Roots of unity in a local field

Question

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?

Answer 1

As you noticed, the first term ${n \choose 1}x$ dominates, unless $p\mid n$. Actually any root of unity $u=1+x$ in $U^{(1)}$ must be of order that is a power of $p$, for otherwise a suitable power of $u$ would have order prime to $p$, and we just saw that this cannot happen.

Such roots of unity may or may not exist in your field. For example the $p$-adic field $\mathbf{Q}_p$ has no roots of unity of order $p$, if $p>2$, because the term ${p\choose 1}x$ dominates (all the binomial coefficients save the last one are divisible by $\pi=p$ exactly once, and also $p\mid x$, so that last term $x^p$ is then also divisible by a higher power of $p$ than the first term).

But surely you can adjoin a $p$th root of unity $\zeta_p$ to $\mathbf{Q}_p$! If you do that you get a ramified extension of $\mathbf{Q}_p$. You have probably seen the same thing happen with the cyclotomic extension $\mathbf{Q}(\zeta_p)/\mathbf{Q}$ of algebraic number fields. There the prime $p$ is totally ramified.

Note also that the above argument showing that $(1+x)^p\neq1$ for a non-zero $x\in p\mathbf{Z}_p$ fails, when there is ramification. The last term $x^p$ may then be divisible by the same power of $\pi$ as the ( in the earlier case dominating) first term $px$. In the case $\mathbf{Q}_p(\zeta_p)$ we can thus deduce that $p$ must be divisible by $\pi^{p-1}$.

Answer 2

As has been observed, the roots of unity inside $U^{(1)}$ are precisely the $p$th power roots of $1$ in your field. In particular, in the case when your field has char. $p$, there are no non-trivial such roots of unity, and you are done.

In the case when your field $K$ is of char. $0$, so a finite extension of $\mathbb Q_p$, then if $d := [K:\mathbb Q_p]$, then $U^{(1)} \cong \mu \times \mathbb Z_p^d$, where $\mu$ is the group of $p$th power roots of unity in $K$.

Since $\mathbb Q_p(\zeta_p)$ is totally ramified over $\mathbb Q_p$, you can bound the size of $\mu$ by bounding the ramification of $K$, or by bounding the discriminant of $K$.

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If you want to compute the exact size of $\mu$, here is one way (there may well be more efficient ways):

first of all, use a ramification or discriminant bound to find a power of $p$, say $p^N$, so that $\mu$ has size bounded by $p^N$. Now we see that $$U^{(1)}/(U^{(1)})^{p^N} \cong \mu \times (\mathbb Z/p^N)^d,$$ and so we can compute the exact size of $\mu$ by computing the size of the LHS of this isomorphism.

Now (by considering the binomial theorem, or similar tools) one can find $M$ such that $U^{(M)} \subset (U^{(1)})^{p^N}$. Thus, if we let $U' := U^{(1)}/U^{(M)},$ then $$U^{(1)}/(U^{(1)})^{p^N} \cong U'/(U')^{p^N}.$$ Now $U'$ is a finite group, and so you can compute the RHS of the preceding isomorphism explicitly in any given case. So we are done.

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If you wanted to try this out, you could try it in the case $K = \mathbb Q_2(i,\sqrt{2}).$

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Incidentally, the above algorithm involves the kinds of computations that I strongly recommend doing if you want to get a good feeling for the multiplicative structure of $p$-adic fields. (I give these sorts of problems to my students when they are trying to learn this material.)