Timeline for Field extension of fields
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2013 at 4:13 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Jul 27, 2013 at 11:51 | comment | added | Venkataramana | I think peter clark is right; what is true is that the only finite order automorphisms of $\mathbb C$ is of order two. If $F$ is a "finite index" subfield of the reals, then it is easy to see that there will be elements of order more than two fixing $F$ and that cannot be. This is the result that I am using. | |
| Jul 27, 2013 at 7:24 | comment | added | Pete L. Clark | In fact the theorem stated here is not true: if it were, then every real-closed field of continuum cardinality would be isomorphic to $\mathbb{R}$. But the real-closure of $\mathbb{R}((t))$ is a counterexample. It follows from another question that I asked here that in fact there are $2^{\# \mathbb{R}}$ conjugacy classes of order $2$ automorphisms on the complex numbers. | |
| Jul 16, 2013 at 15:45 | vote | accept | user46336 | ||
| Jul 16, 2013 at 15:45 | comment | added | Emil Jeřábek | More directly: there is a theorem that the algebraic closure of $k$ is a finite extension of $k$ only if $k$ is algebraically closed or real-closed. | |
| Jul 16, 2013 at 15:29 | history | answered | Venkataramana | CC BY-SA 3.0 |