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

Question

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 $d_K=-3,-4$.)

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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?

Answer 1

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.

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