Timeline for Abelian extensions of number fields generated by torsion points of elliptic curve (as analogy to Lubin-Tate theory)
Current License: CC BY-SA 4.0
24 events
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| Jun 2 at 15:52 | comment | added | KConrad | A (totally) ramified quadratic extension is gotten by adjoining the square root of any uniformizer: $K(\sqrt{\pi u})$ where $u$ runs through the units. Since $K(\sqrt{\alpha})=K(\sqrt{\beta})$ if and only if $\alpha/\beta$ is a square in $K$, and $(\pi u)/(\pi u')=u/u'$, all the quadratic ramified extensions occur as $u$ runs over representatives of $\mathcal O_K^\times/(\mathcal O_K^\times)^2$, which has order $2$: it's represented by $1$ and $r$ where $r$ is any unit that is not a square in the residue field. When $K=\mathbf Q_5$, its quad. ram. extns are $K(\sqrt{5})$ and $K(\sqrt{10})$. | |
| Jun 2 at 13:27 | comment | added | user267839 | uniformizer $\pi'$ and construct the extension as before with Eisenstein pol, but for $\pi'$? If that's the right approach, how can we decide if two uniformizers $\pi, \pi'$ give same or different quadratic tot ramif ext's? A naive guess: These give different ext iff for relation $u \cdot \pi = \pi'$ the unit $u$ contains the lifted $2$-root from residue as factor? Is it correct; how to prove it formally? (...by the way, this would also explain why $L'L/K$ cannot be totally ramified due to this $2$-root $\zeta_2 = \pi'/ \pi$... | |
| Jun 2 at 13:14 | comment | added | user267839 | @KConrad: one nitpick: how can one see in situation above that the are two quadratic totally ramifided extensions? Say $\pi$ is a uniformizer of $K$, then one comes obviously from Eisenstein pol $X^2-\pi$ (or more abstractly via Galois corresp, $\text{Gal}(K_{\pi}/K) \cong \mathcal{O}_K^{\times}$, we can find a unique(!) subgroup of index $2$ inside $\mathcal{O}_K^{\times}/U_1 \cong \Bbb F_q^*$ where we exploit here that the characteristic of residue field is assumed to be odd. But where is the "other" quadratic totally ramif ext? Guess: can we produce the other by choosing different | |
| Jun 2 at 12:45 | comment | added | user267839 | @KConrad: Thanks for elaborating the "mechanism" behind Lubin-Tate theory precisely, I tried to implement it in the question. But the nature of the question is of course confessedly very vague, since I'm to sure which common number theoretic features the fields obtained by adjoining the (coords) of $n$ -torsion points of the ell curve $E/K$ to the base should share, in order to ponder about for which type of Abelian extensions these should naturally "serve as container" in analogy to Lubin-Tate resp Weber's philosophy (presumably, if I understood it correctly) | |
| Jun 2 at 11:45 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 2 at 11:39 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 2 at 11:30 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 2 at 11:23 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 2 at 0:29 | comment | added | KConrad | (contd.) Each finite tot. ram. abelian extn. of $K$ (the quadratic extns in my last comment are abelian) is contained in a Lubin-Tate extn $K_{\pi,n}$ for some uniformizer $\pi$ of $K$ and $n \geq 1$. Since $K_{\pi,n}\subset K_{\pi,n+1}$ for all $n$, the $K_{\pi,n}$'s as $n$ (but not $\pi$) varies are as nice as $\mathbf Q_p(\zeta_{p^n})$, but random totally ramified abelian extensions of $K$ with increasing degree aren't nice under field composites. The union of all $K_{\pi,n}$ as $n$ varies is a complement to the max. unram. extn. of $K$ in $K^{\rm ab}$, but it is not a canonical complement. | |
| Jun 2 at 0:14 | comment | added | KConrad | You write "it generates in that way all finite totally ramified extensions. So [it] is exhaustive wrt the ramified part." You meant to say at the end "wrt the totally ramified part", but keep in mind that totally ramified extensions (in a common field) are not well-behaved under composites: if $L$ and $L'$ are totally ramified over $K$, $LL'$ need not be: when $K$ has odd residue field characteristic, it has 3 quadratic extensions (2 tot. ram. and 1 unr.), and the composite of the 2 quadratic tot. ramified extns is not tot. ramified, since the composite contains the quad. unramified extn. | |
| Jun 1 at 13:32 | comment | added | Will Sawin | $L^+$ is the totally real subfield of a CM field $L$. Alternatively $L^+$ is an arbitrary totally real field and $L$ is a totally imaginary quadratic extension. | |
| Jun 1 at 10:26 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 1 at 10:24 | comment | added | user267839 | @user491858: what do you mean by subfield $L^+$ of $L$; what does the upper plus indicate? Is it a kind special subfied? (have to admit that I never saw such notation with $+$ in this context before before). Does this notation just serve to distinguish it from bigger field $L$? is it a formal symbol to indicate that the considered extension is totally imaginary CM? | |
| Jun 1 at 1:14 | history | edited | GH from MO |
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| Jun 1 at 1:03 | comment | added | user491858 | A vague question deserves a vague answer. If $L/L^{+}$ is a totally imaginary CM extension, one can hope to construct by these methods the part of $G^{\mathrm{ab}}_L$ on which $\mathrm{Gal}(L/L^{+})$ acts by $-1$. | |
| Jun 1 at 0:27 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 1 at 0:24 | comment | added | Lubin | I think you have to be careful in describing the extensions gotten by L & T: these are abelian ramified extensions only. | |
| May 31 at 16:09 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 31 at 15:55 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 31 at 15:32 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 31 at 15:18 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 31 at 15:13 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 31 at 15:02 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 31 at 14:55 | history | asked | user267839 | CC BY-SA 4.0 |