Combining p-adic metrics

**SlipEternal** · May 22nd 2014, 09:38 AM

Let $p,q$ be two prime numbers. Define $d:\mathbb{Q} \times \mathbb{Q} \to [0,\infty)$ by $d(x,y) = \max\{|x-y|_p,|x-y|_q\}$ . Is there an algebraic analogue to the completion of the rationals by this metric similar to the inverse limit of $\Bbb{Z}/p^n\Bbb{Z}$ for the completion of the rationals by a single p-adic metric? If not, is there a way to combine the p-adic and q-adic metrics that is more algebraically intuitive?

**SlipEternal** · Jun 5th 2014, 05:10 PM

Ok, I had another thought about this. Given any integer, we can write it as $x = p^a k$ for some nonnegative integer $a$ and an integer $k$ not divisible by $p$ . Then, the map $x \mapsto (k\pmod{q^n}, p^a)$ along with the maps $(k\pmod{q^n}, p^a) \mapsto (k\pmod{q^m}, p^a)$ allows us to take the inverse limit, which yields $\Bbb{Z}_q \times \{p^n \mid n \in \Bbb{N}^*\}$ . There is a natural injection $\Bbb{Z} \hookrightarrow \Bbb{Z}_q \times \{p^n \mid n \in \Bbb{N}^*\}$ .

Next, define $d: \Bbb{Q}\times \Bbb{Q} \to [0,\infty]$ by

$d(x,y) = \begin{cases}|x-y|_q & \text{if }|x|_p = |y|_p \\ \infty & \text{otherwise}\end{cases}$

It is easy to see that this is a metric. Since $\{x \in \Bbb{Q} \mid |x|_p = 1\}$ is dense in $\Bbb{Q}$ , it seems like this completion gives a copy of $\Bbb{Q}_q$ for each nonzero power of $p$ (or am I mistaken?) and again, there seems to be a natural injection $\Bbb{Q} \hookrightarrow \Bbb{Q}_q \times \left\{p^n \mid n\in \Bbb{Z}\right\}$ (again, unless I am mistaken, and the completion of the rationals with respect to that metric produces something different).

So, this seems like the analytic completion of the rationals and the algebraic completion of the integers with respect to two primes simultaneously that I was hoping for (at least, in some form). Am I correct? It is entirely possible that this does not work as I hope.

**SlipEternal** · Jun 5th 2014, 05:38 PM

I think this still doesn't quite work, but I hope I am getting closer to something that will.

How about this:

Let $K_p = \{x \in \Bbb{Q}: |x|_p = 1\}\cup \{0\}$ . Then, I can define

$f:\Bbb{Q} \to K_p \times \{|x|_p: x\in \Bbb{Q}\}$ by

$f(x) = (x|x|_p, |x|_p)$

Then, I can complete each $K_p$ with respect to the q-adic metric. That might be more what I was looking for.

Perhaps I should explain a little more. I just want to be able to complete the rationals with respect to the $q$ -adic metric (algebraically or analytically), but I want every element to have a defined $p$ -adic norm.

**SlipEternal** · Jun 5th 2014, 06:20 PM

Maybe my attempt is doomed. I tried seeing how addition might work, and with the algebraic attempt, I'm not sure if addition works the way I want (I want normal integer addition and addition in my inverse limit to commute).

**SlipEternal** · Jun 6th 2014, 09:53 AM

I think I figured it out. Let $p,q$ be primes with $p$ a primitive root modulo $q^n$ for all positive integers $n$ . Let $X_1 = \Bbb{Z}\setminus q\Bbb{Z}$ and $X_2 = X_1\setminus p\Bbb{Z}$ . Now, $X_1$ is the set of all integers that are units of $\Bbb{Z}/q^n\Bbb{Z}$ for all $n$ , and $X_2$ is the set of all integers such that for all $x \in X_2$ , $|x|_p|x|_q = 1$ .

The idea is, for each positive integer $n \in \Bbb{N}$ and each $x_1\in X_1$ , there exists $x_2 \in X_2, k\in \Bbb{Z}$ such that $0\le k < (q-1)q^{n-1}$ and $x_1 \equiv x_2p^k \pmod{q^n}$ , and moreover, $x_1|x_1|_p \equiv x_2 \pmod{q^n}$ . So, I am hoping to build a function that will essentially be the p-adic norm on q-adic units. I am still not quite sure how to develop it further, but this looks to be a promising start.

**SlipEternal** · Jun 6th 2014, 12:50 PM

Ok, I had more on paper, and now I realize why. I need a function that takes integers not divisible by p to ones that are divisible by p, but not q. So, for each $1 \le i < p$ , define a function $k_i: \Bbb{Z}\setminus p\Bbb{Z} \to X_1$ by $k_i(x) = (qx+i)|qx+i|_p$ . Then $x = \dfrac{p^r k_i(x) - i}{q}$ , so $q$ divides $p^rk_i(x)-i$ , but won't divide $k_i(x)$ . That's what I was missing. So, for what values of $x$ will $qx+i$ be divisible by p? I need $x \equiv -\dfrac{i}{q}\pmod{p}$ , so I think I need $q$ to be a primitive root modulo $p$ (it doesn't need to be one for higher powers, though).

**SlipEternal** · Jun 12th 2014, 08:35 PM

I now understand that I can define a $q$-adic valuation for $p$-adic numbers that would agree on all of $\Bbb{Q}$, but I likely cannot construct such a valuation. If I consider the algebraic closures of $\Bbb{Q}_p, \Bbb{Q}_q$, their closures are both isomorphic to $\Bbb{C}$.

Hence, if I choose particular embeddings $\Bbb{Q}_p \hookrightarrow \overline{\Bbb{Q}_p} \cong \Bbb{C}$ and $\Bbb{Q}_q \hookrightarrow \overline{\Bbb{Q}_q}\cong \Bbb{C}$, then I can define a $q$-adic valuation on $p$-adic numbers. This valuation will not be unique. It will depend entirely on my choice of embeddings for the p-adics and q-adics into $\Bbb{C}$.

What I was attempting may have yielded a constructive method for generating a $q$-adic valuation on $\Bbb{Q}_p$, but it is simply not worth it, since any such valuation would not be unique.

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