3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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, $L_1\neq L_2$.

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.