Let $k$ be a field of characteristic zero, and let $a, b \in \bar{k}$ ($\bar{k}$ is an algebraic closure of $k$) be two distinct elements, such that $k(a)=k(b)$.
Notice that $k(a)=k(b)$ implies that the degree of the minimal polynomial of $a$ over $k$, $d_a$, equals the degree of the minimal polynomial of $b$ over $k$, $d_b$; denote $d:=d_a=d_b$.
Further assume that $k(ab)=k(a)=k(b)$ (hence the degree of the minimal polynomial of $ab$ over $k$, $d_{ab}$, equals $d$).
(1) Could one find a concrete example to the above situation with, for example, $k=\mathbb{Q}$?
(2) Is there something 'interesting' to say about $ab$?
Same questions (1) and (2) for the following special cases:
(i) $b=a-\lambda$, for some $\lambda \in k^{\times}$? Should it have 'easier/nicer' answers?
(ii) $d$ is a prime number $\geq 3$? Also, should the extension be Galois? See this and this questions.
Notice that this question is not relevant (at least not directly, but maybe something can be obtained from it), since it talks about relatively prime degrees and sum of elements, while here we talk about equal degrees (probably $d>1$) and product of elements.
Perhaps the primitive element theorem can help?
Thank you very much!
In short, I am asking: Assume that the product $ab$ of two distinct primitive elements $a,b \in L$ for the field extension $k \subseteq L$ (namely, $L=k(a)=k(b)$) is also a primitive element (namely, $L=k(ab)$), does this tell something interesting?