For p a given prime number, define the p-adic norm | * |p as follows on...

For p a given prime number, define the p-adic norm | * |p as follows on Q: Given q in Q, we can write it as a product q = (p^m)(a/b) with a,b integers which are not divisible by p, and m an integer which is uniquely determined by q (check that m is indeed uniquely determined by q). Then define |q|p = p^(-m).

Check that Q with distance dp(q1,q2) = |q1 - q2|p is a metric space (here q1-q2 just means the usual operation of subtraction in Q). Show moreover
that |q1 + q2|p <= max(|q1|p, |q2|p).

Solutions

Expert Solution

first to see that it is a metric space,

$Now\ consider\ a,b,c\in Q. Let\ |a-b|_p=p^{-k} and |b-c|_p=p^{-j}.\\ Then,\ let\ m=min(k,j)\ so\ that\ p^m|(a-b) \ and\ p^m|(b-c) \Rightarrow p^m|(a-c) \because a-c=a-b+b-c. \\ Hence,\ the \highest\ power\ of\ p\, say\ it\ l, \such\ that\ p^l|(a-c) \ satisfies\ l\geq min(k,j). \Hence,\\ d_p(a,c)=|a-c|_p=p^{-l}\leq p^{-min(k,j)}=max(p^{-k},p^{-j})=max(|a-b|_p,max(|b-c|_p))\\ \leq |a-b|_p+|b-c|_p=d_p(a,b)+d_p(b,c)$

now for second part,

and we are done.

Colby Messinger answered 1 year ago

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