# Are all totally ramified $\mathbb{Z}_p$-extensions of local fields come from (relative) Lubin-Tate formal groups?

The setup is as follows:

• $k/\mathbb{Q}_p$ is a finite extension, $\mathfrak{p}$ is the maximal ideal of $\mathcal{O}_k$, $q=\#(\mathcal{O}_k/\mathfrak{p})$
• $k'/k$ is a finite unramified extension of degree $d$

It's known that for a relative Lubin-Tate formal group $\mathcal{F}$ relative to $k'/k$ with parameter $\xi$ ($\xi\in\mathcal{O}_k$ with $\mathrm{ord}_{\mathfrak{p}}(\xi)=d$), it gives an abelian extension tower $k'(\mathcal{F}[\mathfrak{p}^n])$ of $k$ with degree $[k'(\mathcal{F}[\mathfrak{p}^n]):k']=(q-1)q^{n-1}$ and each $k'(\mathcal{F}[\mathfrak{p}^n])/k'$ is totally ramified.

My question is, if a tower $\{k_n'\}$ is given such that for any $n$, $k_n'/k$ is abelian, $k_n'/k'$ is totally ramified, and $[k_n':k']=(q-1)q^{n-1}$, then can we find a relative Lubin-Tate formal group $\mathcal{F}$ such that $k_n'=k'(\mathcal{F}[\mathfrak{p}^n])$? If not, are there any criteria for it?

In fact I'm considering the following special case: let $K$ be an imaginary quadratic field, $p$ be a prime split in $K$, $\mathfrak{p}$ be a prime of $K$ above $p$, $H$ be the Hilbert class field of $K$, $w$ be a prime of $H$ above $\mathfrak{p}$. Let $H_n$ be the ring class field of $K$ of conductor $p^n$, then $w$ is totally ramified over $H_n/H$ and we have $[H_n:H]=2(p-1)p^{n-1}/\#\mathcal{O}_K^\times$. I'd like to know the answer of above question for $k=K_{\mathfrak{p}}\cong\mathbb{Q}_p$, $k'=H_w$, $k_n'=H_{n,w}$, i.e. whether the anti-cyclotomic tower $\{H_{n,w}\}$ comes from (relative) Lubin-Tate formal group. (Of course it's true for cyclotomic tower and the $\mathbb{Z}_p$ extension tower unramified outside $\mathfrak{p}$. And it looks like that we are in trouble when $d_K=-3,-4$.)

EDIT: If we assume $k=\mathbb{Q}_p$ (as in the special case I'm considering) and $p\geq 3$, then by the fact $\mathbb{Q}_p^{\mathrm{ab}}=\mathbb{Q}_p^{\mathrm{unr}}(\mu_{p^\infty})$ and that the open subgroup of $\mathrm{Gal}(\mathbb{Q}_p^{\mathrm{unr}}(\mu_{p^\infty})/\mathbb{Q}_p^{\mathrm{unr}})\cong\mathbb{Z}_p^\times$ of index $(p-1)p^{n-1}$ is unique, we can conclude that $\mathbb{Q}_p^{\mathrm{unr}}k_n'=\mathbb{Q}_p^{\mathrm{unr}}(\mu_{p^n})$ for any $n$. Can we obtain more information from this?

• Say, how many formal groups are there, and how many such towers are there? The formal groups will be determined by their logarithm, each of which is a countable number of coefficients, each of which in turn can be approximated by an integer while preserving the field extension tower (I think). As for the number of towers, I've heard the words "wild problem" before. Is it possible that once you allow $\zeta_p\in k$, then there are more than a countable number of towers? Since everything is abelian, maybe it's not possible... – Dror Speiser Aug 10 at 21:16

After discussing with local people, I come out with a proof of my question under the assumption

• $p\nmid\#(\mathcal{O}_k^\times)_{\mathrm{tors}}$.

For example, if $k=\mathbb{Q}_p$ and $p\geq 3$. This assumption ensures that for any $n\geq 1$, the open subgroup of $\mathcal{O}_k^\times$ of index $(q-1)q^{n-1}$ is unique.

First we choose any relative Lubin-Tate formal group $\mathcal{F}_\xi$ relative to $k'/k$ of parameter $\xi$, where $\xi\in\mathcal{O}_k$ with $\mathrm{ord}_{\mathfrak{p}}(\xi)=d$. Then $k^{\mathrm{ab}}=k^{\mathrm{unr}}(\mathcal{F}_\xi[\mathfrak{p}^\infty])$ with Lubin-Tate character $\chi:k^{\mathrm{ab}}/k^{\mathrm{unr}}\to\mathcal{O}_k^\times$ which is an isomorphism.

For any $n\geq 1$, the fields $k^{\mathrm{unr}}/k'$ and $k_n'/k'$ are linear disjoint over $k'$, since one is unramified and another one is totally ramified. Note that $k^{\mathrm{unr}}k_n'$ is a subextension of $k^{\mathrm{ab}}/k^{\mathrm{unr}}$ of degree $(q-1)q^{n-1}$, this forces that $k^{\mathrm{unr}}k_n'=k^{\mathrm{unr}}(\mathcal{F}_\xi[\mathfrak{p}^n])$ and the natural restriction map $\mathrm{Gal}(k^{\mathrm{unr}}(\mathcal{F}_\xi[\mathfrak{p}^n])/k^{\mathrm{unr}})\to\mathrm{Gal}(k_n'/k')$ is an isomorphism. Therefore $k^{\mathrm{unr}}k_\infty'=k^{\mathrm{ab}}$ and the natural map $$\mathrm{Gal}(k^{\mathrm{ab}}/k^{\mathrm{unr}}) \hookrightarrow\mathrm{Gal}(k^{\mathrm{ab}}/k') \twoheadrightarrow\mathrm{Gal}(k_\infty'/k')\qquad(*)$$ is an isomorphism.

The local class field theory assert the following diagram commutes ([1], Chapter I, Proposition 1.8) $$\begin{matrix} \mathcal{O}_k^\times & \hookrightarrow & k^\times \\ \chi^{-1}\uparrow\cong~~ & & ~~~\downarrow\mathrm{Art} \\ \mathrm{Gal}(k^{\mathrm{ab}}/k^{\mathrm{unr}}) & \hookrightarrow & \mathrm{Gal}(k^{\mathrm{ab}}/k) \end{matrix}$$ Note that $\mathrm{ord}_\mathfrak{p}(\xi)=d$, so $\mathrm{Art}(\xi)$ is in fact contained in $\mathrm{Gal}(k^{\mathrm{ab}}/k')$. Hence we can define $\alpha$ to be the image of $\mathrm{Art}(\xi)$ under the composition map $$\mathrm{Gal}(k^{\mathrm{ab}}/k')\twoheadrightarrow \mathrm{Gal}(k_\infty'/k')\xrightarrow{(*)^{-1}} \mathrm{Gal}(k^{\mathrm{ab}}/k^{\mathrm{unr}})\xrightarrow[\cong]{\chi^{-1}} \mathcal{O}_k^\times\hookrightarrow k^\times,$$ and if we define $\varpi:=\xi\alpha^{-1}$ then we have $\varpi\in\ker(k^\times\xrightarrow{\mathrm{Art}}\mathrm{Gal}(k^{\mathrm{ab}}/k)\twoheadrightarrow \mathrm{Gal}(k_\infty'/k))$.

We claim that a relative Lubin-Tate formal group $\mathcal{F}_\varpi$ relative to $k'/k$ of parameter $\varpi$ is what we want to find. In fact, by [1], page 11, paragraph 3, for any $n\geq 1$, the kernel of $k^\times\to\mathrm{Gal}(k'(\mathcal{F}_\varpi[\mathfrak{p}^n])/k)$ equals $\langle\varpi\rangle\cdot(1+\mathfrak{p}^n)$, hence the kernel of $k^\times\to\mathrm{Gal}(k'(\mathcal{F}_\varpi[\mathfrak{p}^\infty])/k)$ equals $\langle\varpi\rangle$. This forces that $k_\infty'=k'(\mathcal{F}_\varpi[\mathfrak{p}^\infty])$, and by the uniqueness of the open subgroup of $\mathcal{O}_k^\times$ of given index, we must have $k_n'=k'(\mathcal{F}_\varpi[\mathfrak{p}^n])$.

# References

[1] E. de Shalit. Iwasawa Theory of Elliptic Curves with Complex Multiplication. Academic Press, 1987.

I still want to know a proof which doesn't need the assumption.