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I read this article "Local class field theory via Lubin-Tate theory" http://arxiv.org/pdf/math/0606108v2.pdf. And I want to find the maximal abelian extensions for quadratic extensions of $\mathbb Q_p,\ p>2$. Of course there are only 3 quadratic extensions: $\mathbb Q_p(\sqrt{p}), \mathbb Q_p(\sqrt{\varepsilon}),\mathbb Q_p(\sqrt{p\varepsilon})$ (where $\varepsilon$ is not a quadratic residue in $\mathbb F_p$). Let $K$ be one of these extensions, and $\mathbb F_q$ its residue field. According to this theory I just need to find a polynomial $f \in \mathcal O_K[T]$, such that $$ f = \pi T\mod deg \ 2, f=T^q \mod \mathfrak m , $$ such that f has "good roots", and $f_m=f^{\phi^{m-1}} \circ ... f^{\phi} \circ f$ too. In the case of $\mathbb Q_p$ there is "good polynomial" $f=(T+1)^p-1$. But I can't create a "good polynomial" for quadratic case. Is there any good polynomials for quadratic cases?

P.S. I have heard something that formal groups are related to elliptic curves, but I don't know much about latter. I'll be glad for any references to this theme.

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  • $\begingroup$ It is my impression that Lubin-Tate Theory was inspired by the theory of Complex Multiplication of elliptic curves. A good reference for this is Silverman's "Advanced Topics in Elliptic Curves". An introduction to the use of formal groups in studying elliptic curves is in Silverman's "The Arithmetic of Elliptic Curves". $\endgroup$
    – Jesse Silliman
    Commented Nov 26, 2013 at 6:28
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    $\begingroup$ What you need is a power series $f$ satisfying the stated properties. But although such an $f$ exists (and is essentially unique), it is hard to write it explicitly. If $\pi$ is quadratic over $\mathbf Q$ and you are given a CM elliptic curve with multiplication by $\mathbf Q(\pi)$, with equation $y^2=x^3+ax+b$, say, then there are rational functions $u$ and $v$ such that $\pi\cdot(x,y)=(u(x,y),v(x,y))$, and one will have $f(-y/x)=-v(x,y)/u(x,y)$; but even that does not give $f$ explicitly — one needs to solve equations formally. $\endgroup$
    – ACL
    Commented Nov 26, 2013 at 7:50
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    $\begingroup$ You can always take $f(T)=\pi T + T^q$ where $\pi$ is a uniformizer and $q$ is the cardinality of the residue field. Any power series satisfying the conditions that you wrote will do. $\endgroup$
    – Laurent Berger
    Commented Nov 26, 2013 at 8:05
  • $\begingroup$ What do you mean by the additional condition “such that $f$ has good roots”? $\endgroup$
    – Lubin
    Commented Mar 17, 2014 at 14:28

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