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Let $K$ be a non archimedian local field of characteristic $p>0$ with residue field $\mathbb{F}_p$ and $l\neq p$ be a prime.

It is known by local classfieldtheory that any abelian Galois extension $L|K$ lies in $L\subset K^{nr}K_\infty$, where $K^{nr}|K$ denotes the maximally unramified extension and $K_\infty|K$ is a fixed Lubin-Tate extension, which is a tower of fields, which are totally ramified over $K$ with galois group $Gal(K_\infty|K)\cong\mathcal{O}_K^\times$.

Can this fact be used to show that any cyclic Galois extension $L|K$ with degree $l^2$ is either unramified or totally ramified, or are there examples of $L|K$ with $Gal(L|K)=\mathbb{Z}/l^2\mathbb{Z}$, such that the inertia degree $f(L|K)=l$ and the ramification index $e(L|K)=l$?

I am searching explicitly for the case that $p=3$ and $l=2$.

Estus
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  • Maybe like this in the situation $l^2\nmid p-1$: Since $\mathbb{F}p$ is the residue field of $K$, we have $Gal(K^{nr}K\infty)=\hat{\mathbb{Z}}\times\mathbb{Z}/(p-1)\mathbb{Z}\times (1+\mathfrak{m}_K)$, where $\mathfrak{m}_K$ is the maximal ideal of the integers of $K$. But $(1+\mathfrak{m}_K)$ is an abelian pro-p-group, so the only continuous morphism to a finite group with order $l^2$ is the trivial one (Is this true?). So the only surjective continuous morphism to $\mathbb{Z}/l^2\mathbb{Z}$ can come from the projection $\hat{\mathbb{Z}}\rightarrow\mathbb{Z}/l^2\mathbb{Z}$. Is this correct? – Estus Jan 26 '19 at 18:02
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    Marius, won't you have non-trivial homomorphisms from $\Bbb{Z}/(p-1)\Bbb{Z}$ to the cyclic group of order $\ell^2$ if $\ell\mid p-1$? – Jyrki Lahtonen Jan 26 '19 at 19:28
  • Yes, exactly one, if $l=2$ and $p=3$ even. So there should be two extensions with cyclic galois group of order $4$. The unramified one given by $(pr,trivial):\hat{\mathbb{Z}}\times\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Z}/4\mathbb{Z}$ and one given by $(pr,embedding):\hat{\mathbb{Z}}\times\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Z}/4\mathbb{Z}$. Maybe that is your example. But I still have the hope that there is one given by an Eisenstein polynomial or something. :) – Estus Jan 26 '19 at 19:40
  • I am approaching the limit of my understanding, but won't adjoining a zero of an Eisenstein polynomial yield a totally ramified extension? – Jyrki Lahtonen Jan 26 '19 at 19:44
  • Yes, adjoining the zero of a eisenstein polynomial to a local field always yields a totally ramified extension. – Estus Jan 26 '19 at 19:47
  • Marius: Yup, working backwards from your description of $Gal(K^{ur}K_\infty)$ made the construction of an example with $e=f=\ell$ quite straightforward. I'm a beginner in this area, so I didn't see it right away. The challenges will come when we need the Lubin-Tate part. Jonathan Lubin actually visits the site relatively regularly. So if we run into difficulties there, we may get lucky! – Jyrki Lahtonen Jan 27 '19 at 05:58
  • Anyway, I think I learned something today. – Jyrki Lahtonen Jan 27 '19 at 05:59

2 Answers2

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I think the following is an example of a cyclic extension of degree $4$ over the field $K=\Bbb{F}_3((x))$ with $e=f=2$.

Let $i$ be a solution of $i^2+1=0$ in the extension field $\Bbb{F}_9$. Let $$ u=\sqrt{(1+i)x},\quad v=\sqrt{(1-i)x}. $$ It follows that $uv=\pm ix\notin K$. Similarly $u^2,v^2\notin K$, but $u^2+v^2=-x\in K$. This implies that $$ p(T)=(T^2-u^2)(T^2-v^2)=T^4+xT^2-x^2\in K[T] $$ is irreducible. None of the zeros $\pm u,\pm v$ are in $K$, ruling out linear factors, and no product of two zeros is in $K$ ruling out quadratic factors.

The coefficients $a=x$ and $b=-x^2$ satisfy the condition described here, part (b). Neither $b=-x^2$ nor $a^2-4b=-x^2$ is a square in $K$ but their product is. Therefore the Galois group of $p(T)$ over $K$ is cyclic of order four.

We easily see that $L=K(u)$ is the splitting field. The element $i=(u^2/x)-1\in L$ as is $v=ix/u$. So $L$ has a subfield isomorphic to $\Bbb{F}_9$. This forces the inertia degree to be divisible by two. On the other hand $u^2=(1+i)x$ forces the ramification degree $e$ to be at least two also. As $[L:K]=ef=4$, the conclusion is that we must have $e=f=2$.

It is a bit unnerving that the leading coefficient of $u$ appears to be $\sqrt{1+i}\notin\Bbb{F}_9$. The explanation is surely that $\sqrt x\notin L$! Mentioning this because something similar may allow us to generalize this construction to other $(p,\ell)$ pairs? A key ingredient seems to be that $L=\Bbb{F}_9((u))$, but $u$ is not really a fractional power of the original local parameter $x$. We have this extra twist coming from the coefficient $1+i$ under the square root. Similar twisting may give us extensions with $e=f=\ell$ whenever $\ell\mid p-1$, but that needs a bit more work.

Jyrki Lahtonen
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Expanding the idea from the last paragraph of my first attempt to an example of a cyclic extension $L/K$ of characteristic $p$ local fields such that $e=f=\ell$. As Marius foresaw in the comments, we need $\ell$ to be a factor of $p-1$.

Let $K=\Bbb{F}_p((x))$. Let $E=\Bbb{F}_{p^{\ell^2}}$ be an extended constant field, and let $M=E((x^{1/\ell}))$. As we assume $\ell\mid p-1$ there exists a root of unity $\zeta$ of order $\ell$ in the prime field $\Bbb{F}_p$. The extensions $E((x))/K$ and $K(x^{1/\ell})/K$ are known to be Galois with cyclic Galois groups of respective orders $\ell^2$ and $\ell$. One is unramified and the other is totally ramified, so they are linearly disjoint. Hence their compositum $M$ is also Galois with Galois group $G=\Bbb{Z}/\langle \ell^2\rangle\times \Bbb{Z}/\langle \ell\rangle$. More precisely, the $K$-automorphism $\sigma_{a,b}$ of $M$ associated to $(a,b)\in G$ is the automorphism fully described by $$ \sigma_{a,b}(z)=z^{p^a}\ \text{for all $z\in E$},\quad \sigma_{a,b}(x^{1/\ell})=\zeta^b x^{1/\ell}. $$ Let us consider the subgroup $H\le G$ generated by $\tau:=\sigma_{\ell,1}$. Clearly $\tau$ is of order $\ell$, and $G/H\simeq \Bbb{Z}/\langle \ell^2\rangle$.

Next I want to identify the fixed field $L=M^H$. Clearly $\Bbb{F}_{p^\ell}\subset E$ is fixed by all the automorphisms in $H$. To get all of $L$ I need the following observation.

Fact. There exists an element $\eta\in E$ such that $\eta^{p^\ell-1}=1/\zeta$.

Proof. Consider the homomorphism of cyclic groups $f:E^*\to E^*$ given by $f(z)=z^{p^\ell-1}$. Because $p\equiv1\pmod\ell$ $$ N:=\frac{p^{\ell^2}-1}{p^\ell-1}=1+p^\ell+p^{2\ell}+\cdots+p^{(\ell-1)\ell}\equiv1+1+\cdots+1\equiv0\pmod{\ell}. $$ The image of $f$ is the unique cyclic subgroup of $E^*$ of order $N$. We just saw that $\ell\mid N$ implying that all $\ell$th roots of unity are in $\operatorname{Im}(f)$. QED

The rest is straightforward. With $\eta$ an element promised by the above observation let $u=\eta x^{1/\ell}$. Then $$ \sigma_{\ell,1}(u)=\eta^{p^\ell}\zeta x^{1/\ell}=\eta(\eta^{p^\ell-1}\zeta)x^{1/\ell}=u. $$ So the field $\Bbb{F}_{p^\ell}((u))\subseteq M^H$. Clearly this is a degree $\ell^2$ extension of $K$, so $L=\Bbb{F}_{p^\ell}((u))$.

Here $Gal(L/K)\simeq G/H$ is cyclic of order $\ell^2$ as prescribed. Equally clearly $e(L/K)=\ell=f(L/K)$.

This is surely a more illuminating and generalizable construction. The methods in the other answer are ad hoc.

Jyrki Lahtonen
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