So $p =ax+p^2y$ and $p^2 =bz+p^3w$
$ax= p-p^2y$ and $bz= p^2-p^3w$
$axbz= p^3-p^4w-p^4y+p^5wy$
$abxz= p^3-p^4(w+y-pwy)$
$abxz+p^4(w+y-pwy) = p^3$
How can I say that $\gcd(ab,p^4) = p^3$ ?
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Here's a MathJax tutorial :) – Shaun Apr 04 '20 at 14:21
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The redaction is really not so good (and not smooth to read). – Surb Apr 04 '20 at 14:26
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It's simply $,\nu(a) = 1,\ \nu(b) = 2,\Rightarrow, \nu(ab) = \nu(a)+\nu(b) = 1 + 2\ $ $\phantom{}$ by the linked additivity property of the $p$-adic valuation. – Bill Dubuque Apr 04 '20 at 15:40
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Just do it. $a=mp$ where $\gcd(m,p)=1$ and $b=np^2$ where $\gcd(n,p)=1$. ANd so $ab=mnp^3$ and $\gcd(mn,p)=1$ so $\gcd(ab,p^k)=p^3$ if $k\ge 3$. – fleablood Apr 04 '20 at 16:10
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$\newcommand{\gcd}{\text{gcd}}$
If $$\gcd(a,p^2)=p\quad \text{and}\quad \gcd(b,p^3)=p^2,$$ then $$a=p\alpha ,\quad \text{where }\gcd(\alpha ,p)=1,$$ and $$b=p^2\beta ,\quad \text{where } \gcd(\beta ,p)=1.$$
Therefore $$ab=p^3\alpha \beta ,$$ where $p\nmid \alpha \beta $. And thus, the claim follow.
Surb
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The problem in what you wrote in your question it's you use a wrong result that is : "if there are $u,v\in\mathbb Z$ s.t. $au+bv=k$ then $\gcd(a,b)=k$". This is unfortunately not correct. Nevertheless, if you can prove that $k$ is the biggest integer s.t. there are $u,v\in\mathbb Z$ s.t. $au+bv=k$, then indeed $gcd(a,b)=k$ and your proof is fine. In your case, it's rather straightforward. Here, I just proposed a way to avoid this problem. @AndiDanuSetiawan – Surb Apr 04 '20 at 14:35
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No, it's not $gcb(a,b)=p^3$ but $gcb(ab,p^4)=p^3$. You have that $abu+p^4v=p^3$. Now, if $gcd(ab,p^3)>p^3$, then necessary, $gcd(ab,p^4)=p^4$, i.e. $p^4\mid ab$. So know you have to justify that $p^4\nmid ab$, and then you can conclude that $p^3$ is the biggest integer s.t. $abu+p^4v=p^3$ for some $u,v\in\mathbb Z$ and thus that $gcd(ab,p^4)=p^3$.@AndiDanuSetiawan – Surb Apr 04 '20 at 14:45
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@AndiDanuSetiawan More generally and conceptually this may be viewed as a special case of the additivity of the $p$-adic valuation, see the linked dupe. – Bill Dubuque Apr 04 '20 at 15:45
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