field theory - When a product of two primitive elements is also primitive? - Mathematics Stack Exchange

let $k \subseteq l$ finite separable field extension. ($k$ infinite. not mind assume $k$ of characteristic zero).

assume $a,b \in l$ 2 primitive elements extension, namely, $l=k(a)=k(b)$.

is there criterion tells when $ab$ primitive element extension, namely, $l=k(ab)$?

remarks:

(1) primitive element theorem, $a+\lambda b$ primitive element, infinitely many $k \ni \lambda$'s, asking product not sum.

(2) in few examples have in mind, $ab$ not primitive element, example: $a=i,b=-i,k=\mathbb{r},l=\mathbb{c}$; $a=\sqrt2,b=-\sqrt2,k=\mathbb{q},l=\mathbb{q}(\sqrt2)$; $a=\frac{-1+\sqrt3 i}{2},b=-\frac{-1-\sqrt3 i}{2},k=\mathbb{q},l=\mathbb{q}(\sqrt3 ,i)$. extensions splitting fields of $t^2+1$, $t^2-2$, $t^2+t+1$, respectively; guess in order find extensions $a,b,ab$ primitive higher degrees polynomials required.

thank much!

edit: (1) have found this paper, may (though deals $k$ finite). (2) perhaps this question relevant. (3) comments this quetion show answer of lubin best answer can get.

i think may have trouble getting useful, general, statement. remember this:

if $k\subset k$, finite separable extension, in rough sense, almost every element of $k$ primitive extension. is, in such extension, don’t have trouble finding primitive element, have trouble finding non-primitive element outside $k$. fact, think it’s still true if $a$ primitive $k\supset k$, every choice of primitive $b$ make $ab$ primitive.