# cyclic extension of prime power of a local field

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 $$l=2$$.

• 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 at 18:02
• 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 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 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 at 19:44
• Yes, adjoining the zero of a eisenstein polynomial to a local field always yields a totally ramified extension. – Estus Jan 26 at 19:47

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.

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.