Timeline for A subfield $R \subseteq \mathbb{C}(x,y)$ with 'many' generators $w$, $R(w)=\mathbb{C}(x,y)$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jul 15 at 18:48 | comment | added | user237522 | @LaurentMoret-Bailly, thank you very much! Great! I hope to upload it to arXiv in a few hours, so hopefully it will appear in 30 hours. | |
| Jul 15 at 18:20 | comment | added | Laurent Moret-Bailly | @user237522 No problem! | |
| Jul 15 at 13:40 | comment | added | user237522 | @LaurentMoret-Bailly, please, do you allow me to quote this post mentioning your name in the acknowledgements of a paper I'm willing to send to arXiv? Is it ok? Thank you. | |
| Jun 29 at 23:47 | comment | added | user237522 | Also, for example, $[\mathbb{C}(t^{\frac{1}{2}}):\mathbb{C}(t)]=2$, $[\mathbb{C}(t^{\frac{1}{3}}):\mathbb{C}(t)]=3$, with $\mathbb{C}(t^{\frac{1}{2}}) \simeq \mathbb{C}(t^{\frac{1}{3}})$, so isomorphic fields may not have equal dimension over same subfield. But perhaps this is not a problem here, with the map $s \mapsto -s$ combined with the 'rare' property. | |
| Jun 29 at 23:22 | comment | added | user237522 | @GerryMyerson, thanks. Truly, it is clear that $L$ is isomorphic to $\mathbb{C}(x,y)$, because $L=\mathbb{C}(s,v)$ with $s,v$ algebraically independent over $\mathbb{C}$. I do not understand why the choice $x=s+v, y=s+2v$ is possible; it implies that $v=y-x$, which may not hold. So even if this was just an example, how we guarantee that there is a valid choice? | |
| Jun 20 at 7:40 | comment | added | Gerry Myerson | I think the last bit of the preceding comment of @Laurent is supposed to be $\mathbb{C}(x,y)\cong L$. | |
| Jun 13 at 17:22 | comment | added | Laurent Moret-Bailly | Well, $(x,y)$ as defined is algebraically free and generates $L$ over $\mathbb{C}$, whence an isomorphism $\mathbb{C(x,y)\cong L$. | |
| Jun 13 at 17:04 | comment | added | user237522 | Please, I do not understand why we can view $L$ as $\mathbb{C}(x,y)$. I just see that $L \subset \overline{\mathbb{C}(u,v)}$, a subfield of an algebraic closure of $\mathbb{C}(u,v)$. An explanation would help. Thank you! | |
| Jun 12 at 20:08 | comment | added | Laurent Moret-Bailly | I don't understand the second comment. | |
| Jun 12 at 20:07 | comment | added | Laurent Moret-Bailly | First comment: yes, but I would not be surprised if your "rare" property were not rare at all. | |
| Jun 12 at 19:56 | comment | added | user237522 | $s$ is a root of $T^2-u \in \mathbb{C}(u,v)[T]$, so $s$ belongs to an algebraic closure of $\mathbb{C}(u,v)$. Perhaps now we divide into two cases: (1) $s \in \mathbb{C}(x,y)$. (2) $s \in \overline{\mathbb{C}(x,y)} - \mathbb{C}(x,y)$. In case (1) $L=\mathbb{C}(x,y)$ and $[\mathbb{C}(x,y):R]=2$. In case (2) perhaps $R=\mathbb{C}(x,y)$. | |
| Jun 12 at 19:36 | comment | added | user237522 | Thank you very much!! Please, if I understand your answer, you proved that for a given $R$ satisfying the condition of my question, we have $[\mathbb{C}(x,y) : R]=2$? | |
| Jun 12 at 19:24 | vote | accept | user237522 | ||
| Jun 12 at 14:04 | history | edited | Laurent Moret-Bailly | CC BY-SA 4.0 |
added 10 characters in body
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| Jun 11 at 15:43 | history | answered | Laurent Moret-Bailly | CC BY-SA 4.0 |