2
$\begingroup$

In his ''Course in Arithmetic'' J.P. Serre defines the p-adic inters $\mathbb{Z}_p$ as the projective limit of the finite rings $\mathbb{Z}/p^n\mathbb{Z}$. Then he shows that $\mathbb{Z}_p$ is an integeral domain and defines the field $\mathbb{Q}_p$ as the field of fractions of $\mathbb{Z}_p$.

My Question is wheter one could define the topology on $\mathbb{Q}_p$ without using valuations and metrics as follows: Endow $\mathbb{Q}_p$ with the finest topology such that the inclusion $\iota: a \mapsto \frac{a}{1}, \mathbb{Z}_p \rightarrow \mathbb{Q}_p$ is continous. Concretey, declare a subset $U$ of $\mathbb{Q}_p$ to be open, if and only if $\iota^{-1}(U)$ is open in $\mathbb{Z}_p$.

  1. Does this give the ''right'' topology on $\mathbb{Q}_p$?
  2. Does this topology turn $\mathbb{Q}_p$ into a topological ring?
  3. Is $\mathbb{Q}_p$ localy compact with respect to this toplogy?

Of course, if the answer to 1 is yes, then the answer to 2 and 3 is also yes.

  • 0
    A sideways approach is to note the map $\mathbb Z_p\to \mathbb p^n\mathbb Z_p$ is a homeomorphism, and thus for each map $f_n:\mathbb Z_p\to\mathbb Q_p$ of the form $f_n(x)=p^{-n}(x)$, the images of $f_n$ are an increasing inclusion of spaces, all homeomorphic to $\mathbb Z_p$, and the topology on $\mathbb Q_p$ is the direct limit of the topological spaces on $p^{-n}\mathbb Z_p$. – 2017-02-22

1 Answers 1

1

No, no and yes.

(1) It is not the usual topology on $ \mathbf Q_p $, because the topology you describe is essentially the topology in which $ \mathbf Z_p $ as a subspace of $ \mathbf Q_p $ has its usual topology; and any singleton which is not in $ \mathbf Z_p $ is open.

(2) It doesn't - it doesn't even make it a topological group. Indeed, pick any $ p $-adic number that is not a $ p $-adic integer, say $ z $, and consider the map $ x \to x + z $. $ \{ z\} $ is open, but its preimage $ \{0\} $ is not.

(3) Yes, because any point not in $ \mathbf Z_p $ is open as a singleton, thus trivially has a compact neighborhood; and anything in $ \mathbf Z_p $ has the compact neighborhood $ \mathbf Z_p $.

  • 0
    @ThomasAndrews The preimage of any such singleton under the given inclusion is the empty set, which is clearly open in $ \mathbf Z_p $. – 2017-02-22
  • 0
    Oh, I misread the definition - I thought he wanted all maps $f_q:\mathbb Z_p\to\mathbb Q_p$ defined as $f_q(x)=x/q$ as continuous. – 2017-02-22
  • 0
    @ThomasAndrews Just requiring it for $ q $ equal to some divergent sequence of the powers of $ p $ is sufficient to ensure that the resulting topology on $ \mathbf Q_p $ is its usual topology. – 2017-02-22
  • 0
    Well, $p,p^2,\dots$ isn't divergent, but I know what you mean. – 2017-02-22
  • 0
    @ThomasAndrews It is divergent in $ \mathbb R $, duh... – 2017-02-22
  • 1
    You actually only need, of course, that the map $x\to\frac{x}{p}$ is continuous from $\mathbb Z_p\to\mathbb Q_p$. – 2017-02-22