4
$\begingroup$

In a comment to this question, David Loeffler asked if one can show that the (local) Artin map $$K^\times \to G_K^{ab}$$ does not have a lift to $G_K$. Probably this wouldn't be canonical, but I can't show that such a lift doesn't exist.

So does there exist such a lift?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

No, there does not, except possibly a few small corner cases. The problem is finding somewhere for the torsion in $K^\times$ to go.

If $K$ is a finite extension of $\mathbf{Q}_p$, then there exists a number field $\mathscr{K}$ and a prime $v$ of $\mathscr{K}$ above $p$ such that $\mathscr{K}_v = K$; and hence we obtain an embedding $G_K \hookrightarrow G_{\mathscr{K}} \hookrightarrow G_{\mathbb{Q}}$. But $G_{\mathbb{Q}}$ is known to have no nontrivial elements of finite order except the conjugacy class of complex conjugation of order 2 (see this question).

So $K^\times$ cannot embed into $G_K$ unless $(K^\times)_{\mathrm{tors}}$ has order exactly 2, which can only happen if $p = 2$ or $p = 3$ (and I am too tired this evening to think about those cases).

Improve this answer
$\endgroup$
2
  • $\begingroup$ thanks for the question and the answer! :) $\endgroup$
    – curious math guy
    Commented Aug 29, 2022 at 19:32
  • 1
    $\begingroup$ You can cover the case of all primes using some finer versions of Artin-Schreier theorem, see e.g. Theorem 3.1 here. If $G_K$ has an element of order $2$ (or any other finite order other than $1$), then its fixed field would satisfy the conditions of the theorem, implying that sums of squares in $K$ don't vanish. But this doesn't hold for any local field, since some negative integers will be squares. $\endgroup$
    – Wojowu
    Commented Aug 29, 2022 at 20:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .