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!

  • $\begingroup$ I count two extensions. Also, there are no finite extensions of $ℂ$ other than $ℂ$ itself because every finite extension is algebraic and $ℂ$ is algebraically closed. $\endgroup$ – k.stm Nov 16 '17 at 20:34
  • $\begingroup$ Counterexample without $\mathbb{C} \subset L_1$: $L_1 = \mathbb{Q}$, $L_2 = \mathbb{Q}[\sqrt{2}]$, and $L_3 = \mathbb{Q}[\sqrt[4]{2}]$. $\endgroup$ – Connor Harris Nov 16 '17 at 20:35
  • $\begingroup$ @k.stm, thanks for your comment. By three extensions I meant: $L_1 \subseteq L_2$, $L_2 \subset L_3$ and $L_1 \subset L_3$. Also, I did not say that $\mathbb{C} \subset L_1$ is finite, only that $\mathbb{C}$ is strictly contained in $L_1$. I thought of $L_1=\mathbb{C}(x,y)$. $\endgroup$ – user237522 Nov 16 '17 at 20:40
  • $\begingroup$ @ConnorHarris, thank you for your counterexample. $\endgroup$ – user237522 Nov 16 '17 at 20:46

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, $L_1\neq L_2$.

  • 1
    $\begingroup$ There's a big family of similar counterexamples all following this pattern: take some $x \in L_1$ and let $a, b$ be integers at least 2, then $L_2 = L_1[\sqrt[a]{x}]$ and $L_3 = L_1[\sqrt[ab]{x}]$. $\endgroup$ – Connor Harris Nov 16 '17 at 20:43
  • $\begingroup$ @ConnorHarris, thank you. I like the fact that in this family we can take $L_1=\mathbb{C}(x,y)$. $\endgroup$ – user237522 Nov 16 '17 at 20:50
  • $\begingroup$ Dietrich Burde, thank you, I voted your answer but not accepted it, since I prefer to have $\mathbb{C} \subset L_1$. $\endgroup$ – user237522 Nov 16 '17 at 20:58
  • $\begingroup$ Thank you. Be sure to write such assumptions next time into your question, and may be even into the title:) $\endgroup$ – Dietrich Burde Nov 16 '17 at 21:00
  • $\begingroup$ I agree with you. I should have written more precisely what are my assumptions, not just 'I prefer' etc. Thanks :) $\endgroup$ – user237522 Nov 16 '17 at 21:04

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.

  • $\begingroup$ Thanks, but I required $L_1 \subset L_2$. $\endgroup$ – user237522 Nov 16 '17 at 20:44
  • $\begingroup$ Sorry, I tweaked the values a bit. It should now be correct! $\endgroup$ – neptun Nov 16 '17 at 21:03
  • $\begingroup$ @user237522 I also made an example where they contain the complex numbers $\endgroup$ – neptun Nov 16 '17 at 21:07
  • $\begingroup$ ok, thanks. I voted your answer but not accepted it, since Connor Harris already brought this idea (I guess you have not seen his comments because you were writing your answer). $\endgroup$ – user237522 Nov 16 '17 at 21:11
  • $\begingroup$ $\mathbb{C} \subset \mathbb{C}(t^2)$ is not finite. $\endgroup$ – user237522 Nov 16 '17 at 21:14

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