Constructing a new monogenic extension of fields Let $K / F$ and $L / F$ be two extension of fields, where $K / F$ is furthermore monogenic, such that $K$ and $L$ are isomorphic (as fields). Can I state that $L / F$ is monogenic too? If not, can I add extra conditions to $K$, $L$ and $F$ to get it? My attempt has no result, but I suppose that $\theta : K \to L$ is an isomorphism and $K = F(x)$, one candidate for $L$ seems to be $F(\theta(x))$. So if I take an arbitrary element $y$ in $L$ and I suppose that $y \notin F$, how can I see that $y \in F(\sigma(x))$? Is it possible? Thank you very much in advance.

No, this is extremely false. For instance, if $k$ is any field and $F=k(X_1,X_2,\dots)$ is a field of rational functions in infinitely many variables, then the fields of rational functions $K=F(Y)$ and $L=F(Y,Z)$ are isomorphic but $K$ is monogenic over $F$ and $L$ is not.

It is true if you assume $K$ and $L$ are isomorphic not just as fields but as fields over $F$. That is, assume there is an isomorphism $\theta:K\to L$ which is the identity on $F$ (or actually, it would suffice to have $\theta(F)\supseteq F$). Then the proof is trivial: if $K=F(a)$, then $L=\theta(F(a))=\theta(F)(\theta(a))=F(\theta(a))$.