I'm studying p-adic analysis and recently I've learned about the p-adic logarithm function but I can't understand very well how the process of calculating the value should be done.
As an exercise I'm trying to find the 7-adic expansion of $\log_7 42$.
From the definition I know that in order to calculate $\log_p x$ for an arbitrary $x\in\Omega$, I have to write $x=p^r\omega(x_1)\langle x_1\rangle$ where $r=ord(x)=\frac{a}{b}$ and $p^r$ is formally a root of $x^b-p^a=0$, $\omega(x_1)$ is a $(p^f-1)$-th root of 1, and $\langle x_1\rangle$ is in the disc around 1 with radius 1, that is $|\langle x_1\rangle-1|_p<1$. Then
$$\log_p x=\sum\frac{(-1)^{n+1}(\langle x_1\rangle-1)^n}{n}.$$
Now, I have that $42=7(6)$ where $7$ is a root of $x-7=0$, the problem I facing now is that I don know how to calculate the adequate root of 1. I mean, which root to choose? Can I choose an arbitrary $f$? Like for example $f=1$ which gives me that $(-1)^{7^1-1}=1$ that is $-1$ is a 6-th root of 1, and after plugging this into the equation I get $42=7(-1)(-6)$ and $\langle x_1\rangle=-6$, hence
$$\log_p 42=\sum\frac{(-1)^{n+1}(-6-1)^n}{n}=-\sum\frac{7^n}{n}$$
I'm not sure that my answer is correct but I've checked the solution on my book and this gives the correct answer at least for the first 4 digits of the 7-adic expansion (the solution asks precisely for the first 4 digits).
I would like to know if the process is correct for the particular problem, but also what should be the approach for the general case.
I have a suspicion that my solution should be correct, because $f\cdot e=n$ where n is the degree of the extension of $\mathbb{Q}_p$ which contains $x$ and $e$ is the index of ramification. In my case (I'm not sure) I have $e=1$ and $n=1$ since $42\in\mathbb{Q}_p$, hence $f=1$ which is preceisely what I've done.
It will be great to have a practical clarification of what $e$ and $f$ could mean for a specific problem like this (more than just the index of ramification and the residue field degree).
The root of unity that you choose is the unique one that’s congruent to your unit modulo the maximal ideal. In your case of $42$, your unit is $6$, and the unique root of unity that’s congruent to it is $-1$. So your calculation is correct. It would have been much more fun if you had chosen $35=7\cdot5$; then you would have needed to find the sixth root of unity that’s congruent to $5$. You may do this computationally without Hensel just by applying $z\mapsto z^7$ repeatedly, starting with $z_0=5$. You get one more $7$-adic place of accuracy in each iteration, so it’s not as fast as Newton-Raphson.
In general if you have a finite extension $k\supset\Bbb Q_p$, then since the $\Bbb Q_p$-absolute value $|\star|_p$ extends uniquely to $k$, you form the set of all elements $z\in k$ with $|z|_p\le1$. This is the ring of integers of $k$, call it $\mathfrak o$. It has the unique maximal ideal $\mathfrak m$, consisting of all $z\in k$ with $|z|_p<1$. The field $\kappa=\mathfrak o/\mathfrak m$ is the residue field belonging to $k$, and it’s a finite extension of the corresponding construct for $\Bbb Q_p$, which is $\Bbb Z/p\Bbb Z=\Bbb F_p$. The field degree $[\kappa:\Bbb F_p]$ is your $f$. In particular, if $k=\Bbb Q_p$, $f=1$, as in your case.
The multiplicative subgroup of the reals, $|k^\times|_p$, of all the absolute values of nonzero elements of $k$, contains the corresponding group for $\Bbb Q_p$, which is the set of all powers of $p$, you might write it $p^{\Bbb Z}$. You can show easily enough that the index $(|K^\times|_p:p^{\Bbb Z})$ is finite; this is the ramification index, $e$. Once more, if $k=\Bbb Q_p$, then $e=1$. Always, no matter what, if $[k:\Bbb Q_p]=n$, you have the nice relation $ef=n$.
Examples: Let’s see what happens when you adjoin a cube root of unity to $\Bbb Q_p$ for all possible $p$’s. The $\Bbb Q$-minimal polynomial of a primitive cube root of $1$ is $\Phi_3(X)=X^2+X+1$. This is irreducible over $\Bbb Q_2$, and I hope you know that in characteristic two, the roots of this polynomial generate the field with $4$ elements. So $e=1$, $f=2$ here. For $p=3$, since the roots of $\Phi_3$ are $-\frac12\pm\frac{\sqrt{-3}}2$, you’re adjoining the square root of $-3$, which necessarily has absolute value $1/\sqrt3$, so you have $e=2$, $f=1$. This is the only $p$ for which $e\ne1$. For all other primes $p\ge5$, it’s just a matter of the congruence of $p$ modulo $3$, since if $p\equiv1\pmod3$, then $p-1$, the number of nonzero elements of $\Bbb F_p^\times$, is divisible by $3$, so that there already are cube roots of unity in $\Bbb Q_p$. You can calculate them by the method I mentioned above, but Newton-Raphson works far faster. On the other hand, if $p\equiv-1\pmod3$, then $3$ does not divide the number of nonzero elements of $\Bbb F_p$, and you need to make a quadratic extension to get them. Thus in this case, $f=2$, $e=1$.