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The multiplicative group of a local field $K$ with valuation ring $\mathcal{O}$ and residue class field of $\overline{K}$ of degree $q=p^f$ splits as

$K=\langle \pi\rangle\times \mu_{q-1}\times U^{(1)}$,

where $\mu_{q-1}$ are the $q-1$ roots of unity in $\mathcal{O}$, $\pi$ a uniformizer and $U^{(1)}$ the principal units.

I was wondering what we can say about the number of roots of unity in $K$? This boils down to finding the possible roots of unity contained in $U^{(1)}$. If $1+x\in U^{(1)}$, then

$(1+x)^n=\sum_{k=0}^n {n\choose k}x^n$,

so showing that this equals $1$ would boil down to proving that

${n\choose 1}x+{n\choose 2}x^2+\ldots+{n\choose n}x^n$

is zero. Is this ever possible? I've tried proving it, but the problem seems to be if $\pi\mid {n\choose k}$ for a lot of different $k$, so we can't necessarily show that one of the above terms would have a larger absolute value than the rest, which would imply that this is impossible.

Can anyone elaborate on the number of roots of unity?

  • 0
    I might add in the characteristic 0 case we have the exponential and logarithm function to use which can be used to show precisely that $U^{(1)}$ contains no roots of unity. In other words, I'm interested in the characteristic $p$ case.2011-09-18
  • 3
    $@$dst: your comment above is false, as Jyrki's answer shows. The simplest counterxample is that for $\mathbb{Q}_2$, $U = U^{(1)}$ so $-1 \in U^{(1)}$. But more generally, in the presence of ramification the $p$-adic logarithm will be defined only on some sufficiently small finite index subgroup of $U^{(1)}$, and this allows $p$-adic fields which are ramified over $\mathbb{Q}_p$ to have nontrivial $p$-power roots of unity.2011-09-18

2 Answers 2