Does $L_1(w)=L_2(w)$ imply $L_1=L_2$?

Question

Let $L_1 \subseteq L_2 \subsetneq L_3$ be inclusions of three fields, where $L_1$ may or may not equal $L_2$, and $L_2$ is strictly contained $L_3$.

Assume that the three field extensions are finite and separable, so there exists a primitive element for each extension.

In particular, there exist $w_1,w_2 \in L_3-L_2$ such that $L_3=L_1(w_1)=L_2(w_2)$.

Further assume that $w:=w_1=w_2$, so $L_3=L_1(w)=L_2(w)$.

Is it true that $L_1=L_2$? If not, it would be nice to have a counterexample; I prefer a counterexample in which $\mathbb{C}$ is strictly contained in $L_1$.

Concerning the (finite) degrees of the field extensions: I do not mind to further assume that $[L_3:L_2]=2$ and $w^2 \in L_2$ (I think I once showed that in that case $L_1=L_2$, but I am not sure).

Any hints are welcome!

Answer 1

Here is a counterexample. Take $L_1=\mathbb{Q}$, $L_2=\mathbb{Q}(i)$ and $w=\zeta_8=e^{2\pi i/8}$. Then $L_3=\mathbb{Q}(\zeta_8)$, and $L_1(w)=L_2(w)=L_3$. On the other hand, $L_1\neq L_2$.

Answer 2

No, consider $L_1=\mathbb Q$ and $L_2=\mathbb Q(\sqrt 3)$ which aren't equal but certainly exist inside $L_3=\mathbb Q({}^4\!\!\!\sqrt 3)$.

Here with $w={}^4\!\!\!\sqrt 3$ we have $L_1(w)=L_2(w)=L_3$ but not $L_1=L_2$.

Edit: since you want the fields to contain $\mathbb C$, we can replace the number with any variable $t$.

Consider $\mathbb C(t^4) \subset \mathbb C(t^2)$ who both are in $\mathbb C(t)$. Clearly $\mathbb C(t)=\mathbb C(t^2,t)=\mathbb C(t^4,t)$ but the fields are not equal.