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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?

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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

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up vote 5 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 http://en.wikipedia.org/wiki/P-adic_exponential_function for more discussion, links, and workarounds.

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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 $\mathbb{C}_p \setminus \{0\}$.

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