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I'm reading Number Theory I, by Kato, et al. The authors claim that for $x$, $y\in \mathbb{Q}$ and $m$ a square free positive integer, if $2x$, $x^2-my^2 \in \mathbb{Z}$, then for any odd prime $p$ $\mathrm{ord}_p(m)+2\mathrm{ord}_p(y) \geq 0$.

I get that $\mathrm{ord}_p(my^2)=\mathrm{ord}_p(m)+2\mathrm{ord}_p(y)$ and $\mathrm{ord}_p(x)\geq 0$. I'm just not sure that I can say that $\mathrm{ord}_p(x^2-my^2)=\min\{\mathrm{ord}_p(x^2), \mathrm{ord}_p(my^2)\}$.

I'm having a bit of trouble seeing this. Does anyone know what to do here?

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up vote 1 down vote accepted

HINT $\rm\ \ 2x\in \mathbb Z\ \Rightarrow\ ord_p(x)\ge 0\ \ $ so $\rm\ ord_p (x^2-my^2) \ge 0\ \Rightarrow\ ord_p(my^2) \ge 0$

Recall the principle of domination: $\rm\ ord_p\ Y < ord_p\ X\ \Rightarrow\ ord_p\:(X+Y) = ord_p\ Y\:$

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I see your reasoning here but I was under the impression that $ord_p(a+b)=min (ord_p(a),ord_p(b))$ only when $ord_p(a)\neq ord_p(b)$ and it seems that this may happen in this situation. In any case, if we are allowed this equality, then I dont think we need $ord_p(x)\geq0$. – Jason Smith Nov 18 '10 at 0:04
@Jason: You do need it, e.g. x = y = 1/p, m = 1 – Bill Dubuque Nov 18 '10 at 0:16
Thanks a lot. Where could I find a good treatment of the basics of Ord_a(x)? – Jason Smith Nov 18 '10 at 0:17
@Jason: That depends much on your background, e.g. do you already know some ring theory? I recommend that you ask that as a new question since it would be helpful to get answers from folks of all different levels. – Bill Dubuque Nov 18 '10 at 0:30

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