Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – dst Sep 18 '11 at 17:46
$@$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. – Pete L. Clark Sep 18 '11 at 19:08

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 $\pi^{p-1}$.

share|cite|improve this answer
Added the necessary condition, $p>2$, for the $p$-adic field not to have primitive roots of order $p$. – Jyrki Lahtonen Sep 18 '11 at 18:47

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 $K$.

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.

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

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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