$p$-adic logarithm is injective if $p > 2$?

Question

Define the $p$-adic logarithm$$\log_p(1 + x) = \sum_{i = 1}^\infty (-1)^{i-1}x^i/i.$$I know that $\log_p$ is a homomorphism from $U_1$ to the additive group of $\mathbb{Q}_p$, where $U_1$ is the subset of elements of $\mathbb{Q}_p$ of the form $1 + x$, with $|x|_p < 1$. How do I see that it is injective if $p > 2$?

Answer 1

This is a really interesting issue when you look at it from a more advanced standpoint. But here’s how you do it when you consider the log to be defined for elements $z$ of $\Bbb Q_p$ with $|z|_p<1$. Since you are restricting to elements of $\Bbb Q_p$, this means that $z=p\zeta$, where $\zeta\in\Bbb Z_p$, the $p$-adic integers.

Keeping this in mind, I’ll substitute $x=pt$ in your series expression, to give $$ \log(1+pt)=\sum_{i=1}^\infty(-1)^{i-1}p^it^i/i = p\sum_{i=1}^\infty(-p)^{i-1}t^i/i=p\biggl[t-\frac{pt^2}2+\frac{p^2t^3}3-\cdots\biggr]\,, $$ in which you see that the powers of $p$ upstairs more than compensate for the possible divisibility by $p$ of the $i$ downstairs. In other words, $\log(1+pt)=pg(t)$ where $g(t)\in\Bbb Z_p[[t]]$ with the form $g(t)=t+$(higher terms). But any such series as $g$ in this form has an inverse $h(t)\in\Bbb Z[[t]]$ in the sense that $g(h(t))=h(g(t))=t$. This is enough to make $g$ injective.

But since I am who I am, I have to tell you that restricting to inputs from $\Bbb Q_p$ misses the interesting behavior of the logarithm. In fact it makes sense to plug in for the original variable $x$ any element $z$ of an algebraic extension of $\Bbb Q_p$ with $|z|_p<1$, and now the logarithm series vanishes whenever $z+1$ is a $p$-power root of unity: the log is most emphatically not injective, even though it still is a homomorphism. For the case $p=2$, the value $z=-2$ gives the good old $2$-power root of unity $z+1=-1$, so the prime two is not so special after all.