Timeline for On rational functions with rational power series
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2012 at 22:55 | comment | added | ACL | To Hugo: You probably want that $L^{Aut_k(L)}=k$. But such an $L$ only exists if the field extension $K/k$ is separable. | |
| May 22, 2012 at 2:18 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Jan 28, 2011 at 0:46 | comment | added | Hugo Chapdelaine | Yes you are right, I use the fact that the fixed field of $Aut(C)$ is $Q$. The fact that $Aut(R)={Id}$ is not a problem since you may work in a suitable algebraic closure and as you know $C$ is an algebraic closure of $R$. I guess that in general if you have a field $k\subseteq K$ then you want to show the existence of a field $L$ which contains $K$ such that $Aut_k(L)=k$. Once you have that the proof works. | |
| Jan 27, 2011 at 22:06 | comment | added | Auguste Hoang Duc | There is something which bothers me. You use the fact that the subfield of $\C$ fixed by $Aut(\C)$ is $\Q$. Why is it true ? For exemple it is false if you replace $\C$ by $\R$. | |
| Jan 23, 2011 at 14:54 | comment | added | Hugo Chapdelaine | Yes, I did not check it but the proof might carry over to power series ring in many variables. | |
| Jan 23, 2011 at 14:47 | history | edited | Hugo Chapdelaine | CC BY-SA 2.5 |
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| Jan 23, 2011 at 12:35 | comment | added | GH from MO | Yes, and this works in Gjergji's more general setting, too. I would emphasize that $K[[x]]$ is a UFD, it is implicitly used in your proof. | |
| Jan 23, 2011 at 0:37 | history | edited | Hugo Chapdelaine | CC BY-SA 2.5 |
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| Jan 22, 2011 at 21:11 | history | answered | Hugo Chapdelaine | CC BY-SA 2.5 |