$p$-adic valuation.

Question

Let $\alpha_1,\alpha_2\in \mathbb Z_p$ such that $v_p(\alpha_1)<v_p(\alpha_2).$ How to prouve that $v_p(\alpha_2-\alpha_1)=v_p(\alpha_1)$ ?

I think this is a stupid question but I'm really confused, Can someone explain this to me. Thanks.

Answer 1

This works for any valuation.

$v(a_1)=v(a_1-a_2+a_2)\ge\min(v(a_1-a_2),v(a_2))=v(a_1-a_2)$ (we can't have $\min(v(a_1-a_2),v(a_2))=v(a_2)$ since then $v(a_1)\ge v(a_2)$, a contradiction). On the other side, $v(a_1-a_2)\ge\min(v(a_1),v(a_2))=v(a_1)$.

Answer 2

$\ v(\alpha_2\!-\!\alpha_1)\, =\, v(\alpha_1) +\!\!\!\!\!\!\!\! \overbrace{v\left(\frac{\alpha_2}{\alpha_1}\! -1\right)}^{\large \ge\, \min(\,>0,\ \, 0)\, \ge\, 0\ }\!\!\!\!\!\!\!\! = v(\alpha_1)\ \ \ $ $\left[ \begin{eqnarray}{\rm for\ else}\ \ &\frac{\alpha_2}{\alpha_1},\!\!\!&\ &\ \frac{\alpha_2}{\alpha_1}-1& {\rm have} \ \ v > 0\\ \Rightarrow\ \color{#c00}{\bf 1}\, =\, &\frac{\alpha_2}{\alpha_1}\!\!&-&\left[\frac{\alpha_2}{\alpha_1}-1\right]&\ {\rm has}\ \ \ \color{#c00}{v > 0}\end{eqnarray}\,\right]$