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<m. 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?

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