Timeline for Ways to order an algebraic extension
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Sep 30, 2014 at 9:20 | comment | added | Emil Jeřábek | This follows from the existence and uniqueness of real closures. If $(K,Q)$ is an order extending $P$, let $S$ be its real closure. Since $K/k$ is algebraic, $S$ is also a real closure of $(k,P)$, hence by uniqueness, there exists a $k$-isomorphism $S\to R$. It restricts to an order-preserving $k$-embedding $K\to R$. By a similar argument, the fact that two different embeddings cannot induce the same order follows from uniqueness of the $K$-isomorphism of two real closures of $K$. | |
| Sep 30, 2014 at 4:48 | vote | accept | mathcounterexamples.net | ||
| Sep 30, 2014 at 4:48 | comment | added | mathcounterexamples.net | Thanks Emil for your response. While I have an idea to prove that to a $k$-embedding of $K$ in $R$ is associated an order of $K$, I don't see how to prove that there is not more orders. Do you have a hint for that? | |
| Sep 29, 2014 at 20:46 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |