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Let $n$ be a integer not equal to zero, $p$ a prime and $$|n|_p = p^{-\operatorname{ord}_p(n)}; |0|_p=0$$

Proposition: $|xy|_p=|x|_p|y|_p$

We want to show: $\operatorname{ord}_p(ab)=\operatorname{ord}_{p}(a)+\operatorname{ord}_p(b)$ So if we suppose there is a largest integral $v$ such that $p^{v}|(ab)$, then we can split up v such that $p^{v_1}|a,p^{v_2}|b$. If we suppose that there exist $v_1,v_2$ with them both being the largest integers such that $p^{v_1}|a, p^{v_2}|b$, with $v_1+v_2= : v $ then $p^v|ab$. (I don't believe this to be a formal proof but rather my best own "idea" I can come up with...)

The script also mentions that the padic valuations' sums are bounded: $|x+y|_p\le \sup\{|x|_p,|y|_p\}$, how to show this?

Edit:

According to Robert Israel and Bill Dubuques answers, the first proposition is proven, if I understood correctly:

Given $ord(a)= v$ and $ord(b) = u$ which is the same as $a:= p^{v}m, b:= p^{u}n$ with $gcd(m,p)=gcd(n,p)=1$, because $ab=p^{u+v}mn$ with $gcd(mn,p)=1$ it follows that $ord(ab)=ord(a)+ord(b)$ and so $|xy|_{p}=p^{-ord(xy)}=p^{-(ord(x)+ord(y))}=|x|_{p}|y|_{p}$

Now to show: $|x+y|_{p}\le sup\{|x|_{p},|y|_{p} \}$

and the second can be tackled like this: set $a:=p^{v}m. b:=p^{u}n$, then checking two cases $1:(u=v)$ and $2:(v<u)$: $(a+b)=p^{v}m+p^{u}n= p^{v}(m+p^{u-v}n)$ so in case of (u=v) it follows that $ord(a+b)= u = sup\{ord(a),ord(b)\}$

and in case of $(v<u)$ because the biggest v is $v:= u-1$: $ord(a+b)\le p^{u-1}(m+pn) \le p^{u}(m+n)= sup\{ord(a),ord(b)\}$

It seems that I did something wrong in both cases...

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Notes: $\newcommand{\ord}{\operatorname{ord}}$Your second part is equivalent to proving that \\[ \ord_p(x + y) \geq \min(\ord_p(x), \ord_p(y)). \\] In fact, equality holds if $\ord_p(x)$ and $\ord_p(y)$ are not equal. See Prof Israel's answer. –  Dylan Moreland Mar 4 '12 at 18:52
    
Thank you Dylan Moreland. However, I don't get to this conclusion. –  VVV Mar 5 '12 at 12:24
    
I don't think you can conclude, in the case of $u = v$, that the order is $u$. It could be the case that $m + n$ is divisible by $p$. For example, with $p = 2$ and $u = v = 0$ one could have $x = y = 1$, whereupon $\operatorname{ord}_p(x + y) = 1$. –  Dylan Moreland Mar 5 '12 at 16:37

2 Answers 2

up vote 4 down vote accepted

It's very simple. If $a = p^u m$ and $b = p^v n$ with $(m,p) = (n,p) = 1$, then $ab = p^{u+v} m n$ with $(mn,p) = 1$.

For the other question, if $u \le v$ then $a + b = p^u (m + p^{v-u} n)$ so $\ldots$

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Thank you Robert Israel. –  VVV Mar 5 '12 at 12:27

Hint $\rm\ (p^i m)\:(p^j n)\ =\ p^{i+j}\:mn\ $ and $\rm\ p\nmid m,n\ \Rightarrow\ p\nmid mn\:$ by $\rm\:p\:$ prime.

This is equivalent to uniqueness of prime factorizations.

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Thank you Bill Dubuque. –  VVV Mar 5 '12 at 12:24

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