Let $K \subseteq L$ be a finite separable field extension. ($K$ is infinite. I do not mind to assume that $K$ is of characteristic zero).
Assume that $a,b \in L$ are two primitive elements for the extension, namely, $L=K(a)=K(b)$.
Is there a criterion which tells when $ab$ is also a primitive element for the extension, namely, $L=K(ab)$?
Remarks:
(1) By the primitive element theorem, $a+\lambda b$ is also a primitive element, for infinitely many $K \ni \lambda$'s, but I am asking about the product not about the sum.
(2) In a few examples I have in mind, $ab$ is not a primitive element, for example: $a=i,b=-i,K=\mathbb{R},L=\mathbb{C}$; $a=\sqrt2,b=-\sqrt2,K=\mathbb{Q},L=\mathbb{Q}(\sqrt2)$; $a=\frac{-1+\sqrt3 i}{2},b=-\frac{-1-\sqrt3 i}{2},K=\mathbb{Q},L=\mathbb{Q}(\sqrt3 ,i)$. Those extensions are splitting fields of $t^2+1$, $t^2-2$, $t^2+t+1$, respectively; I guess that in order to find extensions with $a,b,ab$ primitive higher degrees polynomials are required.
Thank you very much!
Edit: (1) I have just found this paper, which may help (though it deals with $K$ finite). (2) Perhaps this question is relevant. (3) The comments for this quetion probably show that the answer of Lubin is the best answer I can get.
I think you may have trouble getting a useful, very general, statement. Remember this:
If $k\subset K$, say a finite separable extension, then in some rough sense, almost every element of $K$ will be primitive for the extension. That is, in such an extension, you don’t have trouble finding a primitive element, you have trouble finding a non-primitive element outside $k$. From this fact, I think that it’s still true that if $a$ is primitive for $K\supset k$, almost every choice of a primitive $b$ will make $ab$ primitive.