Let $k \subseteq L$ be a finite separable field extension (not necessarily Galois) of degree $3$, and assume that $L=k(u)=k(v)$, for some $u,v \in L$. Clearly, $k \subseteq k(uv) \subseteq k(u)=k(v)$, and since $[L:k(uv)][k(uv):k]=[L:k]=3$, we get that $k(uv)=k$ or $k(uv)=L$. Further assume that $k(uv)=L$.
Denote the minimal polynomial of $u$ over $k$ by $A(t)=t^3+at+b$, and the minimal polynomial of $v$ over $k$ by $B(t)=t^3+ct+d$, where $a,b,c,d \in k$.
What is the minimal polynomial of $uv$ over $k$, in terms of $a,b,c,d \in k$?
Notice that this question is a special case of this recent question.
Remark:
Generlly, the minimal polynomial of $u$ over $k$ is of the form: $t^3+et^2+at+b$, $e,a,b \in k$.
We can assume that $e=0$ from the following reason: $t^3+3\frac{e}{3}t^2+3\frac{e^2}{9}t+\frac{e^3}{27}+(b-\frac{e^3}{27}) -3\frac{e^2}{9}t+at = (t-\frac{e}{3})^3+(a-\frac{e^2}{3})(t-\frac{e}{3}+\frac{e}{3})+(b-\frac{e^3}{27})=(t-\frac{e}{3})^3+(a-\frac{e^2}{3})(t-\frac{e}{3})+[(a-\frac{e^2}{3})\frac{e}{3}+(b-\frac{e^3}{27})]$
Denote $\tilde{a}:= (a-\frac{e^2}{3})$ and $\tilde{b}:=(a-\frac{e^2}{3})\frac{e}{3}+(b-\frac{e^3}{27})$, and $\tilde{u}:=u-\frac{e}{3}$.
Then $\tilde{u}$ is a root of $\tilde{A}:=t^3+\tilde{a}t+\tilde{b}$; indeed, $\tilde{A}(\tilde{u})=(\tilde{u})^3+\tilde{a}(\tilde{u})+\tilde{b}=0$. It is clear that $k(u)=k(u-\frac{e}{3})=k(\tilde{u})$.
Thank you very much!
