## Some more comments on week 2

There are a few more points I forgot to emphasize in class, even though they’re written in the notes.

The zeroth fact is that $x\in {\bf Z}_p$ is a unit if and only if $x\mod p$ is non-zero in ${\bf F}_p$. This is easy to check, and I leave it to you.

The first fact is that a series

$\sum_{i=k}^{\infty}x_i$

is convergent in ${\bf Q}_p$ if and only if $|x_i|\rightarrow 0$. Thus convergence is very easy to test in ${\bf Q}_p$. This is one among many aspects of ${\bf Q}_p$ that are much easier than reals. So, for example, when we proved that every element of ${\bf Q}_p$ can be written uniquely as

$\sum_{i=N}^{\infty}a_ip^i$

for $a_i\in \{ 0, 1, \ldots, p-1\}$, the convergence, that is, the fact that the series defines an element of ${\bf Q}_p$, is essentially trivial.

Another point regarding the valuation is that while

$|x+y|_p \leq max (|x|_p, |y|_p)$

is always true, in fact, if the two valuations are different, $|x|_p\neq |y|_p$, then we have equality:

$|x+y|_p = max (|x|_p, |y|_p)$

This is easy to see if one is zero, so we assume both are non-zero. Then we write

$x=p^nu$

and

$y=p^mv$

with $u,v$ units.

If the valuations are different, then we may assume, without loss of generality, that $|x|_p>|y|_p$. But this is just saying that $n. Now,

$x+y=p^nu+p^m(v)=p^n(u+p^{m-n}v).$

But $u+p^{m-n}v\equiv u \mod p$, so $u+p^{m-n}v$ is also a unit. Hence,
$|x+y|_p=|p^n|_p|(u+p^{m-n}v|_p=p^{-n}=|x|_p.$

What might happen to the sum if the valuations are the same?