K ] = [ L : L ∩ K ] if [ L : L ∩ K ] is prime?

Here K, L are fields of characteristic 0 (definitely not p), but I don't assume [ K : K ∩ L ] is finite (or normal or anything).  I think it doesn't have to be true if that index [ L : L ∩ K ] is not prime and the field extension K ∩ L ≤ K is not normal.  I am hoping it is true when the index [ L : L ∩ K ] is prime.
The motivation is to show either every non-identity p-subgroup of PGL(2, K) is reducible or every non-identity p-subgroup of PGL(2, K) is irreducible.
 A: The answer is no. Let $L = \mathbb{Q}(\sqrt[3]{5})$ and $K = \mathbb{Q}(\zeta_3\sqrt[3]{5}).$ Then $[LK:L] = 2 \neq 3 = [L: K\cap L]$ and $L\otimes_{\mathbb{Q}} K$ is not a field.  
Edit in responce to the comment: For any fields $L$ and $K,$ if $L/L\cap K$ is finite, Galois, then so too is $KL/K$ and $[LK : K]$ divides $[L:L\cap K].$
Consequently, if $[L:L\cap K]$ is prime then $L\neq K$ and $[LK : K] = [L:L\cap K].$ 
Linear disjointness follows, assume momentarily that $K/L\cap K$ is finite and set $E = L\cap K.$ Then $$\mathrm{dim}_{E}(L\otimes_{\mathbb{E}} K) = [L:E][K:E] = [LK : K][K:E] = dim_{E}(LK).$$ Hence, the natural ring epimorphism from $L\otimes_{\mathbb{E}} K$ to $LK$ is an isomorphism of $E$-vector spaces and hence a ring isomorphism. It follows $L\otimes_{\mathbb{E}} K$ is a field. 
For the case where $K/E$ is infinite,
Let $\mathcal{K} :=\{K'\}$ be the collection of fields which are finite extensions of $E$ and contained in $K.$ Then
$L\otimes_{\mathbb{E}} K \cong L\otimes_{\mathbb{E}} (\displaystyle\lim_{\rightarrow}K') \cong \displaystyle\lim_{\rightarrow}(L\otimes_{\mathbb{E}} K') \cong \displaystyle\lim_{\rightarrow} LK' \cong LK.$
where colimits are taken over $\mathcal{K}.$ It follows $L\otimes_{\mathbb{E}} K$ is a field and $K$ and $L$ are linearly disjoint.
