Timeline for Fibers of reciprocity maps and higher dimensional analogs
Current License: CC BY-SA 3.0
7 events
when toggle format | what | by | license | comment | |
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Jan 7, 2018 at 6:05 | comment | added | user95222 | (the comments above are meant to answer a question previously visible on top of them) | |
Jan 7, 2018 at 4:16 | history | edited | user95222 | CC BY-SA 3.0 |
added 59 characters in body
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Jan 7, 2018 at 4:06 | comment | added | user95222 | (final comment: $K_v^{\rm LT}$ still depends on $\varpi_v$, but the composite $K_v^{\rm unr}\cdot K_v^{\rm LT}$ doesn't) | |
Jan 7, 2018 at 4:04 | comment | added | user95222 | The local-at-$v$ Kronecker Weber Thm then tells you $K_v^{\rm unr}\cdot K_v^{\rm LT}= K_v^{\rm ab}$, so you have defined $\rho_v : K_v^{\times}\to\text{Gal}(K_v^{\rm ab}/K_v)$. It turns out this $\rho_v$ agrees with the Artin map defined abstractly in Artin-Tate. This is local class field theory via Lubin-Tate theory. See Cassels-Frolich, for instance. | |
Jan 7, 2018 at 4:02 | comment | added | user95222 | it also turns out that such automorphism group is isomorphic to the Galois group of the Lubin-Tate extension $K_v^{\rm LT}$ of $K_v$. Such isomorphism is the $\rho_v$ in your question. If one takes the composite field $K_v^{\rm unr}\cdot K_v^{\rm LT}$, one can extend $\rho_v$ to $K_v^{\times} = \varpi_v^{\mathbf{Z}}\times\mathcal{O}_v^{\times}$ by declaring that $\rho_v(\varpi_v) = \text{Frob}_v\in\text{Gal}(K_v^{\rm unr}/K_v)$. | |
Jan 7, 2018 at 3:59 | comment | added | user95222 | There exists a unique formal group law $F$ on $\mathcal{O}_v[\![t]\!]$ with the property that $f(t) = t^q + t\varpi_v$, with $q = \#(\mathcal{O}_v/\mathfrak{m}_v)$, is an endomorphism of $F$. $F$ depends on $f$ and $\varpi_v$, a priori, but in fact only on $\varpi_v$. It turns out that for such an $F$, the resulting formal group is actually a formal $\mathcal{O}_v$-module, whose $\varpi_v$-division points generate the so called Lubin-Tate extension of $K_v$. It turns out that the group of formal module automorphisms of $F$ is exactly $\mathcal{O}_v^{\times}$, and | |
Jan 7, 2018 at 3:14 | history | asked | user95222 | CC BY-SA 3.0 |