Finite extension of local fields

Question

Can a (higher) local field have uncountably many finite (seperable) extensions?

Answer 1

Let $k$ be any field and $K=k((x,y))$ the fraction field of the ring $k[[x,y]]$ of formal Laurent series in two variables. Then Harbater and Stevenson proved that the absolute Galois group of $K=k((x,y))$ is quasi-free of rank equal the cardinality of $m={\rm card}(K)>\aleph_0$. In particular, for any nontrivial finite group $G$, there exist $m$ distinct Galois extensions of $K$ of Galois group $G$. In fact, these extensions can be chosen to be linearly disjoint (since the absolute Galois group is even "semi-free", as was shown by Harbater-Haran and myself).

In particular if one takes $k$ to be a finite field, then $K$ is local field of dimension $2$.

Answer 2

Yes. For an example, let $k$ be an uncountable algebraically closed field of characteristic $p>0$, and let $K = k((x))$, the field of fractions of the ring of formal power series $\mathcal{O}_K = k[[x]]$. Consider the extensions $K_a = K[t]/(t^p - t - ax^{-1})$ for different nonzero $a\in k$. These are separable, totally ramified degree $p$ extensions of $K$ with Galois group $\mathbb{Z}/p\mathbb{Z}$. They correspond to etale $\mathbb{Z}/p\mathbb{Z}$-torsors over $\eta := Spec (K)$, which are classified by the etale cohomology group $H^1(\eta_{et}, \mathbb{Z}/p\mathbb{Z})$ (= the Galois cohomology $H^1(Gal(\bar K/K), \mathbb{Z}/p\mathbb{Z})$). To compute this group, we use the Artin-Schreier sequence: $$ 0\to \mathbb{Z}/p\mathbb{Z}\to K \stackrel{1-F}{\to} K \to 0 $$ where $F$ is the Frobenius. Thanks to Hilbert 90, $H^1(K)=0$, and so the associated long cohomology exact sequence shows that $H^1(\mathbb{Z}/p\mathbb{Z}) = coker(1-F:K\to K)$. Via this isomorphism, the class corresponding to $K_a$ is identified with the class of $ax^{-1}$ modulo the image of $1-F$. As these will be different for different $a$, we get uncountably many non-isomorphic extensions.

Answer 3

Here is a proof that $K=\mathbf{F}_q((t))$ admits only countably many extensions in each finite degree. Here $q$ is a power of a prime $p$. It is enough to prove this countability result for minimal extensions $K'\subset K$, i.e., those with no intermediate field between $K'$ and $K$. Indeed, suppose the result is proved for minimal extensions and let $d$ be minimal such that for some $q$, $K$ admits uncountably many non-isomorphic extensions $(K_i)$ of degree $d$. Then all but countably of those admit a minimal subextension $L_i$ of some degree $1<d'<d$, which, by the minimal case have to form countably many classes, and by induction there are only countably many extensions of degree $d/d'$ for $L_i$, which yields a contradiction.

Now let us prove the countability result in the minimal case. Given $K'\supset K$, we have $K'=K[x]$ for some $x$ (because $K'$ is minimal), say with $x$ of degree $d\ge 2$ over $k$, with monic minimal polynomial $P$. If $P$ has the form $Q(X^p)$, then $K[x^p]$ generates a subfield of degree $d/p$ and hence by minimality $d/p=1$, so $K'=\mathbf{F}_q((t^{1/p}))$. Now suppose that $P$ does not have the form $Q(X^p)$. Hence $P'\neq 0$. Therefore since $P$ is the minimal polynomial of $x$, we deduce that $P'(x)\neq 0$ and hence $K'\supset K$ is separable.

Given $d\ge 0$, let $\mathcal{P}_d$ be the affine space of monic polynomials of degree $d$ in $K[X]$. Now I want to prove the following: (*) given any separable irreducible $P\in\mathcal{P}_d$, there exists a neighborhood $\Omega_P$ of $P$ in $\mathcal{P}_d$ such that for every $Q\in\Omega_P$ is irreducible and satisfies $K[X]/Q\simeq K[X]/P$.

This is enough: indeed it shows that the subset $\mathcal{P}_d^{m,s}$ of monic separable polynomials is open and that the map on $\mathcal{P}_d^{m,s}$ defined by $Q\mapsto K[X]/Q$, valued in isomorphism classes of extensions of $K$, is locally constant, and hence has at most countably many values (because an open subset of $\mathcal{P}_d$ cannot have uncountably many disjoint nonempty open subsets).

So we have to prove (*). Write $K'=K[X]/P$, $\hat{K}$ and algebraic closure of $K'$ and let $x=x_1,\dots,x_d$ be the distinct conjugates of $x$ in $\hat{K}$, and $r=\inf_{1\le i<j\le k}|x_i-x_j|$. Let $\Omega_P$ be the set of $Q\in\mathcal{P}_d$ such that $|Q(x_i)|<r^d$ for all $i$; this is a neighborhood of $P$. I claim that every $Q\in\Omega_P$ has a root in the open $r$-ball around $x_i$ for every $i$. Indeed, write $Q=\prod(X-x'_i)$ in $\hat{K}[X]$. Then if we fix $i$ and we assume by contradiction that $|x_i-x'_j|\ge r$ for all $j$, we deduce that $|Q(x_i)|\ge r^d$, a contradiction. Hence $Q$ has a root $x'_i$ with $|x'_i-x_i|<r$, and these roots have to be pairwise distinct, and hence unique because $Q$ has at most $d$ roots. Krasner's lemma (thanks to Vesselin Dimitrov for the link) implies that $K[x_1]\subset K[x'_1]$. This first proves that $K[x'_1]$ has dimension (at least) $d$, which proves that $Q$ is irreducible for every $Q\in\Omega_P$. Since they have the same dimension, this is an equality: $K[x_1]=K[x'_1]$. Thus $K[X]/P$ and $K[X]/Q$ are isomorphic as $K$-fields for all $Q\in\Omega_P$.