0
$\begingroup$

Let $p$ and $\ell$ be distinct rational primes. Note that the unit group of the finite field $\mathbb{F}_\ell$

is of order $\ell-1$, hence there is the probability of finding a $p$-quotient from $\mathbb{F}_\ell^\times$.

When taking the infinite product $\prod_{\ell\neq p}\mathbb{F}_\ell^\times$, how large could its maximal pro-p quotient be? And do we have a comuptable bound for the p-valuation of $\ell -1$ when $\ell$ tends to infinity?

(My original question was about pro-p quotient of the tame inertia group of a p-adic local field, which didn't make sense for obvious reason (pointed out as below), and I propose the replace it with this one)

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ Tame inertia is prime-to-$p$ by definition, and is isomorphic to $\prod_{\ell \neq p} \mathbb Z_{\ell}$. $\endgroup$
    – Emerton
    Mar 3, 2011 at 8:33
  • $\begingroup$ And do we have a comuptable bound... As there is an $l$ which is $\equiv1\pmod{p^n}$ for every $n>0$, how can there be such a bound ? $\endgroup$
    – Chandan Singh Dalawat
    Mar 3, 2011 at 11:55
  • 1
    $\begingroup$ The answer to your new first question is "infinite". An easy computable upper bound for your second question is $\log \ell/\log p$. Standard conjectures about the asymptotic behavior of primes suggest that this bound is sharp up to subtraction by a constant. I am thinking of closing this question. $\endgroup$
    – S. Carnahan ♦
    Mar 3, 2011 at 13:45

1 Answer 1

Reset to default
2
$\begingroup$

I think the tame ramification group is isomorphic to $\prod_{\ell\neq p}\mathbb{Z}_\ell$, not $\prod_{\ell\neq p}\mathbb{Z}_\ell^\times$. See for example, pp.12 of the book "Gauss sums, Kloosterman sums, and Monodromy Groups" by N. Katz.

.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ The structure of the maximal tamely ramified extension of a local field was determined before N. Katz was born. See for example Hasse's Number Theory (Chapter 16) or jstor.org/stable/1992998 $\endgroup$
    – Chandan Singh Dalawat
    Mar 3, 2011 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.