TITLE: Finite extension of local fields
QUESTION [6 upvotes]: Can a (higher) local field have uncountably many finite (seperable) extensions?

REPLY [3 votes]: 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 $Q\in\Omega_P$.