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

Can we make sense of the logarithm of prime in some algebraic extension of $\mathbb{Q}_q$, where either $q \neq p$ or $p = q$ and both prime numbers?

Some reflections: A naive starting point is perhaps something like $$ \log( 1+x) = -\sum\limits_{n=1}^{\infty} \frac{(-x)^n}{n},$$ which converges in the reals in a neighborhood of $|x|_\infty \leq 1$, but $|p+1|_q$ might be small, at least for $q=p$ it is actually $|p+1|_p=1$, hence it lies at the boundary of the circle of convergence, where I assume that the radius remains by $1$ in the $q$ adic world. Are there analogues of Abel's theorem for the convergence at the boundary?

share|cite|improve this question
There is no analogue of Abel's theorem from real/complex analysis in the setting of p-adic power series since the circle |x|_p = r for r > 0 is not the boundary of the disc |x|_p < r, as the latter is already a closed set. In fact if 0 < |x|_p < r and y is sufficiently close to x (taking |y-x|_p < |x|_p is good enough) then |y|_p = |x|_p < r. Note also that, in R or C, a power series converges absolutely inside its disc of convergence but typically just conditionally on the boundary (where it happens to converge). But p-adic infinite series never converge "conditionally" [continued...] – KCd Jun 30 '11 at 9:52
since any rearrangement of a p-adic infinite series converges with the same limit as the original rearrangement. This is vaguely compatible with the meaninglessness of Abel's theorem in the p-adic setting. – KCd Jun 30 '11 at 9:53
up vote 6 down vote accepted

I think that you found the main obstacle for this to work. If you look at the exponential function $$ E(x)=1+\sum_{n=1}^\infty\frac{x^n}{n!} $$ you see that it cannot converge unless $|x|_q<1$, because in this case the denominator makes things worse (in sharp contrast to the archimedean case). Actually we need a little bit more than this, because the $q$-adic value of the factorial tends towards zero. A more careful analysis starting from the fact that $|n!|_q=q^{-t}$, where $$ t=\left[\frac{n}{q}\right]+\left[\frac{n}{q^2}\right]+\left[\frac{n}{q^3}\right]+\cdots= \sum_{k=1}^{\lceil\log_q n\rceil}\left[\frac{n}{q^k}\right] $$ reveals that we need $|x|_q<q^{-1/(q-1)}$ for $E(x)$ to converge.

So if $x\in\mathbf{Q}_q$, then we must have $x\in q\mathbf{Z}_q$ for $E(x)$to converge, and then $E(x)\equiv 1\pmod{q\mathbf{Z}_q}$. Unless I made a mistake, seeking an $x$ from an extension field is not going to change this.

So in order for the logarithm to make sense, the congruence $p\equiv 1\pmod{q}$ that you observed is necessary and sufficient. See also for more discussion, links, and workarounds.

share|cite|improve this answer

This is more or less a footnote to Jyrki's excellent answer. I find the idea of $q$-adic analysis slightly disturbing, so I'm going to switch the variable names and talk about the $p$-adic logarithm.

There is a useful field $\mathbb{C}_p$, the completion of the algebraic closure of $\mathbb{Q}_p$, which is in some sense the natural place to do $p$-adic analysis. This contains all the algebraic extensions of $\mathbb{Q}_p$, obviously.

As Jyrki remarks, the power series $$ \log(x) = \sum_{n \ge 1}\frac{(-1)^{n+1} (x-1)^n}{n} $$ converges $p$-adically for any $x \in \mathbb{C}_p$ whenever $|x - 1| < 1$. All the finite extensions of $\mathbb{Q}_p$ are closed in the topology of $\mathbb{C}_p$, so if $x$ lives in some finite subextension so does its logarithm.

You can extend the log just a bit further by using the group structure. We want to have $\log(xy) = \log(x) + \log(y)$, so any root of unity in $\mathbb{C}_p$ had better go to zero. Now, every $x \in \mathbb{C}_p \setminus \{0\}$ can be written uniquely in the form $x = p^n y z$ where $n \in \mathbb{Z}$, $y$ is a root of unity of order prime to $p$, and $|z - 1| < 1$. Thus once one decides on what $\log(p)$ should be (a "branch of the logarithm"), one has a uniquely determined logarithm map on $\mathbb{C}_p \setminus \{0\}$.

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.