Restricted P-adic Valuation is Valuation

Theorem
Consider $\nu_p^\Z:\Z\to\N\cup\{+\infty\}$ the restricted valuation.

Then the following holds:


 * $(1)\nu_p^\Z \left({n m}\right) = \nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)$.
 * $(2)\nu_p^\Z\left({n +m}\right) \geq\min\{\nu_p^\Z(n),\nu_p^\Z(m)\}$

$(1)$
Let $n, m \in \Z$.

Suppose $m = 0$ or $n = 0$; then $\nu_p^\Z \left({n}\right) = +\infty$ or $\nu_p^\Z \left({n}\right) = +\infty$.

Now $n m = 0$, and hence $\nu_p^\Z \left({n m}\right) = +\infty = \nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)$.

Next, suppose $n m \ne 0$.

Then $p^{\nu_p^\Z \left({n}\right)} \mid n$, and $p^{\nu_p^\Z \left({n}\right) + 1} \nmid n$; also $p^{\nu_p^\Z \left({m}\right)} \mid m$, but $p^{\nu_p^\Z \left({m}\right) + 1} \nmid m$.

It follows immediately that $p^{\nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)} \mid n m$, and $p^{\nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right) + 1} \nmid n m$.

Hence $\nu_p^\Z \left({n m}\right) = \nu_p^\Z \left({n}\right) + \nu_p^\Z \left({m}\right)$; by definition.

$(2)$
Consider $n,m \in \Z$, and assume that $p^\alpha\mid n$ y $p^\beta\mid m$; where $\alpha\geq\beta$

Then $\exists t\in\Z,\ \exists k\in\Z,\ n+m=p^\alpha k+p^\beta t=p^\beta(p^{\alpha-\beta}k+t)$; and in conclusion $p^\beta\mid (n+m)$.

Hence $\nu_p^\Z(n+m)\geq\min\{\nu_p^\Z(n),\nu_p^\Z(m)\}$ by the definition of $\nu_p^\Z$.