Why is $w_p(x+y) \geq m$ and not $w_p(x+y) = m$ for $x,y \in \mathbb{Q}_p$?

[evinda]

Hello! (Wave)

The additive $p-$ adic valuation of $\mathbb{Q}_p$:

$$w_p: \left\{\begin{matrix}
\mathbb{Q}_p \rightarrow \mathbb{Z} \cup \{\infty\}\\
p^m u \mapsto m\\
0 \mapsto \infty
\end{matrix}\right.$$

$$\forall x,y \in \mathbb{Q}: w_p(x+y) \geq \min \{ w_p(x), w_p(y)\}$$

If $w_p(x) \neq w_p(y)$, then the equality stands.

This is the proof, according to my notes:

$$x=p^m u_1 | u_1 \in \mathbb{Z}_p^*$$

$$y=p^n u^2 | u_2 \in \mathbb{Z}_p^*$$

$$m,n \in \mathbb{N}$$

Without loss of generality, we suppose that $m \leq n$.

$$x+y=p^m(u_1+p^{n-m}u_2)$$

$$w_p(x+y) \geq m$$

If $n>m$, then $u_1+p^{n-m}u_2 \in \mathbb{Z}_p^*$

In this case:

$$w_p(x+y)=m=\min \{ w_p(x), w_p(y) \}$$

If $n=m$, $\displaystyle{ w_p(x+y) \geq n=m=\min \{ w_p(x), w_p(y)\} }$

Could you explain me why it is : $w_p(x+y) \geq m$ and not $w_p(x+y)=m$ ? (Thinking)

 

[mathbalarka]

First of all, your notation sucks.

$p$-adic valuation is usually denoted by $\nu_p(\bullet)$, and the $p$-adic norm as $|\bullet |_p$.

We know by definition that $x = p^{\nu_p(x)} \cdot a$ and $y = p^{\nu_p(y)} \cdot b$. Now $x + y = p^{\nu_p(x)} \cdot a + p^{\nu_p(y)} \cdot b$. Without loss of generality, assume $\nu_p(x) > \nu_p(y)$, then $x + y = p^{\nu_p(x)} \cdot (p^{\nu_p(x)-\nu_p(y)} \cdot a + b)$. Now this factor $p^{\nu_p(x)-\nu_p(y)} \cdot a + b$ might be divisible by $p^k$ for some $k > 0$, who knows? So the largest $n$ such that $x + y$ is divisible by $p^n$ is *at least* $\nu_p(x)$, but it might be larger. Hence $\nu_p(x + y) \geq \nu_p(x) = \text{min}(\nu_p(x), \nu_p(y))$. From this one derives that $|x + y|_p \leq \text{max}(|x|_p, |y|_p)$.

 

[evinda]

$p$-adic valuation is usually denoted by $\nu_p(\bullet)$, and the $p$-adic norm as $|\bullet |_p$.

We know by definition that $x = p^{\nu_p(x)} \cdot a$ and $y = p^{\nu_p(y)} \cdot b$. Now $x + y = p^{\nu_p(x)} \cdot a + p^{\nu_p(y)} \cdot b$. Without loss of generality, assume $\nu_p(x) > \nu_p(y)$, then $x + y = p^{\nu_p(x)} \cdot (p^{\nu_p(x)-\nu_p(y)} \cdot a + b)$. Now this factor $p^{\nu_p(x)-\nu_p(y)} \cdot a + b$ might be divisible by $p^k$ for some $k > 0$, who knows? So the largest $n$ such that $x + y$ is divisible by $p^n$ is *at least* $\nu_p(x)$, but it might be larger. Hence $\nu_p(x + y) \geq \nu_p(x) = \text{min}(\nu_p(x) + \nu_p(y))$. From this one derives that $|x + y|_p \leq \text{max}(|x|_p, |y|_p)$.

Could you explain me how it can be that the factor $p^{\nu_p(x)-\nu_p(y)} \cdot a + b$ might be divisible by $p^k$ for some $k > 0$ ?

In which case would it be like that? (Thinking) (Worried)

 

[mathbalarka]

Say $b = p^k$ for some $k \leq \nu_p(x) - \nu_p(y)$. Then $p^{\nu_p(x) - \nu_p(y)} + b$ is divisible by $p^k$.

 

[evinda]

Say $b = p^k$ for some $k \leq \nu_p(x) - \nu_p(y)$. Then $p^{\nu_p(x) - \nu_p(y)} + b$ is divisible by $p^k$.

According to my notes, $a,b \in \mathbb{Z}_p^*$, so shouldn't it be:

$$b \in \{ 1,2, \dots, p-1\}$$
? Or am I wrong? (Worried)

 

[mathbalarka]

Yes, right, my bad. I wasn't paying attention. Forget what I've said above.

But $p^{\nu_p(x) - \nu_p(y)} \cdot a + b$ might still be divisible by a power of $p$. Consider, for example, $3 + 3$. None of the two $3$s are divsible by $6$, but $3 + 3 = 6$ is.